point-set-topology

[uneven bronze]

Is this an alternative solution method? We want to show $\rho$ induces the topology of uniform convergence. Consider for $n>1$,\begin{align*}&{g\in \mathbb{C}^X:\sup_{x\in X}|f(x)-g(x)|<(n-1)^{-1}}\ &\phantom{==}=\left{g\in \mathbb{C}^X:\rho(f,g)<n^{-1}\right}\ &\phantom{==}=B_\rho(n^{-1},f),
\end{align*}
and for $n=1$, $$B_\rho(1,f)={g\in \mathbb{C}^X:\sup_{x\in X}|f(x)-g(x)|<\infty}.$$

[gentle osprey]

psie

[uneven bronze]

Now, is the second set, i.e. the one for n=1, open in the topology of uniform convergence?

[gaunt linden]

Why is sup|f-g| not a metric?

[uneven bronze]

Because it takes on the value infinity (maybe). A metric is a function from X x X to [0,oo).

[gaunt linden]

Oh, there are unbounded functions.

What is your definition of "the topology of uniform convergence"?

[uneven bronze]

The topology of uniform convergence on $\mathbb{C}^X$ is generated by sets of the form $${g\in\mathbb{C}^X:\sup_{x\in X}|g(x)-f(x)|<n^{-1}}\quad(n\in\mathbb{N},f\in\mathbb{C}^X).$$

[gentle osprey]

psie

[gaunt linden]

All those sets are open balls according to your rho, and every rho-ball contains one of them.

[uneven bronze]

Yes, but what about the rho-ball with radius 1? Is it also an open set in the topology of uniform convergence?

[gaunt linden]

That's the second part -- a rho-ball with radius 1 contains a rho-ball with radius 1/2 around each of its points.

(Which is more intricate than it needs to be, really).

[uneven bronze]

Hmm. So you are saying each point in $$A={g\in \mathbb{C}^X:\sup_{x\in X}|f(x)-g(x)|<\infty}$$ contains a $\rho$-ball with radius $1/2$ contained in $A$?

[gentle osprey]

psie

[gaunt linden]

More generally, if (X,d) is any metric space, then for every point y in B(x,r) there is an s strictly between 0 and r such that B(y,s) subseteq B(x,r) -- namely, you can take s = min(r/2, r-d(x,y)).

And s < 1/2 means that each of those B(y,s) is entirely in one of your generating sets for the topology of uniform convergence.
Wait, I confused myself. This is true but irrelevant in context.

The point is that B(y,s) contains a generating set for the topology of uniform convergence, no matter what s is. And the union of all those generators (one for each y) is B(x,r) no matter what r is.

[uneven bronze]

Thank you for explaining. It is clearer now.

[uneven bronze]

Define $\Phi:[0,\infty]\to[0,1]$ by $\Phi(t)=t/(t+1)$ for $t\in[0,\infty)$ and $\Phi(\infty)=1$. If $X$ is $\sigma$-compact LCH and ${U_n}$ are precompact sets that satisfy $\overline{U}n\subset U{n+1}$ and $X=\bigcup_1^\infty U_n$, a metric that induces the topology of uniform convergence on compact sets on $\mathbb{C}^X$ is $$\rho(f,g)=\sum_1^\infty2^{-n}\Phi\left(\sup_{x\in\overline{U}n}|f(x)-g(x)|\right).$$I'm trying to verify this is a metric. Nonnegativity and symmetry are clear I think. However, with $\rho(f,g)=0\implies f=g$ I struggle (the other direction is obvious). What exactly are $\eta_n(f,g)=\Phi\left(\sup{x\in\overline{U}_n}|f(x)-g(x)|\right)$? Are these metrics on $\overline{U}_n$?

[gentle osprey]

psie

[gaunt linden]

It may be easier to think about the contrapositive: if f != g, then rho(f,g) > 0.

[uneven bronze]

Recall Tychonoff's theorem. The way this is proved is by showing that any net $\langle x_i\rangle_{i\in I}$ in $X=\prod_{\alpha\in A}X_\alpha$ has a cluster point. The way this is done is by examining cluster points of the nets $\langle\pi_B(x_i)\rangle$ in the subproducts of $X$ (here $\pi_B(x_i)$ is the restriction of $x_i$ to $B\subset A$). To this end, define $$\mathcal{P}=\bigcup_{B\subset A}\left{p\in\prod_{\alpha\in B}X_\alpha: p\text{ is a cluster point of }\langle\pi_B(x_i)\rangle\right}.$$ $\mathcal{P}$ is nonempty since each $X_\alpha$ is compact and so $\langle\pi_B(x_i)\rangle$ has a cluster point when $B={\alpha}$. I get it. What I don't understand is that $\mathcal{P}$ is partially ordered by extension; meaning $p\leq q$ if $q$ extends $p$ as a mapping (i.e. $p$ is in the product with the index set $B\subset A$ and $q$ in the product with the index set $C\subset A$, and $B\subset C$ and $p(\alpha)=q(\alpha)$ for $\alpha\in B$). Why is $\mathcal{P}$ partially ordered by extension? I'm a bit uncertain what it is I have to check.

[gentle osprey]

psie

[uneven bronze]

(1) Reflexivity: What is it that I have to show here?\

(2) Transitivity: If $p,q,z$ are cluster points of $\langle \pi_A(x_i)\rangle$, $\langle \pi_B(x_i)\rangle$ and $\langle \pi_C(x_i)\rangle$, with $p\leq q$ and $q\leq z$, then do I have to show $z$ is also a cluster point of $\langle \pi_A(x_i)\rangle$?\

(3) If $p\leq q$ and $q\leq p$, then $A\subset B$ and $B\subset A$, so $A=B$, but why would this imply that $p=q$?

[gentle osprey]

psie

[uneven bronze]

Any help would be appreciated. 😔

[mellow basalt]

It's extension, doesn't have anything to do with it being a cluster point

p \leq q means for each a in A p(a) = q(a)

[uneven bronze]

yes, but it says that \leq is a partial order on P, which is a collection of cluster points of nets

[uneven bronze]

But I realize that < pi_A(x_i) > and < pi_C(x_i) > are nets in different spaces. This complicates my attempt above.

[mellow basalt]

P doesn't have anything to do with how <= is defined.

You can remove the condition that p is a cluster point, and show that it's a partial order, then restrict it to P

[uneven bronze]

Hmm, what do you mean by restrict to P? The proof I'm reading says that <= is a partial order on P. Isn't P relevant in that statement?

[mellow basalt]

Say, for transitivity, take the p, q, z you chose. Let a in A, then since q is an extension of p, p(a) = q(a), and since z is an extension of q, q(a) = z(a), so p(a) = z(a), which does not use anything about them being cluster points

[uneven bronze]

ok, thanks for elaborating 🙂 so is <= a partial order on any collection of functions with varying domains? In this case, P is a collection of functions with varying domains (all domains are subsets of A though) and varying codomains.

[mellow basalt]

codomain is cup X_a

[uneven bronze]

Recall Tychonoff's theorem. One proof goes by showing that any net $\langle x_i\rangle_{i\in I}$ in $X=\prod_{\alpha\in A}X_\alpha$ has a cluster point. The way this is done is by examining cluster points of the nets $\langle\pi_B(x_i)\rangle$ in the subproducts of $X$ (here $\pi_B(x_i)$ is the restriction of $x_i$ to $B\subset A$). To this end, define $$\mathcal{P}=\bigcup_{B\subset A}\left{p\in\prod_{\alpha\in B}X_\alpha: p\text{ is a cluster point of }\langle\pi_B(x_i)\rangle\right}.$$ Let ${p_l:l\in L}$ be a linearly ordered subset of $\mathcal{P}$, where $p_l\in\prod_{\alpha\in B_l}X_\alpha$. Let $B^\ast=\bigcup_{l\in L}B_l$ and let $p^\ast$ be the unique element of $\prod_{\alpha\in B^\ast}X_\alpha$ that extends every $p_l$. Why is $p^\ast$ unique?

[gentle osprey]

psie

[uneven bronze]

Any help would be very appreciated. 😔

[quick crane]

it should have to do with the ordering on the set. Do the case where $A = \mathbb{N}$ for visualization

[gentle osprey]

L

[uneven bronze]

Could one simply argue as follows? If there was a $p'\in\prod_{\alpha\in B^\ast}X_\alpha$ that also extended each $p_l$, then for any $\alpha\in B^\ast=\bigcup_{l\in L}B_l$, $\alpha\in B_{l'}$ for some $l'\in L$, but then $p'(\alpha)=p_{l'}(\alpha)=p^\ast(\alpha)$.

[gentle osprey]

psie

[uneven bronze]

Let ${X_\alpha}{\alpha\in A}$ be all compact and consider $X=\prod{\alpha\in A}X_\alpha$ as well as a net $\langle x_i\rangle_{i\in I}$ in $X$. Let $\pi_B(x_i)$ be the restriction of $x_i\in X$ to $B\subset A$ (recall, an element of $X$ is a function from $A\to\bigcup_{\alpha\in A}X_\alpha$).\

Now suppose the net $\langle \pi_B(x_i)\rangle$ has a cluster point $p$, where $B\subset A$ is a proper subset. This means it has a subnet $\langle \pi_B(x_{i(j)})\rangle_{j\in J}$ that converges to $p$. Let $\gamma\in A\setminus B$ and consider the net $\langle\pi_{{\gamma}}(x_{i(j)})\rangle$. This net lives in $X_\gamma$, which is compact, so $\langle\pi_{{\gamma}}(x_{i(j)})\rangle$ has a convergent subnet $\langle\pi_{{\gamma}}(x_{i(j(k))})\rangle_{k\in K}$ converging to $p_\gamma\in X_\gamma$. Let $q\in\prod_{\alpha\in B\cup{\gamma}}X_\alpha$ be the unique extension of $p$ and $p_\gamma$.\

Question: Is it true $\langle\pi_{B\cup{\gamma}}(x_{i(j(k))})\rangle_{k\in K}$ converges to $q$? How does one show this?

[gentle osprey]

psie

[uneven bronze]

@quick crane you don't happen to have any ideas about this? 🙂 My intuition says the net can not converge to anything other than q, but how does one show this with a bit more rigor?

[quick crane]

Yeah this seems correct

[uneven bronze]

Thanks. Is it clear to you why $\langle\pi_{B\cup{\gamma}}(x_{i(j(k))})\rangle_{k\in K}$ would converge to $q$?

[gentle osprey]

psie

[uneven bronze]

Ok, I think I understand now. Yay!

[median sand]

Is the cardinality of the index set of a cube (product space [0,1]^I) an invariant, i.e. [0,1]^I\cong[0,1]^J implies I\cong J? For finite products this works by homology, what about for infinite ones?

[tender halo]

yes because the weight of the Tychonoff cube is equal to the cardinality of I/J for infinite products

[tender halo]

(note that a simple cardinality argument will not do, because 2^\kappa can be equal to 2^\lambda for \kappa < \lambda in the absence of stuff like GCH)

for finite ones, to put forward a point-set-topological argument, look at the lower inductive dimension for example

[granite crane]

Is it from folland?

[rain gyro]

Is it possible to make this sigma algebra verification process less redundant in words?

I mean topological space

Like can I shrink it half

[fringe thorn]

if you know about bases, you can instead prove that the set of all open balls with radii greater than 0 is a basis for a topology on X

[rain gyro]

I’ll study it real quick

I’m making my note I have proved the same thing for 3 times

[fringe thorn]

this amounts to showing two things:

a) the union of all such open balls is equal to X; and
b) that in the intersection of any two open balls, you can find a smaller open ball around each point that fits inside said intersection

[rain gyro]

I just wish my note look complete so I write but this is actually meaningless bcs I’m proving essentially the same thing 😭

Or very identical

[hidden abyss]

I think in b you need to prove that every point in the intersection is contained in such a smaller ball?

[uneven bronze]

Yep. 🙂

[granite crane]

Cool

[fringe thorn]

huh. that's not the basis for a topology criteria I remember

nvm, you're right and I'm being silly

[alpine nest]

That's subbasis, I think?

[fringe thorn]

this is the statement I remember

[alpine nest]

Yeah, you're right

I was misremembering

[tiny obsidian]

yeah so as mentioned you need this smaller ball for every x

[fringe thorn]

ah, yes

[alpine nest]

Of course in this case you'd need to prove that this is the basis specifically for the metric topology

[fringe thorn]

Jussari is correct

[alpine nest]

Since the result of 2.44 just tells you balls would be the basis for some topology

[fringe thorn]

don't you also have to show that every open set is the union of open balls?

that's to Edward

[tiny obsidian]

{empty, X} is a basis for itself as a topology

I thought about it for an extra 5 seconds

[alpine nest]

Which is why I'm more in favour of showing that open sets (in the metric sense) form a topology (in the general topological sense) directly.

It's very straightforward

And fancier tools don't seem to make it much simpler/easier

[rain gyro]

I skim through some of the basis 🤯🤯complicated…

My math study order is completely a mess

[alpine nest]

One thing you could simplify here, but only a bit, is that for the intersection case you could only consider the case of two open sets; since if you show that the intersection of two open sets is open, then it will follow for any finite collection (by an inductive argument that can go without saying at this level, I'd reckon)

[rain gyro]

Thanks so much 🥰🥰🥰

[tiny obsidian]

I would also reword the first sentence; X isn't vacuously open because there are (generally) x in X that you had genuinely had to verify there were open balls for. This is easy (trivial) sure, but not vacuous, unlike the empty set where you immediately have found balls for all 0 points in the set

[rain gyro]

But isn’t the complement of ground set actually empty so there should be no element outside X that was my thinking though

[tiny obsidian]

Sure there's no points outside of X, but that has nothing to do with X being open nvm I might see what you mean, that there are no points that could possibly be in B(x,ε) but not in X and so in that sense something is vacuous

I guess??? maybe? it's very much not the usual meaning of vacuous

I personally wouldn't say it's vacuously open: U is open if 'for all x in U there is a ball etc.' so I would say that 'vacuous' is reserved for the case when this is true because this statement is just 'for all 0 x in U there is a ball etc.' but unsure if that's completely standard

[rain gyro]

You’re right

I am actually not that good at logic

🫣🫣😭😭

[tiny obsidian]

this is more terminology tbh

[rain gyro]

I actually find those wording very hard to be honest

[rain gyro]

I actually thought vacuous truth is like it makes no sense if it’s not true

It’s vague but 😭🫣

[tender halo]

i would call the statement that a subset X of points with certain properties is in fact a subset of X pretty vacuous

[uneven bronze]

Consider \

Arzelà-Ascoli. Let $X$ be a compact Hausdorff space. If $\mathcal F$ is an equicontinuous, pointwise bounded subset of $C(X)$, then $\mathcal{F}$ is totally bounded in the uniform metric, and the closure of $\mathcal{F}$ in $C(X)$ is compact.\

I can't tell from the proof alone where Hausdorff-ness is used? What would your answer be?

[gentle osprey]

psie

[fair stratus]

@uneven bronze could you send your proof? my textbook has a slightly different version and X doesnt need to be Hausdorff there so definitely wierd

[rain gyro]

you can have mine

[green grove]

hi all, i've done the following exercise in Lee's ITM, but I had to resort to examining several cases, and i'm wondering if anyone knows a cleaner/more efficient solution that doesn't require any machinery beyond what is covered up to that point in ITM... here's my solution:

[rancid umbra]

i feel like you can do this with the lifting property for hausdorff spaces relatively quickly

i haven't worked out the details

but this characterization seems like it lends itself to this problem

i think that you are being pretty verbose in your proof tho

so that may be why it seems so long

[uneven bronze]

Sure, it’s kind of been sent already, but I’ll send the whole proof here:

I can't see where Hausdorff-ness is used.

[uneven bronze]

@warm kettle were you trying to say that if X is compact, then C(X) is Hausdorff?

[warm kettle]

I was trying to say that maybe the author uses the convention where a compact space is, by definition, T2, but then, I guess, it wouldn’t make sense to include both words in the theorem’s statement

[uneven bronze]

I see. Hmm, yes. It's incredible that I've thought about this proof now for such a long time I've come to believe the author simply put Hausdorff-ness in there since maybe they didn't want to bother explaining the case otherwise.

[warm kettle]

Maybe it’s to avoid AOC? Idk, but that’s a good question

[uneven bronze]

Perhaps one should simply say Hausdorff-ness is assumed because that's how the theorem is usually applied, i.e. to Hausdorff spaces. 😔

[uneven bronze]

If anyone knows if the Arzelà-Ascoli theorem holds for non-Hausdorff spaces and can provide a source, I'd be grateful. The proof above makes me uncertain why include the Hausdorff assumption at all if it's not used.

[unreal stratus]

There are lots of things on stack exchange/overflow about this - it isnt necessary, it seems

Though i think if X is at lrast compact then C(X) is isomorphically isomstric to C(Y) for compact hausdorff Y

[tender halo]

yeah, in fact you can easily reconstruct the topology of X from C(X) in that case

[rapid escarp]

Does anyone know if the following topology has any nice properties? Has anyone seen something like it before? If

Let $\mathcal{X,Y}$ be a top spaces with (whatever property) and $f: \mathcal{X}\rightarrow \mathcal{Y}$ be cont. Define a new topology on $\mathcal{X}$ via quotienting w.r.t. the relation $\sim_A:={(x,x’)\in \mathcal{X}^2 | \exists y\in A \text{ s.t. } x,x’\in f^{-1}(y)^\circ}\cup \Delta$ for some $A\subset \mathcal{Y}$.

[quartz horizon]

Yes this is $X / \text{ker} f$ right

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[rapid escarp]

Never seen kernel in context of top spaces

[quartz horizon]

Ah you just take the kernel pair

Which is always an equivalence relation

And then quotient by that

[rapid escarp]

I mean the important part is that I’m not quotienting by the fibers

But rather by their interior

[quartz horizon]

oh, interesting

then I haven’t seen this before

[rapid escarp]

I’m trying to see if I can distinguish when the infimum is attained by more than one element

[tender halo]

you have y in two different places, it should be x1 x2 or something

[rapid escarp]

But if there is an open set that avaluates to the infimum then that breaks things

[gentle osprey]

mandaly

[rapid escarp]

Sorry for so many edits.

If you have any ideas please let me know. I can’t think of anything nice immediately.

[tender halo]

isnt it the same as just quotienting out by a collection of open sets

[rapid escarp]

Yeah pretty much

I guess f being cont. can add some more structure but at the root that’s pretty much it

[rancid umbra]

it could be the case that all of them are empty, e.g., f : R^n —> R, f(x) = |x|. so if you are trying to detect extrema, this might not give you any useful info.

[tender halo]

not really right, if you quotient out by a collection of open sets, then the quotient function will be the continuous function in your construction

[rapid escarp]

That is the nice case

[rancid umbra]

how so? your quotient is trivial

[rapid escarp]

Exactly. You get that X has same properties as X/~

[rancid umbra]

but ~ is just the diagonal

[rapid escarp]

Exactly

X is homeo to X/~

In that case

So that’s nice

[rancid umbra]

i don’t see how that’s useful for classifying extrema?

like for your use case

[rapid escarp]

Cuz I’m trying to seperate them

Question is can I seperate the points using information about f

Let f be lsc

Then if the interior of the fiber at infimum is empty and X is T0 I can look at how the nbhd filters interact

Haven’t fully checked if that goes but seems promising

But main question is, when do I get that X/~_{inf f} is T0

[rancid umbra]

i don’t understand the language of filters, unfortunately. so that might explain my why i don’t fully understand. but it seems cool. gl with it

[rapid escarp]

Cheers

Will need plenty luck lol

[quartz horizon]

Filters are goated

[rapid escarp]

Here’s a better explanation. If the interior of the fiber of the infimum is non-empty, then the points inside the interior of said fiber are essentially indistinguishable.

Big sad

So my solution was to quotient the interior out

But then new topology might be very bad

[uneven bronze]

Arzelà-Ascoli Theorem I. Let $X$ be a compact Hausdorff space. If $\mathcal F$ is an equicontinuous, pointwise bounded subset of $C(X)$, then $\mathcal{F}$ is totally bounded in the uniform metric, and the closure of $\mathcal{F}$ in $C(X)$ is compact.\

Arzelà-Ascoli Theorem II. Let $X$ be a $\sigma$-compact LCH space. If ${f_n}$ is an equicontinuous, pointwise bounded sequence in $C(X)$, there exist $f\in C(X)$ and a subsequence of ${f_n}$ that converges to $f$ uniformly on compact sets.\

Proof. By a previous proposition, there is a sequence ${U_n}$ of precompact open sets such that $\overline{U}k\subset U{k+1}$ and $X=\bigcup_1^\infty U_k$. By the first Arzelà-Ascoli theorem, there is a subsequence ${f_{n_j}}_{j=1}^\infty$ of ${f_n}$ that is uniformly Cauchy on $\overline{U}_1$. ...\

Why is there such a subsequence? We have a sequence ${f_n}$ with domain $X$ and $\overline{U}_1\subset X$ is compact Hausdorff. Are we applying the first theorem to $\overline{U}_1$ somehow?

[gentle osprey]

psie

[uneven bronze]

Hmm, I'm puzzled by the application of the first theorem.

[balmy briar]

if X is some totally ordered set and we have a bounded set S with no supremum, can i consider X U sup(S)?

[alpine nest]

If S has no supremum, then what is sup(S)?

Potentially more helpful answer: if you have a set with no supremum in your space, you can try to "add an element to X that would act as the supremum", but it requires some consideration.

The construction of the reals from the rationals via Dedekind Cuts works essentially along those lines.

[paper wedge]

or the completion of a metric space

[balmy briar]

What is the consideration I need? When can I not do this?

[alpine nest]

That one doesn't take order into account, though (although it indeed gives you the same, or at least isomorphic, result when applied to the rationals)

[paper wedge]

yeah Q and R is what i mean

[alpine nest]

I mean you have to explicitly define the element you're adding, and how it fits into the order.

[paper wedge]

just meant that this is not the only construction 😄 (i am biased to the cauchy sequences construction)

[alpine nest]

It is a fine construction, to be sure.

[paper wedge]

yeah 😄

[balmy briar]

I want to define the element as the element that acts as the supremum. Also, Isn't how it fits into the order specified by definition of the supremum?

[paper wedge]

I am not convinced one can do this without asssuming X lives in some ambient space

S = { x | x^2 <2} in Q

if you want to say "oh add sqrt(2)"

[alpine nest]

Is it? That tells you it's greater than all elements of your set, but what about the other elements of your space?

[paper wedge]

u first need to define what sqrt(2) is

[alpine nest]

I mean, you can, this is what Dedekind cuts do

[paper wedge]

yes 😄 this is exactly the construction

but what i meant to say this, there must be some ambient space constructed

[tender halo]

as opposed to...?

we are considering how to embed into some nicer space

[alpine nest]

Well yeah, if you add an "extra" element to your set, you get a larger ambient space in which your original set is a subset

But you construct this ambient space rather than assuming there is one

[balmy briar]

the supremum is exactly between the set S and the rest of the set, no? If I have some other element y that's greater than S it's either sup(S) (Which does not exist by assumption) or greater than sup(S).

[alpine nest]

Of course if you have a useful ambient space already, things are potentially easier

[alpine nest]

Sure, if you can define the relation between your new element and all the other elements of a set in a way that preserves total order (which I agree should be possible), then it's fine, especially if you only "add the supremum" for one set.

[balmy briar]

Okay thank you, I can also show you the question if you'd like I was using it to show something by contradiction.

[tender halo]

i mean there is only the one way to add a supremum, the definition of sup does uniquely specify the gap where the sup should go

[alpine nest]

Yep

[alpine nest]

Sure, why not

[tardy carbon]

maybe the converse relies on X being Hausdorff

[uneven bronze]

Ok. Still thinking on this one... 😔

[balmy briar]

Just a note: after doing this unncessarily long proof I realized that (-infty,sup(S)) = union of (-infty,s) where s in S) which could complete the proof in a couple lines (If its correct). But I'm just trying to understand if this original one is correct.

[alpine nest]

I don't understand this bit, if X is your space, then it can't not be dense.

Unless you have some ambient space which you didn't mention.

Every topological space is dense in itself, that's a fully general statement with no assumptions needed.

[paper wedge]

I htink

OP means dense as in orderly dense

[alpine nest]

Ah

[balmy briar]

yeah i meant that

sorry

[alpine nest]

Fair enough, probably an acceptable shorthand in the context.

Just took me by surprise.

[paper wedge]

yeah same 😄

[tender halo]

that is very wordy

i guess you dont want to refer to the completion of an order?

because otherwise you can just say "take the completion of the order, if its bigger than the original then you can split it in half"

[tardy carbon]

Yeah $f_n \in C(\overline{U_1})$, are bounded and equicontinuous, so Arzela-Ascoli applies

[gentle osprey]

L

[tender halo]

its a little muddled, what you really are proving in the middle is that an order is complete iff it has no non trivial dedekind cuts

it would be better as a separate lemma

[balmy briar]

I don't know what the completion of an order is, I just wanted to know if this proof is correct.

[tender halo]

every order is uniquely embedded in a (smallest) complete order

if you have a bounded subset with no supremum you already won, the space is obviously disconnected, there is no need to extend the order in any way

the down closure of S and its complement are disjoint open sets whose union is X

[balmy briar]

Yes I added the note that (-infty,sup(S)) is open because its the union of all (-infty,s) where s in S (I think) and so we can find a separation of disjoint open sets whose union is X, i just wanted to know again if this proof is correct

[tender halo]

the second half is superfluous

otherwise its correct

[balmy briar]

okay thank you

[balmy briar]

sorry, even though I know it could be done in an easier way I still want to know if the argument I provided in the second half is correct. @alpine nest

[real notch]

Is there some super nice/natural metrization of the Poincaré disk?

(By which, I really would want some nicely describable metric on the open unit disk in R^2 that makes it that disk, if that’s possible)

[gaunt linden]

Here's what Wikipedia says:

Hmm, that's a Riemannian metric rather than a distance function, though.

Oh, it gives an expression for the distance function too. Unfortunately far from "super nice".

[real notch]

It can be written out in terms of relatively simple functions, so that’s nice enough for me

I mean, that’s about as nice as one could hope yeah?

(And a bit of simplifying when doing d(0, v) can help)

[tribal thicket]

Working on Riemann Mapping Theorem currently in a course. Professor asked as a starting point why open unit disc D = {z : |z| < 1} cannot be biholomorphicly mapped to C.

Is the problem not just already with being a continuous bijection? If f bijects D to C, then f^-1(C) = D. Yet C is closed (with topology of C) and D is not closed (with topology of C), so f cannot be continuous. Am I misremembering my topology badly here?

[alpine nest]

A homeomorphism between the open unit disc and all of C does exist

(unless we're talking C as the Riemann sphere, with the point at infinity, but I doubt that)

[tribal thicket]

Is it one you can give an example of

[gaunt linden]

No, you can have a continuous bijection, for example $$z \mapsto \frac{|z|}{1-|z|}\cdot \frac{z}{|z|}$$

[gentle osprey]

Troposphere

[alpine nest]

And yes, the issue with your reasoning is that D is a closed subset of itself, which is what matters for the purpose of your map being a homeomorphism.

[gaunt linden]

(The |z|s cancel, but make it clearer what's going on -- just stretch each open line out from 0 to a point on the unit circle, such that it goes out to infinity).

[tribal thicket]

Right. change to a different topology

[gaunt linden]

My immediate answer would be that ||a holomorphic surjection C -> D would violate the little Picard theorem,|| but perhaps that's too heavy artillery for that question.

[tribal thicket]

the simplest other I think answer is Liouville

the inverse of the biholomorphism would be entire and have bounded values

thus be constant by liouville, which can't be the case if it's bijective

[gaunt linden]

Ah, that is less heavy.

[lucid ocean]

let L be [0,1]^omega1 be a LOTS under the lex order

let L' denote L disjoint union L, glued end-to-end such that (1,1,...) is identified with (0,0,...)

is the following a homeomorphism from L' to L, H: L' -> L, f((n, x)) = [(i |-> x(i)/2 for i <= alpha1 and x(i) otherwise) if n = 0, (i |-> x(i)/2+1/2 if i <= alpha2 and x(i) otherwise) if n = 1]?

where alpha1 = sup{beta in omega1 | forall gamma in beta, x(gamma) = 1}

and alpha2 = sup{beta in omega1 | forall gamma in beta, x(gamma) = 0}

[atomic totem]

quick qn, how is the definition that a subset of a top space is saturated if it is the intersections of open subsets of the top space equivalent to the definition that a subset C is saturated over a function f if C = f-1(f(C))?

theres an alternate definition with equivalence relations (R an equivalence relation in a top space, then a subset of the top space is saturated over R if its a union of equivalence classes) that i can see the relation to the second def, but im not seeing it for this one

mostly because im not seeing what the equivalent mapping is supposed to be for the first def

[hidden abyss]

Uhh it isn't? The C=f^-1(f(C)) definition obviously depends on f (take f=identity and f=constant) but being the intersection of opens just depends on the topology of the space

[atomic totem]

😔 it being on the same page on wiki confused me lol

[steady gorge]

for 2a, i know we should use the identity map but i need a little push on where to go. i was looking at the conditions for a quotient map and am not totally sure on how to use $p\circ f$ here

[gentle osprey]

ushygushytoes

[ruby delta]

what is the definition of a quotient map you are working with here

[steady gorge]

let $X$ and $Y$ be topological spaces; let $p \colon X \to Y$ be a surjective map. the map $p$ is said to be a quotient map provided a subset $U$ of Y$ is open in $Y$ if and only if $p^{-1}(U)$ is open in $X$.

[gentle osprey]

ushygushytoes

[ruby delta]

all right, first of all can you show that p is surjective

(this is just set theory, no topology needed for this step)

[steady gorge]

given an arbitrary $y \in Y$, $p \circ f$ equals the identity map of $Y$ so $(p \circ f)(y)=y \implies p(f(y))=y$ for all $y \in Y$, right? then $p$ would be a surjective map

[gentle osprey]

ushygushytoes

[steady gorge]

then i need to do the next part of the definition and check that for an open set $U$ of $X$, $p(U)$ is open in $Y$ if and only $p^{-1}(U)$ is open in $X$

[gentle osprey]

ushygushytoes

[steady gorge]

ok i think i got it now

[violet mirage]

So i been trying to show whenever if X is a first-countable space, and x in cl(A), where A is some subset of X, then there is a sequence of points in A converging to x.

Its uhh

https://cdn.discordapp.com/attachments/1207121030326259732/1395683878483791922/1206754479228067880.png?ex=687b5767&is=687a05e7&hm=3769f280b530f0859c926011c6df3066cd61efdd367a2ddcd9c0a4a15ce6c091&

Basically i'm just saying if ${ U_n}$ is the countable basis, then for each k $U_k \in { U_n }$ satisfies that $U_k \cap A \neq \emptyset$

[gentle osprey]

Prelude to archbishop

[violet mirage]

Now is the main part of the proof that i'm kinda stuck on

I know that $U_n$ sort of "grow small" as sets in the sense that $U_{n+1} \subset U_{n}$

[gentle osprey]

Prelude to archbishop

[violet mirage]

So i was wondering like

Is it possible to do zorn's lemma on some kind of collection of intersections? Like doing zorn lemma on $U_k \cap A$ and then proving that maximal element would be one s.t the intersection equals itself and is bigger than other such sets

[gentle osprey]

Prelude to archbishop

[opaque scroll]

Maybe remind yourself of the definition of local basis

[violet mirage]

Ah

Thanks

That might be helpfull

[opaque scroll]

The solution should mostly consist of comparing the definition of basis with the definition of convergence

[violet mirage]

Yea

I just realized

I misread the definition of countable basis at x

I thought it meant like $U_{n+1} \subset U_n$ lol

[gentle osprey]

Prelude to archbishop

[violet mirage]

Well actually this is true, but they meant that there exists some positive integer N s.t for all $n \geq N$ we have that $U_{n} \subset U_x$ where $U_x$ is any non-empty ngbh of x that need not necessarily be a part of the collection

[gentle osprey]

Prelude to archbishop

[violet mirage]

Well actually it might be true*

[violet mirage]

Depends on how the sets are indexed*

[opaque scroll]

So the definition is that for any neighbourhood U of x, there is an n with Un < U. And you can choose the basis such that Un+1 < Un if you like

[granite crane]

Are local basis same as subbasis?

[opaque scroll]

No

[granite crane]

Ohh

[opaque scroll]

It's a basis, but just around a single point

Aka a basis of neighbourhoods

[granite crane]

Oh got it

B is local basis for X at x if for every neighborhood of X contains some element of b of B such that x ∈ b ∈ B

[violet mirage]

No, but i think if X is first-countable then it has countable local basis at each point which can be used to construct a basis for entire X by arbitrary union

[violet mirage]

yes, but i can choose the indexing set so that
...Un+2 subset Un+1 subset Un

[violet mirage]

So i can say this

[granite crane]

oh okk

[violet mirage]

Tho i am curious

Is subbasis for like

Subspaces

Nvm

[granite crane]

subbasis B is just a collection of subsets of top space X such that collection of union of finite intersections of elements of B is a top on X

called topology generated by B and surely B is subbasis

[violet mirage]

A bit ohio in general, but in R standard topology i get it

[violet mirage]

Does union mean that

I can like

$U = (B_1 \cap B_2) \cup (B_3 \cap B_4)$

[gentle osprey]

Prelude to archbishop

[violet mirage]

Where B_i are subbasis elements

Or is that not allowed?

[granite crane]

yes

[violet mirage]

Well then i got some ohio for R^n

https://cdn.discordapp.com/attachments/1207121030326259732/1395727631315374160/1206754479228067880.png?ex=687b8026&is=687a2ea6&hm=8d82cb6e10e8cf6998af18b79319496f8cc6a7e40fda7a71bc3a18f8c84bae08&

[granite crane]

even you can take $B_1\cap (B_1\cap B_2) \cdots $

[violet mirage]

Actually

I can fix it possibly

But too lazy

[violet mirage]

Like instead of connecting the intersection of rectangles

One could do similarily to R case i think instead, but i dont know how to make the rectangle have a circular/spherical shape at the bottom/top

[full merlin]

Question: The annulus embedded in R^3 without a twist is not ambient isotopic to the annulus embedded without a twist (right?). What if we put them into R^n, n >> 3? Will they be ambient isotopic?

Followup: Is there some general theorem that any homeomorphic embedded nice manifold become ambient isotopic after increasing the dimension of the ambient R^n ?

[uneven bronze]

Theorem. Let $X$ be a noncompact LCH space. If $\mathcal{A}$ is a closed subalgebra of $C_0(X,\mathbb{R})=C_0(X)\cap C(X,\mathbb{R})$ that separates points, then either $\mathcal{A}=C_0(X,\mathbb{R})$ or $\mathcal{A}={f\in C_0(X,\mathbb{R}):f(x_0)=0}$ for some $x_0$. \

Above, $C_0(X)$ are the complex-valued functions vanishing at infinity on $X$. To prove this theorem, I know I have to use the one-point compactification somehow, but I don't know how exactly. Grateful for any help.

[gentle osprey]

psie

[gentle osprey]

psie

[uneven bronze]

Is this correct so far? How do I continue here? I'm confused by the fact that I have not considered the functions that vanish at infinity yet.

[gaunt linden]

This makes sure all your functions extend continuously to Y, doesn't it?

[uneven bronze]

Well, if f in C(X \ {x0}), then f = g + c extends continuously to Y by f(oo) = c and where g in C_0(X \ {x0}). Currently I have my eyes on the notation of the theorem, so I don't know how C(X \ {x0}) and C_0(X \ {x0}) relates to C(X) and C_0(X).

[uneven bronze]

By "all your functions", do you mean all f defined on X and in A?

[gaunt linden]

I mean all the functions in C_0(X,R).

If you didn't have the condition of vanishing at infinity, you could have something like X=R and A={all polynomials}, and then a nonconstant polynomial wouldn't extend continuously to infinity.

[uneven bronze]

Hmm, then I don't see why they would extend continuously to Y. Y is the one-point compactification of X \ {x0}, not of X. Unless I'm misunderstanding something.

[gaunt linden]

But each of the functions already go to 0 both for x->x0 and for x->the original infinity.

I assume "vanishing at infinity" means that for each eps>0 there is a compact set H such that |f(x)|<eps when x notin H, right?

[uneven bronze]

Correct. I'm also familiar with the following proposition, which I think will be useful here:\

Proposition. If $X, X^\ast$ and $\mathcal T$ are LCH, the one-point compactification and the topology on $X^\ast$ respectively, then $(X^\ast,\mathcal{T})$ is compact Hausdorff and the inclusion map $i: X \to X^\ast$ is an embedding. Moreover, if $f\in C(X)$, then $f$ extends continuously to $X^\ast$ iff $f=g+c$ where $g\in C_0(X)$ and $c$ is a constant, in which case the continuous extension is given by $f(\infty)=c$.

[gentle osprey]

psie

[uneven bronze]

And of course the regular Stone-Weierstrass theorem, i.e. for compact Hausdorff spaces:

[gaunt linden]

So if you can show that each f in A extends continuously to Y, then you just need to apply Stone-Weierstass to Y, right?

[uneven bronze]

Hmm, yes. So applying the S-W-theorem to Y, the conclusion is that A = C(Y, R) or, if there is no x0 such that f(x0) = 0 for all f in A, that A = {f in C(Y, R): f(x0) = 0}, correct? But you see, the conclusion should be with C_0(X, R) instead of C(Y, R).

Meeh. I don't even see how the assumptions of Stone-Weierstrass are satisfied for Y. 😔

[gaunt linden]

"Closed subalgebra" means closed with respect to the sup-norm, right?

[uneven bronze]

Yes.

[gaunt linden]

So the key points in the exercise must be

• that C0(X,R) and C(Y,R) are "morally" the same set -- you get from f in C0(X,R) to the other space by forgetting f(x0)=0 and defining f(infty)=0, and vice versa,
• that A still separates points in Y
• that A is still closed when seen from Y.

In fact, the first of these items can be understood as saying that X and Y are really the same set, just with x0 declared to be "infinity" and the topology adjusted to fit that. That means that C0(X,R) and C(Y,R) are literally the same functions, just with different topologies on their shared domain. Which again means that they separate points just as well, and are closed under sup-norm just as well, because neither of those two properties care about the topology of the domain anyway.

(Though, whenever I said C(Y,R) in the above it should only be those functions in C(Y,R) whose value at infinity is 0).

[uneven bronze]

Ok. Let me try to summarize what you have said. 🙂 Either X contains an x0 such that f(x0) = 0 for all f in A, or it doesn't. If it does, each f in A is also a function on Z = X \ {x0}. I assume Z is still locally compact if X is (since Z is an open subset of X). And so Z has a one-point compactification Y, and each f in A (which is a function on X) has the same values on Y if we define f(infty) = 0 for all f in A (since f(x0) = 0).

Now, A as a subset of C(Y, R) still separates points (since we only "relabeled" x0 as infty) and is closed (since the sup-norm over Y or X is the same anyway). Thus A as a subset of C(Y, R) satisfies the hypothesis of the regular Stone-Weierstrass theorem, and A = C(Y, R) or A = {f in C(Y, R): f(z) =0} for some z in Y. The first alternative holds iff A contains the constant functions, which can not happen in the case we are considering, since f(infty) = 0 for all f in A. Thus A = {f in C(Y, R): f(z) =0} for some z in Y.

Getting back to X, there can only be one such z in Y since A separates points. So z = infty when we consider A = {f in C(Y, R): f(z) =0}, and z = x0 when we consider A = {f in C_0(X, R): f(z) =0}, since A as a subset of C_0(X, R) also separates points.

If X does not contain x0 such that f(x0) = 0 for all f in A, then we just let Y be the one-point compactification on X, but how do we define f(infty) for all f in A this time? We want to reach the conclusion that A = C_0(X, R), so we want A as a subset of C(Y, R) to contain the constant functions. I don't see how to do that.

[gaunt linden]

That's okay as far as it goes, but I think you're skipping too fast across the crucial point that the extended A is actually a subset of C(Y,R) -- that is, you need a more explicit argument that the extended f: Y->R is still continuous.

If X does not contain x0 such that f(x0) = 0 for all f in A, then we just let Y be the one-point compactification on X, but how do we define f(infty) for all f in A this time?
Still 0, since f vanishes at infinity. If we set f(infty) to anything else it wouldn't be continuous on Y.
Actually I would be tempted to handle that case by adding an new artificial isolated point x0 to X and extending every f in A with f(x0)=0, which reduces it to the general case above. The extended A still separates points because there wasn't already such an x0.

[rain gyro]

Guys help

Is this proof correct and sufficiently concise yet establishing the argument

I felt the only way to reduce the length of the argument is through dropping unnecessary clarification of definitions

[gaunt linden]

You're askied to prove "if and only if" statements, but your offered proofs seem only to to in one direction.

[quick crane]

The English is really bad

[rain gyro]

If and only if so it’s like (A is open then every x is interior) cap (all x of A are interior then A is open)

I used a bit demorgan here

[rain gyro]

I agree 🫣😭 I will refine it so please criticize 😭

[gaunt linden]

It should be

(if A is open then every x is interior) and (if all x of A are interior then A is open)
"Cap" is for sets, "and" is for statements.

(Whoops, wrong reply).

[rain gyro]

Yes

[gaunt linden]

What I meant to say here is that we need to know which definitions you're working with, even if you don't repeat them in your proofs.

[rain gyro]

The complement of and is or so one of the statement s negation should establish the statement right if I proved the contradiction

Since I would have to take the complement once more

[gaunt linden]

Proving an "and" statement by contradiction is really inadvisable.
The negation of your goal becomes an "or" statement, and the only real way to use that is by a case analysis -- which means you'll be doing the same work if you proved the two sides of the "and" separately, but in a more obscure wrapping.

[quick crane]

Try asking AI to just fix the English for you without changing any of the math. If that doesn't work, I'll fix it and show you.

[rain gyro]

Okay 🫣🫣 I will try for this one usually it’s too much macros and it output nonsense
And thanks 🥰

[rain gyro]

Got it then I will be fixing it now

[worn mortar]

“If and only if” indicates a biconditional statement…for the first statement (i), you need to prove

(i) A subset A of X is open if and only if every point of A is an interior point of A.

This means that to prove (i) alone, you need to prove

“If every point of A is an interior point of A, then A is open”

AND

“If A is open, then every point of A is an interior point of A”

[rain gyro]

Yes I just realized it’s not single arrowed statement 😭

[gaunt linden]

The second of those is almost immediate no matter which definition you're working with, but still deserves at least a sentence in the proof.

[rain gyro]

I should delete that 🫣

How’s now I was having a silly detour earlier with the sets

[gentle osprey]

Emmaaaaa

[lucid ocean]

that's a weird thing for them to ask you to prove if they take that to be the definition

presumably they've defined closed to mean is the complement of open

[gaunt linden]

Um, you don't claim that is a complete proof right? You're just stating what you need to prove in each direction of (i), but not actually arguing for it.
In (ii), I strongly suspect you must be using a different defintion of "closed" than the problem author intended.

[rain gyro]

Oh sorryc ChatGPT

🥹🥹I’ll fix it right away

[hidden abyss]

Dont use chatgpt 😭

[teal linden]

how did they define open and closed? cuz its weird that its straight from the definition, no?

[rain gyro]

I was using it to refine language

It changes my statement

[rain gyro]

But it’s equivalent right? I proved this

Already

[rain gyro]

So now I think it’s okay for me to refer it as a definition 🫣🫣

[hidden abyss]

I'd refer to it as Lemma 2.2

[rain gyro]

🥰🥰🥰I’ll do it

[teal linden]

wait what do you mean by x \in A \subset A? is that a typo?

[hidden abyss]

Assuming that your definition of closed is "contains its limit points"

[rain gyro]

A is an open neighborhood if it is a neighborhood for every point of it… ChatGPT changed this part to an incomplete statement almost…

Is my definition sometimes

[teal linden]

oh i see! okay that makes sense

[rain gyro]

Quite an equivalent I feel

Or I messed up again

[teal linden]

whats the whole context for this btw? are you working through a book?

[rain gyro]

Self study I will make note for retention

Then probably won’t be at this scale

The two statements are quite different actually

Yes now references are attached why it’s lemma 2, I really hate coding

[hidden abyss]

I'd explain the first part of claim (i) a bit more. Its not really clear why x ∈ A ⊂ A makes x an interior point

[rain gyro]

Is it sufficient for me to clarify that since A is open it is a neighborhood for every point of itself, hence every point of A is an interior point

That immediately gives for every x in A there exists an open neighborhood for x such that contained within A

[hidden abyss]

I'd write the definition of interior point out

[rain gyro]

So Becasue for every x in A we have a in A subset A, by definition that is for every x in A we have an open neighborhood of x such that it’s a subset of A

Or the method is wrong then I’ll use another one

[hidden abyss]

You could also use this and the message immediately after

Makes it easier to follow the logic

[rain gyro]

How’s now?🥰

And is the language okay I kinda concern about this, I want it to be as clear as possible maintaining a balance between conciseness

[quick crane]

The English is still not all correct. But what is your definition of open?

[rain gyro]

The Union of arbitrary collection of open sets in A

I will fix it 🫣🫣

How’s now I tried to refine it a bit🫣

[gaunt linden]

Could you somehow be convinced to reveal what your definitions are?

[quick crane]

I mean, what is your definition of an open set? What does it mean for $A$ to be open? We need to know the definition because that determines which proofs are correct.

[gentle osprey]

L

[quick crane]

same for closed

[rain gyro]

Open set is the element in the topology and the close set is the complement

I made a couple more 🫣🫣

[quick crane]

ok, so you are using the general definition of topology

[rain gyro]

Should be this one, I had a bad habit to use convenient definition, I only fixed that habit recently🫣

[hidden abyss]

How can you have prop2.5 before defining the induced topology on a metric space

[rain gyro]

I used them quite mixed

I ll send the whole thing

I’m driving

So here

But it’s not like the newest one

[hidden abyss]

What I meant was that prop2.5 doesnt make sense before a definition/proposition about the induced metric on a metric space, and that doesnt make sense before the definition of a topology. So I'm confused on how prop2.5 can immediately follow def2.4

[rain gyro]

I was using a bit mixed definitions like I wanted to see how it works to generalize definitions

So I kinda made it starting from metric

I’ll reorganize the whole thing

[queen prism]

it's not necessarily bad but it is important to understand how much in math depends on a metric and how much requires just a topology

[rain gyro]

I do find my notes too messy

Should I completely rewrite it? I really want that I just feel it’s so bad every once in awhile

Like when I wrote them they are fine, a couple days later I just find them unbearable

[rancid umbra]

why do they feel messy?

[rain gyro]

Not concise, some propositions are equivalent so redundant

Topics aren’t uniformed

[rancid umbra]

after looking at this, it looks like you just need some organization and maybe some more words explaining the direction of the notes.

for example, it just starts off with proposition 1.1 without giving any background defs or reasons for why we should care about it.
it seems like you can group some stuff together using subsections or subsubsections, maybe mimicking the grouping in an existing topology textbook

btw, the front page is cute and i like the style

[rain gyro]

I think I should acquire a textbook , for the metric part I learned completely from YouTube

The part for general I used folland’s which is very short in topology and highly disjointed. so I am considering

And some vids on YouTube’s are so… disjointed

[lucid ocean]

different circumstances but found myself doing something like that

(then things evolved)

[median sand]

Quick sanity check: if the topology of X is initial wrt maps into completely regular spaces (points can be separated from closed sets by continuous functions), then X itself is completely regular, right? Pretty much the same proof as for product spaces should work.

[rain gyro]

Omg you wrote this

You’re literally a god

[tender halo]

yeah, that works

[quick crane]

Ignoring the bad English, in the proof of the converse direction of (i), how exactly are you concluding that $A$ is open? From which definition?

[gentle osprey]

L

[rain gyro]

I think theee is a proposition like if for every x in A, A is a neighborhood of that element the A is open 🫣

This one 🫣

I almost can’t distinguish this between definition.

[quick crane]

ignoring all the problems with this, what you need is the last sentence of this to complete the proof of (i)

[rain gyro]

Because it is an iff statement so I had to prove the converse. That is why I used the definition of interior point. Which is, if x is interior then there exists an open neighborhood U_x such that U_x is a subset of A
The Union of those neighborhood is also open and will be A.

It’s kinda my thought process 🫣

[quick crane]

yeah this is correct, and it all should be written in the proof of (i)

[rain gyro]

For the first direction I just want to express since A is open then it is for every x in A it is a neighborhood of that point.. I kinda want it to be concise but language and general lack of terminology made it awfully bad I guess 😢😢

I’ll do that right away 🥰

[rancid umbra]

that's insane! love the figures. i might look through some of this later

[uneven bronze]

I've been trying to think a bit more about why the Stone-Weierstrass theorem holds for LCH spaces. For me there are two cases; either $X$ contains an $x_0$ such that $f(x_0)=0$ for all $\mathcal{A}_X\subset C_0(X,\mathbb{R})$, or it doesn't. I still struggle understanding both of these cases.\

When there is such an $x_0$, then we use the one-point compactification $Y$ of $Z=X\setminus{x_0}$, with $\mathcal{A}_Z = { f|Z : f \in \mathcal{A}_X}$ and $\mathcal{A}_Y = { f : f|Z \in \mathcal{A}_Z\text{ and }f(\infty)=0}$. $\mathcal{A}_Y$ still separates points and is closed, since all we did is swap $x_0$ with $\infty$, a point at which every function equals $0$ anyway. Applying Stone-Weierstrass to $\mathcal{A}_Y$ we get that $\mathcal{A}_Y = C(Y, \mathbb{R})$ or $\mathcal{A}_Y = {f \in C(Y, \mathbb{R}): f(z) =0}$ for some $z$ in $Y$. The first alternative holds iff $\mathcal{A}_Y$ contains the constant functions on $Y$, which if it did would contradict $f(\infty) = 0$ for all $f \in \mathcal{A}_Y$. Thus $\mathcal{A}_Y = {f \in C(Y, \mathbb{R}): f(z) =0}$ for some $z\in Y$ and since $\mathcal{A}_Y$ separates points, $z=\infty$.\

Now ${f \in C(Y, \mathbb{R}): f(\infty) =0}$ are precisely those functions whose restriction to $Z$ are in $C_0(Z,\mathbb{R})$. In other words, $C_0(Z,\mathbb{R})={f|Z:f\in \mathcal{A}_Y}$. I'm stuck here. How does $C_0(Z,\mathbb{R})$ relate to ${f\in C_0(X,\mathbb{R}):f(x_0)=0}$?

[gentle osprey]

psie

[gaunt linden]

Beware that it's a bit subtler than "Stone-Weierstrass holds for LCH spaces" -- that description doesn't contain the assumption that the functions in A vanish at infinity. Without this assumption, we could for example take X = R and A = the algebra of polynomials (which is closed under the sup-norm).

[uneven bronze]

Yes, indeed. We need to consider a subalgebra of C_0(X, R).

[gaunt linden]

I think your answer is that you shouldn't go from C(Y,R){f(oo)=0} to C0(Z,R) to C0(X,R).
Instead go directly from C(Y,R){f(oo)=0} to C0(X,R).
That lets you see that when you set f(x0)=0, you still get a continuous function on X.
Namely, f^-1((-eps, eps)) in Y is an open eighborhood of oo. It contains a (punctured) X-neighborhood of x0 (the Y-complement of the Y-neighborhood is compact and doesn't contain oo, so it is a compact subset of X and therefore closed in X, and certainly doesn't contain x0) so f^-1((-eps, eps)) is a neighborhood of x0, which is what we needed.

[uneven bronze]

With regards to the notation I introduced above, any $f\in \mathcal{A}_Y$ has the same values as some $g\in\mathcal{A}_X$. How can one make this correspondence more precise? Is there some function between the sets? If so, how is it defined?

[gentle osprey]

psie

[gaunt linden]

$$f\in \mathcal A_Y \rightsquigarrow \Bigl( f|_{X\setminus{x_0}} \cup {(x_0, 0)}\Bigr) \in \mathcal A_X$$
where the $\cup$ views a function as a set of ordered pairs?

[gentle osprey]

Troposphere

[uneven bronze]

Ok, makes sense. 👍

[pseudo marsh]

The map pi’ is proper, how does this imply the preimage has finitely many connected components, since U*_a is not compact

What am I missing here?

[iron bolt]

maybe that proper coverings automatically have finite degree?

[pseudo marsh]

How does that help

[uneven bronze]

Is it obvious to you that this map is a bijection (if it is)?

[gaunt linden]

Um, yes.

Restricting to X\{x0} just forgets the value of f(oo), but that is always 0, so two functions in Ay that were different cannot become equal by being mapped -- it's injective.

On the other hand, if we have some g in Ax, it will (by assumption) satisfy g(x0)=0, so it is hit by the function in Ay we get from replacing x0 by oo in the domain of g.

[uneven bronze]

Ah, ok. 👍

[gaunt linden]

The meat is in convincing oneself it actually maps into Ax, which I feel I've sketched several times without really getting much response from you -- so it's unclear to me whether that's simply obvious to you or you haven't noticed the need for that argument.

(Perhaps "several times" was only in my mind and I just actually posted it once ...)

[uneven bronze]

Well, I would say I haven't noticed the need for that argument, since we constructed Ay from Ax. So by design kind of, any function in Ay corresponds to a function in Ax, and vice versa.

Anyway, the reason I'm somewhat suddenly interested in this map is because I'm hoping it let's us conclude from Ay = {f in C(Y, R): f(infty) = 0} that Ax = {f in C_0(X, R): f(x0) = 0}, though it's not crystal clear to me yet. Yes, f in Ay corresponds uniquely to a g in Ax (by this map we've been talking about), but what worries me is that when we send an f in Ay = {f in C(Y, R): f(infty) = 0} to g in Ax, couldn't there be some h in C_0(X, R) with h(x0) = 0 that we are missing? At this point, we don't know yet that Ax = {f in C_0(X, R): f(x0) = 0}, this is the very thing we want to prove.

[gaunt linden]

Hmm. Suppose you start with some g in {f in C0(X,R) | f(x0)=0}, and you don't know yet is in Ax.
If you remove g(x0)= and add g(oo)=0, you get something in {f in C(Y,R) | f(oo)=0}. But we know that is the same as Ay.
So now remove g(oo)=0 again and put in g(x0); that gives ups something in Ax. But that "something in Ax" is the same as the g we started with.

The point is that the correspondence I wrote down works both as a correspondence between Ay and Ax, and as a correspondence between {f in C(Y,R) | f(oo)=0} and {f in C0(X,R) | f(x0)=0}.

Which would be true even if A were not closed, such that those sets were genuinely different.

[uneven bronze]

Ok, makes sense. 😅 Thank you.

[rain gyro]

Is this proof making sense with a little external set justification? For commutation between preimages and complement. I didn’t this yesterday, felt muddled due to improper justification for orders of complements and inversion; however i don’t know if it’s necessary though

Or this is also okay 🫣.. or I say use the lemma above to pass the properties to closed sets

[uneven bronze]

Let $X$ be compact Hausdorff. An ideal in $C(X,\mathbb{R})$ is a subalgebra $\mathcal{I}$ of $C(X,\mathbb{R})$ such that if $f\in\mathcal{I}$ and $g\in C(X,\mathbb{R})$ then $fg\in\mathcal{I}$.\

Exercise: If $E\subset X$, let $k(E)={f\in C(X,\mathbb{R}):f(x)=0\text{ for all }x\in E}$. Then $k(E)$ is a closed ideal in $C(X,\mathbb{R})$, called the kernel of $E$.\

I see why $E$ is an ideal, but I don't see why it is closed. I've been trying to write $k(E)$ as the inverse image of some function, but without success.

[gentle osprey]

psie

[gaunt linden]

Perhaps just show directly that if f is an accumulation point of k(E), then f vanishes on E too.

[uneven bronze]

Question; I was thinking, if the evaluation/coordinate map ev_x: C(X, R) --> R is continuous, then k(E) = intersection of ev_x^{-1}({0}) over all x in E. However, C(X, R) is equipped with the uniform metric and I'm uncertain if this induces the product topology, the (smallest) topology for which ev_x would be continuous?

[gaunt linden]

That works too -- you don't even need to worry about the product, just that an arbitrary intersection of closed sets is closed.

[uneven bronze]

Well, I was more concerned about the sets ev_x^{-1}({0}) themselves. They are closed if ev_x is continuous. 🙂 And that continuity is determined by the topology/metric, or?

[gaunt linden]

Right, you'd need to know that f --> f(x) is continuous, which might be just as involved as the task you're actually aiming for.
Continuity would be according to the sup-norm (uniform metric) on C(X), and the usual topology on R.

[uneven bronze]

Hmm, ok. I have to think about it.

[gaunt linden]

To answer explicily, though: the product topology is finer than the uniform topology (with equality iff X is a finite set).

[uneven bronze]

Ok, I think I got it. It's actually Lipschitz with constant $1$. Indeed, given $\epsilon>0$, for $|g-f|_u <\delta=\epsilon$, we have $$|\pi_x(g)-\pi_x(f)|\leq |g(x)-f(x)|\leq|g-f|_u<\epsilon.$$Bingo.

[gentle osprey]

psie

[uneven bronze]

Thanks for the help.

[hexed steppe]

coarser*

[gaunt linden]

Ohshit, did I get it wrong again? Somehow the fine/coarse terminology doesn't seem to come very naturally to me ...

[hexed steppe]

idk the conventions around this are annoying

anyway product topology will have fewer open sets

[unreal stratus]

I guess way to remember it is that the indiscrete topology is definitrly coarse than discrete

[uneven bronze]

Let $X$ be compact Hausdorff. An ideal in $C(X,\mathbb{R})$ is a subalgebra $\mathcal{I}$ of $C(X,\mathbb{R})$ such that $f\in\mathcal{I}$ and $g\in C(X,\mathbb{R})$ then $fg\in\mathcal{I}$. The hull of an ideal is $h(\mathcal{I})={x\in X:f(x)=0\text{ for all }f\in\mathcal{I}}$, which is a closed subset of $X$. The kernel is $k(E)={f\in C(X,\mathbb{R}):f(x)=0\text{ for all }x\in E}$, which is a closed ideal in $C(X,\mathbb{R})$.\

I have shown $h(k(E))=\overline{E}$ and $k(h(\mathcal{I}))=\overline{\mathcal{I}}$. The next claim is that there is one-to-one correspondence between the closed subsets of $X$ and the closed ideals of $C(X,\mathbb{R})$. What is this one-to-one correspondence?

[gentle osprey]

psie

[hexed steppe]

what do you think

[uneven bronze]

Hmm, first I thought k and h don't look like inverses to each. But if we restrict k to the closed sets, then h(k(E))=E, but I still have problems verifying the other composition, namely k(h(I))=I. Why would that be true?

[hexed steppe]

didnt you say you showed that it’s true

[uneven bronze]

yes, but k(h(I)) should return the closure of the ideal I (that's what I showed), not I itself (that's what I want to show now).

[hexed steppe]

ok, why is it true that h(k(E)) = E for closed sets E?

[uneven bronze]

because E equals its closure when E is closed

[hexed steppe]

right

so…

[uneven bronze]

Hmm, all I can observe is that k only maps to closed ideals. I still don't see k(h(I))=I.

[hexed steppe]

what if you replace the letter E with the letter I

[uneven bronze]

Ok, I think I get it now. 😅 Makes sense. k(h(I))=I for I a closed ideal.

So k restricted to the closed subsets of X has inverse h?

[hexed steppe]

i think at this level you should spend longer thinking about details like this on your own before seeking help on this discord server/elsewhere

you need to develop intuition for basic technical arguments, and also confidence in your own ideas

here you framed the question as “what is the correspondence?” but your next message revealed that you actually had some sense what it ought to be. you would learn more by trying to make your intuition precise, even when your initial attempts seem to fail

[opaque cloud]

Neam's general topology arc (insert sotrue)

[rain gyro]

How can one solves the third one , I am completely confused by index set for the finite intersection

[thorny agate]

What is the definition of $\tau_\mathcal{B}$

[gentle osprey]

Spamakin🎷

[rain gyro]

A set such that it satisfies for every open set x in U there exists B in B such that x in B subset U

And another axiom

Suppose x in U_1 in B and U_2 in B there is U_3 so x in U_3 in B

Am I doing it completely wrong or just the third part? Should I stick to index sets, how to manipulate this 😵‍💫😵‍💫😵‍💫

[thorny agate]

Is this your proof or from a textbook?

[rain gyro]

My own

I got so confused with index sets

[hidden abyss]

the set ${U_{\lambda}}{\lambda \in \Lambda}$ should probably be a subset of $\tau{\mathcal B}}$, not $X$?

[gentle osprey]

Jussari
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[thorny agate]

Ok so I think what would help is if you were more explicit in what you are proving. 
Here's an example.
You write ``take a finite collection $\{U_\lambda\}_{\lambda \in \Lambda} \subseteq X$.''
This is incorrect notation.
$\{U_\lambda\}_{\lambda \in \Lambda}$ is not a subset of $X$, it is a \emph{collection of subsets of} $X$.
What do you want $\{U_\lambda\}_{\lambda \in \Lambda}$ to be a subset of?

[gentle osprey]

Spamakin🎷

[rain gyro]

Yes I was doing it and I thought maybe it’s a bit circular because that was things I was proving? I was so confused so I changed it to X and back to tau and X

In the beginning it was in tau but at one point I got too confused

But just assume that part is fine for now

[thorny agate]

And then when you say ``there exists $\{ \mathcal{D}_i \}_{I_\lambda \cap \Lambda} \subseteq \mathcal{B}$...'' what are the $\mathcal{D}_i$, what is $I_\lambda$, and why do these sets exist?

[gentle osprey]

Spamakin🎷

[thorny agate]

you should be very explicit in your reasoning

you will be less confused, and your proofs will be clearer

[rain gyro]

By the definition of the basis

Becasue B is a basis of X so the collection exists?

[thorny agate]

What is I_\lambda?

Go back, rewrite what you have so far for that third part

being more explicit

[rain gyro]

A arbitrary subcollection of \Lambsa

[thorny agate]

see if that helps you figure out the proof

[rain gyro]

Okay but still how should I deal with this index set in the end

I rewrote this four times already

[thorny agate]

I feel if you try to be more explicit, you'll either resolve how to deal with your indexing issues

or it'll be at least clearer where you're stuck and you can come back and ask again

[hidden abyss]

I'm also confused at what the $C_i$ are. Your definition of $\tau_\mathcal{B}$ is presumably something like $U \in \tau_{\mathcal B}$ iff $U = \bigcup_{i\in I} B_i$ for some index set $I$ and $B_i \in \mathcal {B}$

[thorny agate]

I think you genuinely have the right idea of how to prove this

[gentle osprey]

Jussari

[thorny agate]

which is why I'm being pushy for you to try to again but be more explicit

[rain gyro]

You’re right! I’ll be working on it for once more… but the index has always been confusing for whatever the question I have

[thorny agate]

you'll get better with working with indexing sets as you struggle with it more

I promise

[rain gyro]

C_I is a subcollection of sets such that for every x in U_lambda we have x in B in C_i_lambda

This is very confusing but I try to keep track with the ground index set so I don’t get totally confused 😵‍💫😵‍💫

Because by definition of basis U_lambda is open for every x in U_lambda there exists C_{i_lambda} such that x in C_{i_lambda}

So I made the complete collection of i_lambda for every point in the single open set U_lambda into one index set and it has to be the subcollection of the ground set Lambda

But in the end I found the indexes got increasingly confusing

[hidden abyss]

then what are the B \in C_i?

[rain gyro]

Becasue the C_I is a subsets of the basis which contains a set such that x in B subset U

Sorry there’s a bit mistakes for my earlier message I am still too confused for this 😵‍💫😵‍💫😵‍💫

[hidden abyss]

the C_i should probably be elements of \mathcal{B} (and subsets of U_lambda)

[rain gyro]

Yes that would make things easier

I would actually making a maze for myself 😭😭😭

This thing starts to get not as nice as a couple days back where the arguments were still simple

[rain gyro]

Is there anything that I can learn specifically about the index sets

[hidden abyss]

You dont really need anything specific about the index sets for this

For the intersection part of the proof, it might be helpful to first consider the intersection of two open sets U, V

From there, you can either generalize the argument for n open sets, or use induction

[tardy carbon]

I think you mean ${U_{\lambda} : \lambda \in \Lambda}$ when you write ${U_{\lambda}}{\lambda \in \Lambda}$. At the core, we have a function (set theorists please don't flame me) $\lambda \mapsto U{\lambda}$.

[gentle osprey]

L

[rain gyro]

Yes exactly that thing I never really understood it properly

[hoary breach]

think of how it generalizes sequences in some infinite (possibly uncountable) tuple way

[rain gyro]

It’s not that I want to study this in particular, but I kinda felt if I did manipulate the index right it should work, but Becasue of confusion I lost all..

[rain gyro]

You’re right I should do it

I just realize that🫣

[rain gyro]

I tried many times but this concept in particular had been super super hard for me

[uneven bronze]

Let $I=[0,1]$ be the unit interval. If $X$ is a topological space and $\mathcal{F}\subset C(X,I)$, we say that $\mathcal{F}$ separates points and closed sets if for every closed $E\subset X$ and every $x\in E^c$ there exists $f\in\mathcal{F}$ such that $f(x)\notin\overline{f(E)}$. In regards to the above theorem, what does $\mathcal{F}$ separate points mean by itself, and why does it imply injectivity? \

Edit: Each nonempty $\mathcal{F}\subset C(X,I)$ canonically induces a map $e:X\to I^{\mathcal{F}}$ by the formula $\pi_f(e(x))=f(x)$, where $\pi_f:I^{\mathcal{F}}\to I$ is the coordinate map.

[gentle osprey]

psie

[uneven bronze]

Ignore the above, I think I understand now.

[rain gyro]

Help, I revised my proof. I realized I made a crucial mistakes by mistaking the concepts of topology and induce topology. In essence, I should’ve tried to use the property of induced topology to show a set is open. So if U is open then for every x in U there exists B in cal B such that x in B subset U.

However this concept was too abstract I felt it should be done by index set (I genuinely think it’s possible but it’s some of a complication)

Is this version correct, I took the advice to study the case for two sets intersecting with one another

[rancid umbra]

you don't need index sets to carry out this proof, if that alternative route is more clear for you at the moment

[rain gyro]

The previous one has a bit of a gap

[rain gyro]

I definitely have to learn a bit about the index set, if manipulated well I felt it’s quite decent.

I actually felt there’s something deep about this index set and that function L mentions..

I looked online it’s a choice function? Not sure, but not necessarily AoC…

[rancid umbra]

i don't think its anything deep. formally, if X is a set, then an indexed family of elements of X with index set I is the image of a function f : I -> X, im(f) = {f(i) in X : i in I} = {x in X : exists i in I with f(i) = x}.

[rain gyro]

It’s abstract because how to reorder all the index 😵‍💫😵‍💫

[rancid umbra]

why do you need to order it at all?

[rain gyro]

And it becomes very complicated I felt if it goes arbitrary.. not like order

But I have to say there exists B in a subset of Basis B

And that language is weird for my brain

Label

I just realized this is actually just a projection

Not the other way around

[rancid umbra]

im not sure what you mean by projection

[rain gyro]

Well not probably a projection projection but somewhat of a projection

[rancid umbra]

i suppose informally you can think of a function as some projection of its domain onto its image

[rain gyro]

Those projectors and choices, index sets are highly unnatural for me

Since I studied (at least I survived) real analysis, I never truly understood those stuff

[rancid umbra]

i think in time you will get used to them

[rain gyro]

Hopefully, I just don’t know why those stuff on the very first chapter on folland’s book turn out to be harder to surviving the whole course… hope when the semester starts I can understand a bit more

[rancid umbra]

what is your definition of an open set in \tau_{\mathcal{B}}?

i can't find it in the conversation above

[rain gyro]

If U is in Tau_b it satisfies for every x in U there exists B in cal B such that x in B subset U

2, if x in B_1 cap B_2, where B_1, B_2 are both in cal B
Then there exists B_3 in cal B such that
xin B_3 subset B_1 cap B_2

[rancid umbra]

cool.

no need for \mathcal{B} \subset \tau in the first sentence.

there is no need for the index \beta on the basis element B. \beta is the index of the U_{\lambda}'s such that x in U_{\beta}.

otherwise, it looks good

[rain gyro]

I took note of it! I’ll fix that tomorrow Inam now too sleepy thanks so much 🥰

[uneven bronze]

If A is nonempty, the product space $I^A$ is called a cube, where $I$ is the unit interval. The claim is that any compact Hausdorff space is homeomorphic to a closed subset of a cube. Indeed, by compact Hausdorff spaces being normal and Urysohn's lemma, we can take $A=C(X,I)$. \

I'm trying to understand this claim. I know there's an embedding from $X\to I^{\mathcal{F}}$ where $\mathcal{F}=C(X,I)$, given by the canonical map $e$ defined via $\pi_f(e(x))=f(x)$. I struggle seeing why in the claim they say a \textbf{closed} subset of a cube. Why is that?

[gentle osprey]

psie

[pulsar lagoon]

Im guessing cuz the graph will be closed

[uneven bronze]

ah ok, because e is simply continuous?

[pulsar lagoon]

yeah

[uneven bronze]

Do you have any idea why mention Urysohn's lemma? Seems superfluous when we have normality (i.e. a compact Hausdorff space), since then it is T1 and admits a family of continuous functions subset of C(X,I) that separate points and closed sets, so e the function is an embedding. No need to mention Urysohn's lemma.

[pulsar lagoon]

I dont think T1 is enough to say that it seperates closed sets

Im not sure why tho

[uneven bronze]

Ok. Since the whole space is both closed and open, won't the image of X be open and closed too?

[pulsar lagoon]

the image will be open in the subspace topology

not in the topology of the cube

cuz like the graph of the constant function 1 is an embedding of R into R2 and the graph is closed but not open

[uneven bronze]

Hmm, ok. I don't think I've learnt the terminology of what it means for a graph to be closed/open.

[pulsar lagoon]

in this case open and closed is in the space where the graph lives

yeah topology is kinda wack when it comes to this

[uneven bronze]

Suppose X is completely regular, i.e. T_{3 1/2}. Then (Y, e) is a compactification of X, where e is the embedding X --> [0,1]^F for some F subset C(X,[0,1]) and Y is the closure of e(X). Now it is claimed that every continuous f in F has a unique extension to Y. Why is this extension unique?

My reasoning goes as follows; we want to show that if distinct f,g in F have the same continuous extension, then f = g on X. I'm a bit confused by a slightly related idea; if f = g on X, then they agree on Y as well since X is dense in Y. Are we using this idea or does what I want to show follow more directly?

[uneven bronze]

I worked it out I think.

[rain gyro]

I completely refined how I prove the topology generated by a basis at a cost I used another definition though equivalent but more constructive.. is this actually correct and rigorous? And is it clearer in argument

Oops a typo

[rain gyro]

And I introduce my old definition as a lemma

[uneven bronze]

Let $I=[0,1]$ be the unit interval. If $X$ is completely regular, if $\mathcal{F}\subset C(X,I)$ separates points and closed sets, $e:X\to I^{\mathcal{F}}$ is the associated embedding, and $Y$ the closure of $e(X)$ in $I^{\mathcal{F}}$, then $(Y,e)$ is a compactification of $X$. It has the property that every $f\in \mathcal{F}$ continuously extends to $Y$ (where we identify $e(X)$ with $X$) and this extension is unique.\

We also have if $f,g$ are bounded continuous on $X$ that extend continuously to $Y$, obviously so are $f+g$ and $fg$, and if $(f_n)$ is a uniformly convergent sequence of functions on $X$ that extend continuously to $Y$, their extensions converge uniformly on $Y$ since $X$ is dense in $Y$, so $f=\lim f_n$ also extends continuously.\

Questions; I'd be grateful if someone could explain the second paragraph in more detail. Why require that $f,g$ be bounded? Can their sum and product be extended because function restriction respects addition and multiplication? I don't see why $f=\lim f_n$ also extends continuously to $Y$ and what this has to do with denseness?

[gentle osprey]

psie

[uneven bronze]

The uniform convergence bit seems clear now, though I would have expected we need boundedness in this very claim (i.e. that their extensions converge uniformly on Y, since we are using the uniform metric). I'm still not sure why we mention that f,g be bounded.

[gaunt linden]

It's a bit weird, because once you assume that f, g extend continuously to Y, you get boundedness for free anyway -- a continuous real function on a compact space is always bounded, and bounded on Y trivially implies bounded on X.

[uneven bronze]

I was thinking that it is weird to even extend continuously an unbounded, continuous function. Is that even possible?

[gaunt linden]

No, for the same reason. If there's a continuous extension to Y, then the image of Y is a compact subset of R, hence bounded, and those bounds work in particular for the values of the original function on X.

(Except in completely different settings where we extend to a space that is not compact. E.g. 1/x on (0,1) is unbounded, but can extend to (0,1] just fine. But that is a boring case!)

[uneven bronze]

by the way, f,g are functions on X and Y is the closure of the embedding of X in a cube (i.e. [0,1]^F where F subset C(X,[0,1])). Of course, we identify X with e(X) where e is the embedding. Since f,g are functions on X, I struggle with interpreting the domains of their extensions. What is the counterpart to Y as a set of points like X is?

[gaunt linden]

Not sure I understand that question. Y is a certain subset of [0,1]^F, and you simply declare its elements to be "points"..

[uneven bronze]

Yes, but I think X is not related to [0,1]^F, or? I.e. X and Y are not subsets of each other. It's true that e(X) is a dense subset of Y.

[gaunt linden]

Well, the 'e' mapping makes it related.

If you have f: X->R, then "g: Y->R is an extension of f" means in this context that g(e(x)) = f(x) for all x.

[livid narwhal]

Is it true?

[gaunt linden]

I don't think it is necessary for p to be continiuous, as long as it is surjective and open. And R can be replaced by any connected space.

[livid narwhal]

What is the meaning of open map?

Take open sets to open?

[livid narwhal]

Yes if this is the definition then continuous is not needed. I misread the question i did not notice that open word.

[unreal stratus]

Is the point that if X = U u V (disjoint) then for each y in R, you can write f^-1(y) as union of intersection with U and intersection with V?

[gaunt linden]

It's not true without "open"; a counterexample would be to take X = (-infty,0) union [1,infty) with the subspace topology and p(x) = x when x<0, p(x) = x-1 when x>0.

[unreal stratus]

Or R with discrete topology mapping to R with usual topology

[uneven bronze]

Let $X$ be a completely regular space. A subalgebra $\mathcal{A}$ of $BC(X)$ is called completely regular if \
(i) it is closed and contains the constant functions, and\
(ii) $\mathcal{A}\cap C(X,[0,1])$ separates points and closed sets (a family $\mathcal{F}\subset C(X,[0,1])$ is said to separate points and closed sets if for any closed set $E\subset X$ and $x\in E^{c}$, there exists $f\in\mathcal{F}$ such that $f(x)\notin\overline{f(E)}$).\

\textit{Attempt}: If we take $f_c \in C(Y)$ equal to constant $c$, then $f_c \circ e$ is the constant $c$ function on $X$, which is thus in $\mathcal{A}_Y$. \

To see that $\mathcal{A}_Y$ separates points, let $x \notin E$ and $E \subset X$ closed. If $e(x) \in \overline{e(E)}$, then $e(x)\in e(\overline{E})=e(E)$ also, implying $x\in E$, contradiction. So $e(x)\notin\overline{e(E)}$. Since $Y$ is completely regular (even normal), separate these by some $f \in C(Y,[0,1])$. Compose $f$ with $e$ and use that $\overline{f(e(X))}=f(\overline{e(E)})$. So $f\circ e\in \mathcal{A}_Y$ separates $E$ and $x$.\

Why is $\mathcal{A}_Y$ closed?

[gentle osprey]

psie

psie

[copper quiver]

I^\mathcal{F} is T2, and X is compact, so the image of the map is a compact in a T2, hence closed.

[uneven bronze]

Proposition 4.56. Suppose that $\mathcal{F}\subset C(X,I)$ separates points and closed sets. Let $(Y,e)$ be the compactification of $Y$ associated to $\mathcal{F}$, and let $\mathcal{A}$ be the smallest closed subalgebra of $BC(X)$ that contains $\mathcal{F}$. Then every $f$ has a continuous extension to $Y$. \

If $X$ is completely regular and $\mathcal{F}=C(X,I)$, the compactification of $X$ is called the Stone-Cech compactification, denoted $\beta X$. Now, I don't understand why when $\mathcal{F}=C(X,I)$, every function from $BC(X)$ extends continuously to $\beta X$. This seems to be saying that $BC(X)$ is the smallest closed subalgebra of $BC(X)$ containing $C(X,I)$. Why is that true?

[gentle osprey]

psie

[tender halo]

remind me what are you reading btw?

[uneven bronze]

It's called Real analysis by Folland. 🙂

[tender halo]

ah i see

thanks

[uneven bronze]

do you understand why every function from BC(X) extends continuously to ßX?

[uneven bronze]

Consider a completely regular space Y. Its Stone-Cech compactification is a pair (ßY, e) where e is an embedding and Y is identified with e(Y), which is supposed to be dense in [0,1]^C(Y,[0,1]). Now is it true that if Y is compact Hausdorff, then e(Y) = ßY?

[gentle osprey]

psie

[uneven bronze]

Anyone who's got any idea about this? 😔

[rain gyro]

Hi, is a restriction equivalent to an inclusion map

[velvet salmon]

That depends a lot on the context

For example restricting functions on a disconnected space to some connected component will usually not be an inclusion

On the other hand restricting to dense opens is oftentimes an inclusion if the space is sufficiently nice

What exactly are the restrictions you're talking about?

[rain gyro]

The maps in context of topology, I was trying to answer a question about restrictions and metric but I didn’t quite know restrictions so I thought maybe there’s a way using the inclusion map view point
Like the inclusion maps on my note

[velvet salmon]

I mean to me restriction refers to restricting a function defined in X to some subset U of X. What does restriction mean to you?

[rain gyro]

For now I only understand it as a composition of a mapping f and inclusion map

[velvet salmon]

Yes that's the same thing

[rain gyro]

Because, I saw his metric problem interesting but I didn’t know inclusion wanted to make sense of it🫣🫣

I will try to prove another corollary like if f X to Y such that for every partition U_i such that sqcup U_i =X. Where for all i f|_{U_i} is continuous then f is continuous

I feel with this statement I might be closer to the core idea for solving his proposed question

[velvet salmon]

Maybe what you wanted to ask was: is the restriction of an inclusion also an inclusion. In that case, the answer is yes

[rain gyro]

That’s such a neat way to expressing my question 🥰🥰🥰

[uneven bronze]

To start with, I'm quite uncertain of how to show injectivity when the map $\Phi$ is defined via a composition, namely $\pi_g(\Phi(p))=\pi_{g\circ\phi}(p)$. We want to show $$\Phi(p)=\Phi(q)\implies p=q.$$But I don't know how $\Phi$ is defined explicity...

[gentle osprey]

psie

[rain gyro]

Is this proof okay ish?

I created that notation $i_{U_\lambda}$ don’t know how to write it properly it means the inverse of $i:U_\lambda\hookrightarrow X$

[gentle osprey]

Emmaaaaa

[rain gyro]

Refined again

[sick elbow]

Hello, Friends:

Can someone please motivate this theorem's proof? (of g in case of X-A). I can verify the proof, but how to think of it.? Is there some general theory, or technique used behind the presentation? Some analogies or similar proofs in other domains of analysis etc will be appreciated. I am clear with Tietze's theorem and it's necessity and importance, what bugs me is the proof.

I don't see any modern book that does it. Most books proof Tietze using semi constructive process of constructing a series of g_i's ( sort of separating functions) and then by uniform convergence, they conclude the result. This one by Dieudonne is constructive.

Many people before Dieudonne too have given similar proofs, but unfortunately all are in French/german and not tralsated. I tried with google tralsate and chatgpt, both are VERY bad trabslators.

why are inf/sup so important here? I mean how does one sense these in the question? I can see they want them to converge to points of A(A is closed), but still things are hazy?

In Case of X=real numbers, I have some intution, that if A is a closed set we can see that it's complement will be union of open intervals and we can define linear maps on those such that the extension is continuous (this is similar to the popular proof of modern books). But these ideas seem to use different appraoch and sense.

[tender halo]

they are important in the sense that they exist

i.e. the function is bounded

the usual proof operates in a normal space, not metric space

you just kinda nudge the function of distance from A and the function f together until it becomes continuous

[sick elbow]

Yeah, this idea and construction is hard to generalise in Normal spaces. I don't know of any such extension. Maybe this is why they stress on that mordern proofs of uniformly convergent sequences of fucntions to be the limit and the extension. But since this is an explicit construction, i wanted to know the motivation.

[tender halo]

motivation for what exactly?

[sick elbow]

the construction of g(x).

[tender halo]

what you need to do is to make sure you converge to f(y) when you take a sequence converging to y

the d(x, y) / d(x, A) will converge to 1, f(y) is the constant factor

you can rewrite the expression as inf(f(y)d(x, y))/inf(d(x, y))

this is why its important that f is between 1 and 2 and not between 0 and 1 for example, its makes it so the f(y) factor does not change what y you will pick as the infimum

i think 1.2 in your second screenshot is the best illustration of the method actually

[sick elbow]

Ok Ok, Thank You.

[sick elbow]

Could you also elaborate on the proof technique of Dieudone's proof? I see he is using epsilon delta proof, case wise dealing of points in or not in A converging to something in A and that he reduces the question to the first case in the last part.

[tender halo]

what i said also goes for dieudonne's proof

they all do the same thing

they just go about it slightly differently

[sick elbow]

Ok Ok. Those arguments seem to have some picture, that's why I asked. The choices of bounds and radius seem very random and still perfect

[tender halo]

the trick is that if you take a sequence that converges to a point of A, lets call it x, you need the function to take the values of f from around x

so you make it so the values around x have the most "weight" for you to choose them

lets say you have a sequence of yn converging to x, and some yi is from radius r around x

then (lets take 1.2 for example), you will have to choose the a for your infimum from the radius 2r at most from x

(i am pretending that there is a value where the infimum is achieved which is strictly speaking not true if the space is not locally compact, but its better for illustration)

[alpine hound]

An n-manifold, by definition, is locally Euclidean of dimension n.

Meaning every point in the manifold has an open neighborhood that is homeomorphic to some open subset of R^n.

My question is:

Is it true or false that not only every point has such an open neighborhood, but every compact set has an open neighborhood that is homeomorphic to an open subset of R^n?

hmmm yep, ok.

[rain gyro]

Is my proof for the corollary correct? Those are the technical lemmas I used

[rain gyro]

Had a typo, and I added that proposition 1.7 which is basically saying A is closed if and only if A=\cl A

It’s quite intriguing though those inclusion are generally false when f is not continuous

[rain gyro]

Help

[rancid umbra]

how are you using minimality? minimality of what?

[rain gyro]

That part on the third direction was a bit unjustified, but ${f^{-1}(\overline{B})}$ is closed after doing a bit set arithmetics should be fine

[gentle osprey]

Emmaaaaa

[rain gyro]

I will refine that part on third paragraph tomorrow, I was a bit tired so kinda didn’t see that glitch but it’s minor

For first minimality

On the first paragraph, Becasue if $f^{-1}(\overline B)$ is closed then it certainly contains the smallest closed set containing

[gentle osprey]

Emmaaaaa

[rain gyro]

$\overline{f^{-1}(B)}$

[gentle osprey]

Emmaaaaa

[rain gyro]

That’s because the monotone property of closure (I know it doesn’t commute with preimage operator and I didn’t assume that) but certainly $B\subset \overline B$

So immediately it justified the part 1

The second part was really just leveraging the order preserving property of preimage operator

The third part where I did have some slips in sealing off the technical details but that is very minor I think if my logic is right I will fix it tomorrow probably rewrite… however I am quite a clueless person so🫣🫣🫣

[gentle osprey]

Emmaaaaa

[rancid umbra]

but you don’t know that f^-1(B) is closed in the first paragraph

[rain gyro]

I do that’s the assumption

Becasue look at the proposition

Proposition 1.7 and that I assumed B to be closed

[rancid umbra]

you are assuming that cl f^-1(B) ⊆ f^-1(cl B), not that f is continuous

[rain gyro]

I know

[rancid umbra]

that isn’t what i am drawing issue with

[rain gyro]

Yes you’re right

I no I can just reverse the inclusion

Since the closure of the preimage must be the smallest set containing it

So immediately I would be able to justify that

I was too sleepy I didn’t notice

Becasue closure itself must be closed

I should just conclude from there

Oops I think I was typing a supset🫣 for the first part

Many mistakes indeed 😢😢

I always make those ultimately silly mistakes. Still need to make my mind more sharp to command all those proposition more fluently and

[rancid umbra]

oh, i see, the first paragraph looks fine. it’s written a bit confusingly tho, which is why i misunderstood what you were saying

[rain gyro]

I was so sleepy and made a couple of mistakes I probably following a sup set for double inclusions but my brain was kinda dead

[rancid umbra]

you should first state the assumption,
“Assume for all B ⊆ Y, we have cl f^-1(B) ⊆ f^-1(cl B),” and then start the proof, “let B be a closed subset of Y,” so that we know what the assumptions are before starting the proof.

[rain gyro]

Yes I actually struggle a lot with proof writing

Lack of formal math education

That’s the 14th questions of the chapter 4 on folland’s real analysis for pointset topology

Finally only 77-14 more

That’s 63

😭😭😭😭😭😭😭😭😭

Why does it have to be this subtle

For all the questions it includes

I thought that kuratowski operator was hard enough and wanted some relaxed moments even this inclusion question was a bit of a teaser since I firstly was like huh it’s probably asking the equality condition or when operator preimage commutes with closure

[rain gyro]

You’re right though I should’ve never made such a sloppy work…

[rancid umbra]

ur getting better!

[rain gyro]

Still it’s easy to think of sets

But some textbooks abstract them further (though fundamentally unchanged) but started to study the properties of those as operators
guess I would just do a couple intro analysis problem to relax a bit and sleep

[gentle osprey]

psie

[uneven bronze]

I struggle with this problem. 😔 Any help is appreciated.

[uneven bronze]

In regards to the diagram above, does it make sense to speak of ßY if Y is a compactification of X? I.e. the Stone-Cech compactification of a compactification?

[rain gyro]

omg this one I’ll be studying this part by the end of summer break 😵‍💫😵‍💫 I just realized we are using exactly the same book.. chap 4 right?

[uneven bronze]

yes 🙂

[rain gyro]

How hard is it the few big theorem there on the book

[uneven bronze]

Well, it's all relative, but I struggle quite a bit. 🥲 There are many instance where the author doesn't spell out, in my opinion, important details.

[rain gyro]

I felt that too… though my basics are bad. I found hard that those basic sets operations are treated as operators and analyse the property of those abstract stuff

[uneven bronze]

Let $(Z,e)$ be a compactification of $X$, and define $\mathcal{A}Z={f\circ e:f\in C(Z)}$. Let $Z'$ be another compactification of $X$, and define similarly $\mathcal{A}{Z'}$. Suppose $\mathcal{A}{Z'}=\mathcal{A}{Z}$. How do I show there's a homeomorphism from $Z\to Z'$? \

I'm tempted to just say $\phi=e'\circ e^{-1}$ according to the diagram below, though I'm not sure this is a homeomorphism, and I'm not sure how to use the assumption $\mathcal{A}{Z'}=\mathcal{A}{Z}$.

[gentle osprey]

psie

[granite crane]

Folland?

[novel stump]

im a bit confused as to why {0} is considered open but {1} isnt, bc isnt the complement of {1} also closed, the same as the complement of {0}? maybe im misunderstanding what they mean?

[ruby delta]

well we just define {1} to be not open

[novel stump]

that makes sense, but then why would {0} be open?

or is that just how its defined

[hoary breach]

yes

[gaunt linden]

You can declare anything you want to be open or closed, as long as the totality of your decisions satisfy the axioms for a topology.

[novel stump]

interesting, so there isnt rly a rationalization behind why the empty set, {0}, and X are open, its just bc thats how theyre defined?

[ruby delta]

well the rationalization is that it is the simplest nontrivial topology you can have

[novel stump]

(or i guess that could be the rationalization)

[hoary breach]

how did you deduce that the complement of {1} is closed

[novel stump]

i realized that this was wrong just now i apologize

[gaunt linden]

Yeah, it is: Let T = {{}, {0}, {0,1}} and notice that this particular T satisfies the conditions.

[hoary breach]

why u apologizing 😭

[novel stump]

ah

[novel stump]

i tend to overapologize 😭

[alpine nest]

Yeah, of course normally which sets are open and which ones are closed follows from some properties, most topologies don't just come out of nowhere. But on the other hand once we have a definition of what "topology" is, we can just create one out of nowhere, so long as it satisfies the required properties.

It won't make much intuitive sense necessarily, but it will be a correct topology.

[novel stump]

that makes sense, ty!!

[hoary breach]

is {emptyset, {1}, X} also a topology on X?

[novel stump]

i dont know unfortunately, i would imagine it could be

[gaunt linden]

Check whether it satisfies the rules!

[hoary breach]

right so do you have a definition of topology yet

[novel stump]

well i know it satisfies that empty set and whole set is open, which is property 1

i know that the union of open sets emptyset \cup X is open from definition, but i dont know how to check if the intersection is open

(i just started the topology chapter yesterday so i have not fully internalized how to check the properties yet)

[hoary breach]

Remember you need to check all the different unions

that would be emptyset cup {1}, emptyset cup X, {1} cup X....

[novel stump]

i would think the only open set would be emptyset cup X since the others would be clopen?

or neither open nor closed

[hoary breach]

no; check as in are those also in T

given elements in T, are the different unions of elements of T also in T

[novel stump]

oh, then yes theyre also in T

[hoary breach]

and then you check the finite intersections which would be...

[novel stump]

emptyset cap {1}, {1} cap X, emptyset cap X

which i think would also be in T

at least it looks obvious to me but i could be oversimplifying it?

i guess i overcomplicated the initial checking of the properties though bc i thought that i had to check if every one of them were open by the definition of being open instead of just loooking if they were in T

[hoary breach]

whether they are in T is what defines them as open

[novel stump]

ohhhhh

okay that makes much more sense

[hoary breach]

unless you were given a different definition of open here which would be uhhh

[novel stump]

i think i have 2 definitions of open

being an element of a topology and then the usual open ball contained in a subset for every point in the subset

i just misunderstood the definition in terms of being an element of the topology

[hoary breach]

ok so how would you define a ball given the definition of a topological space

[novel stump]

the ball is an element of the topological space

i dont rly know how i would define it otherwise

[ruby delta]

open balls are only defined in metric spaces

otherwise in general topological spaces there is no meaningful way to talk about what the ball around a point is

[novel stump]

ohhhhhhhhhhhhh

that makes much more sense

[ruby delta]

and, in a metric space, we define the induced topology to be the topology generated by the open balls

hence why they are called open (for one of various reasons)

[hoary breach]

and not all topologies come from metrics either. The T we defined earlier is one example

[novel stump]

i remember reading this yes

like the indiscrete topology if X has more than one point iirc?

[ruby delta]

the indiscrete topology on a set X is the topology with only 2 open sets

namely ∅ and X

[alpine nest]

The indiscrete topology on the emptyset has just one open set

[novel stump]

yeye, my book says "The indiscrete topology does not come from a metric if X has more than one point, again because points are not closed"

maybe theyre talking about a different topology bc "points are not closed" doesnt rly make sense to me

wait no theyre not

nvm

[ruby delta]

recall the definitions of a metric space

if two points are not the same, their distance is necessarily ____?

[novel stump]

greater than 0

but equal to 0 if and only if the two points are the same

[ruby delta]

so, if a set has at least 2 points, at least how many distinct open balls are there

that aren't ∅ and the whole space?

[novel stump]

at least 1?

[ruby delta]

so can it be the indiscrete topology?

[novel stump]

no

[ruby delta]

so that's your answer

[novel stump]

ohhhhhhhhh

ty!!

[granite crane]

can i know the name of the book if you dont mind

[novel stump]

unfortunately no bc it will doxx me, its my professors own notes

i'm using Gamelin as supplemental reading though

[granite crane]

thats fine

[novel stump]

i love your pfp by the way afzal!

[granite crane]

Thank you Tabby

[novel stump]

of course!!

[uneven bronze]

Let $X$ be a Tychonoff space and $(Y,e)$ a Hausdorff compactification of $X$. The function $e$ is an embedding from $X$ into $Y$, such that $\overline{e(X)}=Y$. Define $\mathcal{A}_Y$ as all the bounded continuous functions on $X$ that can be continuously extended to $Y$.\

$e$ has a unique, continuous and surjective extension to $\beta X$, the Stone-Cech compactification. Denote this extension by $\beta e$.\

Claim. A bounded continuous function $f$ on $X$ belongs to $\mathcal{A}_Y$ if and only if its Stone-Cech extension $\beta f: \beta X \to \mathbb{C}$ is constant on the fibers of the map $\beta e$ (the fiber of $y$ is the set of all $x$ such that $g(x)=y$).\

Could someone explain why this claim is true?

[gentle osprey]

psie

[tender halo]

do you want a proof or a moral justification

[uneven bronze]

well, a proof would be great 🙂

a moral justification would also be great

whatever floats your boat

[tender halo]

a proof is that \betaX -> Y is a quotient map and a function \betaf can be lifted along \beta e iff the aforementioned condition

[uneven bronze]

hmm ok, I have to look up what lifted means

[tender halo]

this is what i mean exactly, you dont need to look up the category theoretic language

the moral justification is thus

\beta X is a space where you have X where all possible limits are added and are distinct from each other. any other compactification is gluing together some of those limits, and an extension of a function is only possible if all those limits agree on their image under f

fibers of \beta e are what limits you are gluing together into a single point

[uneven bronze]

ok

[tender halo]

(note that fibers of points of X are just the exact same point again, and the fibers of points Y \ X are closed subsets of \betaX \ X)

[pulsar lagoon]

Here I think O is open and these sets are open since they are unions of sets of the form $Os$ or $sO$ and these are open since left and right translations are homeomorphisms so images of open sets are open

[gentle osprey]

Khush

[pulsar lagoon]

I dont seem to understand why we need the topology to be T2

[queen prism]

wtf

[pulsar lagoon]

idek

I asked in #groups-rings-fields and someone suggested open subset

so thats what im taking it to be

[viral atlas]

We don't need that here, yeah

Never saw that notation for an open subset before, but it looks reasonable in context

(I tend to write "op" or "cl" over the subseteq sign for open and closed, respectively)

[plush folio]

Lol, that's from the lecture notes I sent you I can confirm, that notation means open subset

There might be some mistakes in those notes btw, so be mindful of that

[pulsar lagoon]

yeah if no one were to respond I was gonna link you

[loud harbor]

Is it okay if I ask for help here, it's not a specific question rather just something I would like for someone tol explain to me specifically

One of the problem sets I was assigned for my topology class has us computing one point compactifications for specific topological spaces, the only one im familiar with is the n sphere minus a point being homeomorphic to R^n

We are asked to compute it for the naturals and omega_1

is there a particular way to approach this?

[tender halo]

no particular way i guess, i would take the space that you expect to be the answer, prove that its compact and thats the other space embeds into it in the natural way

[ruby delta]

The intuitive explanation of one point compactifications is that you fold all the infinities and open edges into the same point

So usually most one point compactifications will look like some kind of inversion

Ie in the naturals case can you somehow “fold” infinity into a convergent point

[tender halo]

you need not actually prove that it is the one point compactification because you actually get it for free

if you have X \cup {y} such that it is Hausdorff and compact then it can only be the one point compactification

[uneven bronze]

Consider the Stone-Cech compactification of $X$, i.e. $\beta X$, and two other compactifications $Z,Z'$ of $X$. $\beta X$ is compact Hausdorff, so that $C(\beta X)$ separates points by Urysohn's lemma. Moreover, $Z,Z'$ can be realized as quotient spaces of $\beta X$. Is it true that if the two equivalence relations defining $Z,Z'$ are respected by the exact same subset of $C(\beta X)$ (i.e. $\mathcal A\subset C(\beta X)$ and $x\sim y$ iff $f(x)=f(y)$ for each $f\in \mathcal A$), then the equivalence relations must be identical?

[gentle osprey]

psie

[lucid ocean]

is the statement that every disjoint pair of compact sets may be separated by a function to [0, 1] equivalent to complete Hausdorffness, or is it strictly stronger

[uneven bronze]

I think I get it.

[uneven bronze]

Actually, I still don't understand. We want to reach the conclusion $$\mathcal{A}Z=\mathcal{A}{Z'}\implies \beta e(p) = \beta e(q) \iff \beta e'(p) = \beta e'(q).$$But $\mathcal{A}_Z$ is not a subset of $C(\beta X)$. Hmm.

[gentle osprey]

psie

[lucid ocean]

if a counterexample exists, it cannot be the expansion of topology of a normal Hausdorff space, because it is a separation property in the sense I've defined before

[tender halo]

yeah

you can completely separate a point and a compact subset and then you can completely separate two compact subsets

[rain gyro]

Is the example correct? I tried to make sense the relations between the inverse of choice function and coordinate operations

So basically if the product is arbitrary I got something like $U_\gamma \times \prod_{\alpha\ne \gamma} X_\alpha$

[gentle osprey]

Emmaaaaa

[tender halo]

\Lambda doesnt have an innate ordering usually

[rain gyro]

Or I messed up again 🫣

[tender halo]

so it doesnt make sense to take the "first" coordinate

[rain gyro]

So I just let an element be there and $\alpha\ne \gamma$

[gentle osprey]

Emmaaaaa

[rain gyro]

Is this better

[tender halo]

the last part doesnt make sense, \alpha should run through \Lambda \ {\gamma} maybe?

[rain gyro]

It was macros for the (x_alpha)

Was it the mistake?

[tender halo]

sure, that is more or less correct, although I would just say that $(x_\lambda) \in U_\lambda \cross \prod_{\alpha \ne \gamma}X_\alpha$

[rain gyro]

The coordination maps’ inverse is restricting the sets to U_lambda for the coordinates lambda the rest is unchanged. Should i vaguely understand it that way

[gentle osprey]

bussy beaver

[tender halo]

yea

[rain gyro]

Thanks 🥰

[rain gyro]

Is the derivation correct

[quick crane]

why is any derivation needed? By definition, $\pi_{\alpha}^{-1}(U_{\alpha})$ is open for every $\alpha$ since these sets generate the product topology, and then a finite intersection of open sets is open.

[gentle osprey]

L

[rain gyro]

You’re right

I didn’t think about that 😭😭😭😭

I was thinking huh it wasn’t easy…

[tender halo]

to me its not a derivation even that just how the product topology is usually defined?

you just wrote down the expression for the standard basic sets for the product topology

[rain gyro]

Kinda of, I just felt a bit unclear because I was proving that the basis of product topology is a basis and the second axiom requires that there exists an set in the basis contains the intersection

And I was trying make the argument hold for that there exists an open set in basis B… I just overall found it too hard so I tried to kinda make sense for each part 😢😢

[rain gyro]

Thanks you two 😢 for answering all the dumb questions for me

[lucid ocean]

a proof of that would look like taking two points x, y in C and finding a function f_y where f_y (y) = 0, f_y (x) = 1

then over all y in C, taking the infimum f_y to obtain a continuous function f where f(C) = {0} and f(x) = 1, then repeat

but I'm not fully convinced this works as stated

what's a more fleshed out proof of the equivalence

[lucid ocean]

(I believe it to be true, but I haven't been able to come up with an airtight, detailed argument)

[rain gyro]

Is this proof correct or my idea is correct 🫣🫣
2.4 is really just composition of continuous maps remains continuous

[tender halo]

lemma: two sets A and B are completely separated iff they are contained in disjoint zero sets (exercise), moreover you can choose the zero sets such that they are nbhds of A and B respectively

[lucid ocean]

that's relatively innocuous, if f and g have zero sets containing A and B (and the zero sets are disjoint), then f/(f+g) suffices

[tender halo]

lemma 2: if you have a compact set A and a set B such that B is completely separated from every point of A, then A and B are completely separated.
proof: for each point of A take the nbhd zero sets U_x and V_x that contain x and B respectively, take the finite subcover U_s of U_x, then \cup U_s and \cap V_s are disjoint zero sets containing A and B respectively

if you have two compact subsets A and B, first apply lemma 2 to A and each point of B, then you apply it to B and A

[lucid ocean]

right, hold on

to "expand" any zero set by a real function f, I guess do max(0, min(1, (2a + 1)f - a))

right I see

that works out, I was wondering how the patchwork could work

at the end you only have finitely many functions to min over

[tender halo]

wdym expand, do you want to prove that zero sets are closed wrt (countable) intersection and (finite) union?

[lucid ocean]

as in if A is the zero set of f, one can make the zero set larger to be an entire neighbourhood of A

[tender halo]

just take f^-1([0; a])

it cointains the nbhd f^-1([0; a))

[lucid ocean]

that's not the zero set f^(-1) ({0}), but basically the idea

[tender halo]

i mean its still a zero set

just of some other function

[lucid ocean]

the function f'(x) = max(0, min(1, (2a + 1)f(x) - a)) has f'^(-1) ({0}) = f^(-1) ([0, a])

yeah

well anyway, that checks out then

[tender halo]

for countable intersection you take the f = 1/2^i f_i

for union you take uhh