point-set-topology

[uneven bronze]

Could someone clarify why the subspace topology of $X\times{y_0}$ as a subspace of $X\times Y$ is the same as the product topology on $X\times{y_0}$ when ${y_0}$ is equipped with the subspace topology inherited from $Y$?\

Context; I'm working an exercise where I'm showing $X\times{y_0}$ is homeomorphic to $X$ for $y_0\in Y$, and it is mentioned $X\times{y_0}$ has the relative topology from $X\times Y$.

[gentle osprey]

psie

[unreal stratus]

This actually holds more generally for any product and sunspaces I.e. if you have subspaces A of X and B of Y, then the subspace topology on A x B (as a subspace of X xY) is the same as the product topology (where A,B are given the subspace topology from X and Y). The easiest way to see this, imo, is to write out bases for the topology of each and observe they're the same

There is also a slightly fancier way by universal properties etc but the above is nice

[uneven bronze]

Ok, I was looking here, and the answer is just overwhelming me. 😔

Perhaps that wasn't what you meant.

[rancid umbra]

that might be the longest mse post i’ve ever seen

[unreal stratus]

Lol that answer seems very over the top

The proof is quite short

[untold lily]

damn, the universal property proof for that is slick as hell, wouldn't have thought of proving like that

though I should have because you have two spaces defined by universal properties

[quick crane]

You can use nets for an easy proof

[tender halo]

i think this one is best proven by the method of "looking at it very hard"

[topaz hazel]

Hi I've been stuck on the (b) part of this. It's clear to me why dim X \ge sup dim U_i but not the other side.

It's from hartshorne btw but i think it's more of a point set Topology problem. Let me know if I am wrong I'll move it

dim X of a topological space X is defined as the supremum length over the ascending chain of irreducible closed sets.

And a subset is irreducible if it can't be written as the union of two proper closed subsets of itself (not necessarily disjoint).

[quartz horizon]

Yeah universal properties are very convenient for proofs like these

[uneven bronze]

Consider the above exercise. Could someone explain the last sentence in parenthesis to me? I don't really know what a "category-theoretic product" is, but isn't X simply a Cartesian product? What makes it into a "category-theoretic" product?

[quartz horizon]

this is to do with universal properties

namely that $X$ satisfies the universal property of being a product of the $X_\alpha$'s

[gentle osprey]

Pseudonium

[quartz horizon]

the way i like to phrase it is

for any other topological space $Y$, continuous maps $f : Y \to \prod_\alpha X_\alpha$ naturally correspond to families of continuous maps $f_\alpha : Y \to X_\alpha$

[gentle osprey]

Pseudonium

[quartz horizon]

in that you can (coherently) interconvert between the two

by packaging/unpackaging the continuous maps

[uneven bronze]

Ok, thanks for this alternative description 👍 I'll think about it.

[uneven bronze]

Consider the statement that a countable product $\prod_{i\in\mathbb N}X_i$ of first countable spaces $X_i$ is first countable. Let $x=(x_1,x_2,\ldots)$ be an element in the product. Each $X_i$ is first countable, so there is a countable neighborhood base $\mathcal B_i$ at $x_i$. This means each element in $\mathcal B_i$ contains $x_i$ and if $U$ is a neighborhood of $x_i$, there is an element in $\mathcal B_i$ contained in this neighborhood. \

Now they claim that the collection of sets of the form $\prod_{i\in\mathbb N}U_i$, where $U_i\in\mathcal B_i$ for finitely many indices $i$ and $U_i=X_i$ otherwise, is a countable neighborhood base at $x$, since every neighborhood of $x$ contains some elemnt of this collection. \

I don't see why \
(a) sets of the form $\prod_{i\in\mathbb N}U_i$ are a neighborhood base at $x$?\
(b) this is a countable collection?

[gentle osprey]

psie

[queen prism]

yes metric spaces are ok

[main ore]

does anyone here know where the original first proof of the fact that the distance from point to subset of a metric space is continuous?

or who did it first?

Could someone who reads german perhaps point me to where that is in here? if it is here at all?

[tardy carbon]

For (a), recall that the product topology is the smallest topology containing $\pi_i^{-1}(U_i)$ for all $i \in \mathbb{N}$ and $U_i$ open in $X_i$. Then recall that the smallest topology containing some collection $\mathcal{C}$ is the set of all arbitrary unions of finite intersections of sets in $\mathcal{C}$. You can use these to prove (a)

[gentle osprey]

L

[hidden abyss]

Page 291, (note that he denotes the distance by δ(A,x), and uses d(A,x) for the supremum of all distances)

[hidden abyss]

^

[main ore]

thanks so much

but supremum? not infimum?

is it the proof labeled (1)?

[hidden abyss]

Yeah δ is infinum, d is supremum

Proof is (2) implies (4), so if b_n is a sequence converging to c, then δ(A,b_n) converges to δ(A,c)

[main ore]

the introduction to this chapter seems to say something like "a metric space has something that can only be described in a purely topological way with difficulty, namely the relationship between the neighborhoods of different points, mediated by the same radius."

is this accurate?

[hidden abyss]

Sounds right to me

German is not my first language though

[main ore]

what does it mean? i know topology and metric spaces but like what does it mean?

[rancid umbra]

let Z_0 ⊂ Z_1 ⊂ … ⊂ Z_n be an ascending chain of distinct irreducible closed subsets of X.

Since the collection of U_i’s covers X and Z_0 is non-empty, let U be some U_i containing a point of Z_0.

Show that the collection U ∩ Z_k is an ascending chain of distinct irreducible closed subsets of U.

[robust arch]

What are the requirements that a map f: X -> Y preserves straight lines/geodesics? I'm pretty sure X and Y have to be diffeomorphic but is that it?

[gaunt linden]

That's more a #diff-geo-diff-top question.
But "diffeomorphic" cannot be enough. The euclidean and hyperbolic planes are diffeomorphic, but a map between them cannot preserve lines. In the hyperbolic plane there are lines k, l, m, such that each of k and l is disjoint from m, and k,l intersect in a single point. This configuration is not possible in the euclidean plane.

(At least if you're implicitly wanting your maps to be bijections. You can make a map that maps the entire hyperbolic plane to a single line in R^2 such that every hyperbolic line maps to the entire image -- at the cost of being very non-injective even on each line separately.)

On the other hand, the projection from R^2 to the flat torus R^2/Z^2 preserves geodesics, but the spaces are clearly not diffeomorphic.

[uneven bronze]

Just trying to make sense of an exercise in my book.\

Exercise. If $(X,\mathcal T)$ is completely regular, then $\mathcal T$ is the weak topology generated by $C(X)$.\

A completely regular space $X$ (or Tychonoff space) is a $T_1$ space and for each closed $A\subset X$ and for each $x\notin A$ there exists $f\in C(X,[0,1])$ such that $f(x)=1$ and $f=0$ on $A$. Now the author has used $C(X)$ to denote the space of continuous complex-valued functions on $X$, so in regards to the definition of a completely regular space and the notation for $C(X)$, does $C(X)$ still denote the continuous complex-valued functions in the exercise or perhaps only the continuous real-valued functions?

[gentle osprey]

psie

[gaunt linden]

I don't think it really matters. You can generate T by just looking at real-valued maps, but you don't get any additional open sets by considering complex-valued maps too.

[uneven bronze]

I wonder why that is so...

[gaunt linden]

It doesn’t have anything in particulare to do with complex numbers. All the maps you're considering are already continuous according to T, so once you have enough of the maps to generate all of T, wanting even more maps to be continuous will not change which topology you're generating.

(It does matter that the usual topology on R is also the subspace topology relative to C, so a map that happens to take values in R is continous X->R iff it is continuous X->C).

[uneven bronze]

After carefully reading through your messages again, I still don't see why $C(X,\mathbb{R})$ and $C(X,\mathbb{C})$ would generate the same weak topology. How would one actually prove this?

[gentle osprey]

psie

[uneven bronze]

Under the assumption, of course, that X is completely regular.

[tardy carbon]

First, since $C(X, \mathbb{C}) \supset C(X, \mathbb{R})$, we have $\mathcal{T}(C(X, \mathbb{C})) \supset \mathcal{T}(C(X, \mathbb{R}))$

[gentle osprey]

L

[tender halo]

continuous functions from X to anywhere generate topology that is (not necessarily properly) weaker than X's

so you need to take enough continuous functions so that the topology is equal to X

continuous functions to R are already enough, so continous functions to C (which already include the R ones) wont change anything

[tardy carbon]

Now suppose $T$ is a topology on $X$ such that all $f \in C(X, \mathbb{R})$ are continuous. Suppose $f \in C(X, \mathbb{C})$. Then $f = a + ib$, where $a, b \in C(X, \mathbb{R})$. By assumption, $a, b$ are continuous with respect to $T$. Hence $f$ is continuous with respect to $T$. Thus all functions in $C(X, \mathbb{C})$ are continuous wrt $T$, so $\mathcal{T}(C(X, \mathbb{C})) \subset T$.

[gentle osprey]

L

[tardy carbon]

Thus $\mathcal{T}(C(X, \mathbb{C})) \subset \mathcal{T}(C(X, \mathbb{R}))$

[gentle osprey]

L

[uneven bronze]

Ok, thank you all. 👍 I get it now.

[uneven bronze]

Though where did you use the completely regular assumption of X, if anywhere?

[tardy carbon]

I didn't, this works for any topology on X.

shows that all functions in C(X, R) are continuous iff all functions in C(X, C) are continuous

[uneven bronze]

Ok, so if I understand things correctly, C(X,R) and C(X,C) always generate the same weak topology, but under the additional assumption that (X,T) is completely regular, we have that T is actually this topology generated by C(X,R) or C(X,C).

[tardy carbon]

Yes. For example, for any metric space X, the metric topology on X is the one generated by C(X, R)

[tender halo]

in fact, any Y and Y^\kappa for a cardinal \kappa would generate the same topology

[gaunt linden]

You'll need this assumption to show that C(X,R) actually generates T.

[uneven bronze]

Ah, ok.

[gaunt linden]

C(X,R) and C(X,C) always generate the same weak topology
That's true, I think (as L argued), but it is a stronger claim than you need here.
In this particular case, you only need (from "completely regular") that C(X,R) generates T and that C(X,C) is some superset of C(X,R).
If we have two families of continuous functions A and B with A subseteq B, then the weak topology generated by B is between the weak topology generated by A and the original topology. But in this case, we already know that the topology generated by A is the same as the original topology, so that's what B has to generate too.

[uneven bronze]

I have another slight notational confusion. When we say that f in C(X,R) is continuous, it is with respect to the usual topology on R and some fixed topology on X, right? Now, what does "the weak topology generated by C(X,R)" actually mean? It's supposed to be the weakest topology on X such that all f in C(X,R) are continuous. Thinking about this makes me feel slightly weird, because when we write f in C(X,R), aren't we already assuming it is continuous?

[gaunt linden]

In the exercise you already have some topology T on X, and it is according to that topology you pick the functions in your C(X,R).
The exercise says, more or less, "if you forget where that particular family of function came from, and just look for the weakest topology for X that make them continuous, the topology you get will be the same as the topology you started with".

[quartz horizon]

Genetically the way this works is

You have the set of all functions from X to R

[gaunt linden]

In other words, for a completely regular space, the answer to "which functions are continuous?" is enough to reconstruct the answer "which sets are open?".

[quartz horizon]

You choose some subset of this to be “continuous” functions

One way of choosing such a subset is by picking a topology on X

But you can also just, like, pick functions you like :3

[quartz horizon]

Anyway, once you’ve chosen a subset S of such functions

You can define a topology on X by - a function f : Y -> X is continuous iff all the composites g o f : Y -> R for g in S are continuous

Here Y is an arbitrary topological space

It turns out this notion of continuity can be described by a topology on X, which is called the weak topology

(Note that proving this takes some effort)

[uneven bronze]

Ok, thanks. 😅

[quartz horizon]

-# hope my explanation was understandable…

[uneven bronze]

Kind of. In my book, the weak topology is the weakest topology on X (note, the domain) such that a family of functions f_a : X -> Y_a are continuous.

[quartz horizon]

Yes, so you can show this follows from such a topology I described

The identity X -> X is always gonna be continuous

This implies that all the functions g : X -> R for g in S will be continuous in the weak topology

It being the weakest topology such that this works turns out to translate into this condition on when a function Y -> X is continuous

[uneven bronze]

Ok.

[uneven bronze]

Proposition: If $X_\alpha$ is connected, then $\prod_{\alpha\in A}X_\alpha$ is connected.\

Proof: If $X=\varnothing$ then $X$ is connected. Otherwise choose $x \in X$ and let $C \subset X$ be the connected component of $x$. Suppose that $C \neq X$. Then there exists $y \in C^c$, and $C$ is closed, so there exist $\alpha_1, \ldots, \alpha_n \in A$ and $U_{\alpha_1} \subset X_{\alpha_1}, \ldots, U_{\alpha_n} \subset X_{\alpha_n}$ open such that $y \in \cap_{k=1}^n \pi_{\alpha_k}^{-1}\left(U_{\alpha_k}\right) \subset C^c$. Define $z_k \in X$ by $$\pi_\alpha\left(z_k\right):= \begin{cases}\pi_\alpha(y), & \alpha \in\left{\alpha_1, \ldots, \alpha_k\right} \ \pi_\alpha(x), & \alpha \notin\left{\alpha_1, \ldots, \alpha_k\right}\end{cases}$$
for each $k \in{0,1, \ldots, n}$, so that $z_0=x$ and $z_n \in \cap_{k=1}^n \pi_{\alpha_k}^{-1}\left(U_{\alpha_k}\right)$. Given $k \in{1,2, \ldots, n}$ define $i_k: X_{\alpha_k} \rightarrow X$ by
$$\pi_\alpha\left(i_k(p)\right):= \begin{cases}p, & \alpha=\alpha_k \ \pi_\alpha\left(z_k\right), & \alpha \neq \alpha_k\end{cases}$$ so that $\pi_\alpha \circ i_k$ is continuous for all $\alpha \in A$ and $i_k$ is continuous. And the proof continues.

[gentle osprey]

psie

[uneven bronze]

I'd be really grateful if someone could explain why the composition is continuous for all alpha and i_k is continuous?

[gaunt linden]

[responding to a now deleted post] Each z_k differs from z_(k-1) in exactly at most one component.

[uneven bronze]

Indeed.

[gaunt linden]

It follows fairly directly from the definition of the product topology that a function f: Y->X is continuous iff all the compositions pi_a o f: Y -> X_a are continuous. Since i_k is defined by giving all these compositions (as either constant functions or the identity), i_k is continuous.

[rancid umbra]

there is a really nice proof of this using the fact that a space is connected if and only if for every open cover, there is a finite chain between any two elements in the cover

[gaunt linden]

The intuitive point about the i_k maps is that if you map X_a into X by letting all the other components be constant, then that map is a homeomorphism between X_a itself and the subspace topology on its image in X. So you can create opens sets in X_a by intersecting an open set in X with the "hyperplane" with all-but-one components fixed.
(The textbook proof will probably take a shortcut and be satisfied with having argued that i_k is continuous, but homemorphism is what you should really have in mind).

[uneven bronze]

Consider again; a product $X=\prod_{i\in I}X_i$ of connected spaces $X_i$ is connected. Here's a slightly different proof (the same idea, but phrased slightly differently).\

Proof: Fix points $p_i \in X_i$ for each $i$, and define $$Y = { (x_i) \in \prod_i X_i: {i \in I: x_i \neq p_i } \text{ is finite }}.$$ Now for each fixed finite subset $F \subset I$, define $Y_F = { (x_i) \in \prod_i X_i: \forall i \notin F: x_i = p_i }$.By the obvious homeomorphism, $Y_F$ is homeomorphic to $\prod_{i \in F} X_i$, which is connected (assuming we have proved the finite case already). So all $Y_F$ are connected, all contain the point $(p_i){i \in I}$ of $\prod{i \in I} X_i$, and their union (over all finite subsets $F$ of $I$) equals $Y$. So as the union of connected subsets of a space, $Y$ is a connected subspace of $\prod_{i \in I} X_i$. ... \

My question; is $Y$ the connected component of $(p_i){i\in I}$, i.e. the maximal connected subset that contains $(p_i){i\in I}$, or is $Y$ simply a subset of the connceted component?

[gentle osprey]

psie

[uneven bronze]

I think the answer is negative. Y is a proper subset of the connected component of p.

[gaunt linden]

Well, since the conclusion is that X is in fact connected, "the connected component of p" will turn out to be all of X, of which Y is (except in degenerate cases) a proper subset.

[uneven bronze]

Ah true.

[amber musk]

let the Euclidean sphere $\mathbb{S}^n \setminus { e_{n+1} }$ be deprived of its north pole $e_{n+1}$ and $Y={ y \in \mathbb{R}^{n+1} (y/e_{n+1})=0 }$. So the sphere deprived of its north pole is homeomorphic to $Y$ and we have what is called the stereographic projection but I don’t understand the notation $(y/e_{n+1})$, does it mean the scalar product?

[gentle osprey]

tm

[gaunt linden]

That notation looks pretty weird.

[amber musk]

fr there’s no further explanation in the book

about this notation

[gaunt linden]

Can you show a picture of how it looks in the book ideally with a bit of context?

[rancid umbra]

y/e_{n + 1} is the (n + 1)th component of y

its like dividing out by the last component to get the scalar attached to it

[rancid umbra]

horrible abuse of notation and i hate it

[amber musk]

so Y is just {y in R^n+1 , y_n+1=0}

[rancid umbra]

pretty sure

it makes sense in context too

[gaunt linden]

Ah, you were missing the semicolon in the set builder, which made it look even more horrible.

[rancid umbra]

lol i saw that

[gentle osprey]

tm

[amber musk]

if u need the translation

[gaunt linden]

I've often thought there should be a notation for that, but this is not one I love ...

[rancid umbra]

yea, this the usual homeomorphism onto R^{n - 1} x 0

[amber musk]

alright thanks

[rancid umbra]

its a bit annoying to mention dual bases if you don't have to, but those are unambiguous

or maybe just like pi_k for F^n

[main ore]

Assume B is compact, we immediately get that is it bounded and closed. This is from another well-known theorem. Now all we need to do is show it is equicontinuous. Let $\epsilon > 0$ be given. By compactness, there's a finite number $f_{i} \in B$ such that $b \subseteq B(f_{i},\epsilon)$ that is, there is a finite cover of B. Thus $\forall f \in B, \exists f_{i} \in B, f_{\infty}(f,f_{i}) < \frac{\epsilon}{3}$

[gentle osprey]

Ameoba 💃🏼

[main ore]

this is my attempt so far at a proof of Arzela-Ascoli's theorem. what i dont get is that last part Thus $\forall f \in B, \exists f{i} \in B, f{\infty}(f,f_{i}) < \frac{\epsilon}{3}$

[gentle osprey]

Ameoba 💃🏼

[main ore]

for reference this is my theorem: Let X be a compact metric space, and let $\mathcal{C}(X,\mathbb{F})$ be the set of all continuous function from X to either $\mathbb{R}$ or $\mathbb{C}$. Equip $\mathcal{C}(X,\mathbb{F})$ with the $d_{\infty}$ metric, then $B \subseteq \mathcal{C}(X,\mathbb{F})$ is compact iff B is bounded, closed and equicontinuous.

[gentle osprey]

Ameoba 💃🏼

[main ore]

nvm yall i got it 👍

[pallid comet]

Is there a countable totally ordered set whose order topology can't be embedded in the interval?

[tardy carbon]

section 3.4 and the exercises here: https://mtaylor.web.unc.edu/wp-content/uploads/sites/16915/2018/04/anal1v.pdf

[prime elbow]

I don't get this statement, P^m is m dimensional projective space

[ruby delta]

what part don't you get

[prime elbow]

P^m has precisely two representative on the sphere S^m

[ruby delta]

what's your definition of P^m?

[prime elbow]

P^m is the equivalence classes on R^m+1 defined by a, b in R^m+1, aRb iff a = rb, for some r in R

[ruby delta]

well, by looking at the equivalence class on R^m+1 you can see that it is the same as considering the equivalence class on S^m u {0} (ignore the origin for now)

what are the equivalence classes of that relation on S^m?

[ruby delta]

also it should be some nonzero r in R

[prime elbow]

Yes my bad

Does it make change?

Yes it will be

[hidden abyss]

Don't you normally remove the origin before quotening out?

[ruby delta]

you dont have to, it still remains in its own equivalence class as long as you require r != 0

[hidden abyss]

yeah but then you have an extra (non hausdorff) point

[prime elbow]

I got it it has two representative

[ruby delta]

how so?

[prime elbow]

How is P^m being obtained from the sphere S^m after identification of each point with its negative?

[hidden abyss]

the only neighbourhood of [0] is the whole space

[ruby delta]

I don't see how this differs from the normal construction of taking it away and readding it

[hidden abyss]

what do you mean by readding it?

[ruby delta]

this

oh, wait, does projective space even have the origin

I could swear it does

if no then I was mistaken

[hidden abyss]

I've never seen it defined with the origin

[ruby delta]

huh, it doesn't

that's... weird

I just assumed by default it had to have a zero element

or else how do you get the ring correspondence

[hidden abyss]

is that an AG thing or something?

[ruby delta]

yes, that's where I initially learned about them

just checked hartshrone really quckly

they exclude the origin as well

[hidden abyss]

at least topologically its nice to have it be a manifold

and to have S^m as the universal cover

[ruby delta]

yeah I forgot about that as well

[iron bolt]

From what I've seen the unit group Mˣ of a topological monoid M is usually equipped not with the subspace topology coming from the inclusion Mˣ → M, but with the topology coming from the embedding Mˣ → M ⨯ M, g ↦ (g,g⁻¹) - is there a reason for that? I find it hard to imagine that there are counterexamples where inversion fails to be continuous with respect to the former topology, but also can't see a reason why such counterexamples can't exist

... and of course, just after I sent that I realised that the existence of such counterexamples is equivalent to the existence of topological monoids in which inversion exists but is discontinuous 😅

I haven't seen an example of those yet, but it feels much more plausible that they exist - after all, if not that would mean that continuity of inversion was entirely automatic

[quartz horizon]

https://math.stackexchange.com/questions/151878/why-metrizable-group-requires-continuity-of-inverse

[gaunt linden]

How about (R,+) with the Sorgenfrey topology?

[iron bolt]

that seems to work, yup

[gaunt linden]

Ah, that was exactly the example in Pseudonium's MSE link too. :-D

[iron bolt]

yep. thank you both ^^

[hidden abyss]

Is this proof that a double cover $p: E\to X$ admitting a continuous section $s$ is trivial correct?

The "flip map" $\tau: E\to E$ which swaps the points in each fiber is continuous: for $e\in E$ we can choose a locally trivial neighbourhood $U \subseteq X$ of $p(e)$, then $\tau$ is continuous on the neighbourhood $p^{-1}(U)$ of $e$. Since $\tau$ is an involution, it follows that its a homeomorphism.

Now we can define a continuous map $f: X\sqcup X\to E$ by sending the first copy of $X$ via $s$ and the other via $\tau \circ s$. Pre/post-composing via $p\sqcup p$ and $p$ respectively gives the identity maps, so $f$ is a homeomorphism and we are done.

[gentle osprey]

Jussari

[hidden abyss]

I'm mostly concerned because this answer claims that this is true assuming X is "sufficiently nice" and connected. I don't see where I'm making any assumptions on X https://math.stackexchange.com/a/2490785/877430

[quartz horizon]

I guess this follows from viewing a double cover as a principal $C_2$-bundle right

[gentle osprey]

Pseudonium

[quartz horizon]

It’s true generally that principal $G$-bundles with global sections are trivial

[gentle osprey]

Pseudonium

[quartz horizon]

I think

[hidden abyss]

oh that's cool

[quartz horizon]

Basically cause a global section gives you a global choice of “identity element” on each fiber

[hidden abyss]

I'm reading the definition for the first time but that makes sense

[quartz horizon]

It’s basically a bundle whose fiber is iso to G

But not canonically so

A choice of element in the fiber is precisely a choice of iso to G

Think affine spaces for example - a choice of “origin” is a choice of iso to R^n

[hidden abyss]

the proof should then be a straightforward generalization of the last paragraph

[quartz horizon]

Yeah

You can send $X \times G$ to $(x, g .s(x))$ roughly

Using $s(x)$ as your choice of “identity element”

[gentle osprey]

Pseudonium

Pseudonium

[quartz horizon]

Actually wait this would just be $(x, g) \mapsto g . s(x)$

[gentle osprey]

Pseudonium

[quartz horizon]

This is a bundle iso to the trivial bundle

[hidden abyss]

do you know of any natural covering maps which aren't principal G-bundles? I know you create counterexamples by taking a disjoint union of two distinct covers of X and then its obviously not transitive

[quartz horizon]

I don’t actually know algtop but

A covering map is a bundle of discrete spaces

That’s locally trivial

However a principal G-bundle is a fiber bundle

Where, in particular, all the fibers are iso

So any covering map where the fibers aren’t of stable cardinality should work

If they’re all of the same cardinality though it just comes down to whether a group structure exists on a set with that cardinality, I’d think

[hidden abyss]

hmm yeah, I was more curious about ones which have

like does there exist a covering map which admits a section but is not trivial

[quartz horizon]

Well its like

[hidden abyss]

the cover should be connected ofc

[quartz horizon]

If each fiber has all the same cardinality, and this is countable, then no

Because you can just make it a principal C_n-bundle in the finite case

Or a principal Z-bundle in the infinite case

[hidden abyss]

oh

[hidden abyss]

how can you guarantee that you can define the group action in a continuous manner

[quartz horizon]

I think defining it locally should work

[hidden abyss]

oh right the section allows that

[quartz horizon]

No you don’t need the section for this

[quartz horizon]

Cause a covering map is locally trivial

Ah I guess you would want some nice compatibility condition though

So that the actions are coherent

[hidden abyss]

yeah

[quartz horizon]

In this case maybe the section actually helps lol

[hidden abyss]

section should allow it though because you can define the action on one fiber and then just extend by multiplying each section with it

[quartz horizon]

Mhm

In general you need choice to get a group structure on a set with arbitrary cardinality

But then otherwise yeah it should work

(Idk AT very well so I may just be yapping)

[hidden abyss]

this one from hatcher (as a cover of S^1 v S^1) should be non-regular

I don't think I really understand whats happening here

[quartz horizon]

yeah i think i no longer believe this works generally

what does seem to be true is that you can split your covering into two parts

a 1-fold cover (so a homeo) and a (n-1)-fold cover

for a double cover this just gets you two homeos

i think i wrongly assumed you could have, like, a consistent way of indexing the elements of the fiber, to allow for applying the principal G-bundle argument

but that probably doesn't hold generally, even if you have a continuous section

2 is special case 2 = 1 + 1 lol

apologies for misleading you earlier

[hidden abyss]

this is without assuming a section?

[quartz horizon]

ah no this is with a section

the section gives you the 1-fold cover

[hidden abyss]

oh right

[quartz horizon]

and then taking the "complement" gives you the (n-1)-fold cover

[hidden abyss]

I don't see why this would fail

is the action not continuous?

[quartz horizon]

wait why is that text red

[hidden abyss]

huh? it looks normal to me

[hidden abyss]

I think that failed to send

[quartz horizon]

hm say, imagine taking the mobius strip bundle over the circle (so a nontrivial double cover)

right

[hidden abyss]

I don't see the red message on my screen at all

[quartz horizon]

and then disjoint union with a trivial cover S^1 -> S^1

[hidden abyss]

right of course

[quartz horizon]

so you get a 3-fold cover (S^1 u M) -> S^1 where M is the mobius one

then this has an obvious global section

but it doesn't work as a principal C_3-bundle, so the bundle isn't fully trivial

however it does split

into the S^1 -> S^1 one, and the M -> S^1 one

[hidden abyss]

at least there's no connected n-covers with a section for n>1

[queen prism]

red means it didn't send properly

you can right-click the message to resend it

[quartz horizon]

yeah that wouldn't work

[quartz horizon]

i feel like you would need this

ah nice

welp i think i can count that as my algtop for today~

back to topos theory :3

[hidden abyss]

yeah I think I'll get back to orientations tomorrow too

[indigo nymph]

is it true that if we take a compact hausdorff space Y
and remove a dense subset X homeomorphic to Rⁿ
Y\X becomes connected for n ≥ 2 and consists of at most 2 components for n=1

it's actually true, but i wonder what's the key idea to prove this

[rancid umbra]

Y is just any compact hausdorff space?

[indigo nymph]

yes

[untold lily]

damn, that theorem sounds crazy

[rancid umbra]

it makes sense tho. Y is a compactification of R^n

[untold lily]

oh, that's right

so I guess to prove this theorem you have to study the compactifications of R^n

[brave lake]

hey, if you have X=R and τ={empty set,R)union{(α,infinity) | a€R} then if S=(0,1), why is S not an open set?

[rancid umbra]

Y is almost a manifold. its separable, but might not be second countable and fails to be locally euclidean on a meagre set

[untold lily]

because it's not an element of tau

[brave lake]

yes, but if α=-1 then isnt S a subset of (-1,infinity) ?

[tribal palm]

so i’ve heard this thing “there are too many continuous spaces for any kind of general classification,” but i’ve never seen much elaboration… is there some kind of theorem that clarifies the situation in the negative, or is it more of a vague sense, for example informed by understanding some of the complications even just for manifolds?

[untold lily]

being a subset of an open set doesn't make you open

the entire space is always open, so by your reasoning every set would be open in any topological space

[brave lake]

but being a subset of an open set which itself is a subset of tau makes you an open set

[untold lily]

open sets can't be subsets of tau

open sets are elements of tau, by definition

[brave lake]

honestly i just learned topology in 20 mins so its hard for me to follow much lol

[gaunt linden]

Hmm, handwavy intuitive suggestion:
Each point in Y\X is connected to the origin in X by a path that's completely in X except for the endpoint. Perhaps if we have two such paths we can interpolate between them, and so show that Y\X is path-connected?

[brave lake]

thats interesting

[rancid umbra]

the collection of all (a,oo) is closed under intersections and unions, so intersecting and unioning a bunch of intervals of the form (a,oo) just gets you another interval of the form (a,oo) where a can be infinite or finite

[untold lily]

tau is a collection of sets satisfying the axioms, open sets are by definition elements of this set

[brave lake]

tbh i didnt understand much of this sorry

[rancid umbra]

(0,1) can't be in tau because it is not of the form (a,oo)

those are the only things that are in tau

[brave lake]

so, how do we prove that S is a subset of tau but not an element of it?

[untold lily]

do you understand that a topological space is a set along with a collection of its subsets (tau) satisfying certain axioms

S is not a subset of tau

[brave lake]

ohhhh wait that makes sense actually. but is the wording right? it confused me when i saw it

[untold lily]

tau is a set of sets

[rancid umbra]

what is confusing?

[untold lily]

a subset of tau must also be a collection of subsets of X

[brave lake]

im having a moment here one sec

[brave lake]

i kinda do, the formality of the definition is what makes my head hurt tho

[brave lake]

yes i learned that much thank god

[untold lily]

do you know metric spaces? they are great motivators of topological spaces if the axioms seem abstract

[brave lake]

firstly, the fact that we gotta do a whole inception movie on maths to explain four types of simple sets lol

[rancid umbra]

maybe after sitting with the definitions for a bit, you will see that they aren't too complicated. its just closure under unions and finite intersections

[brave lake]

not really lol, havent learnt abt them yet

[untold lily]

yeah the axioms themselves are pretty simple, why the axioms are as they are needs motivation (generally provided by metric spaces)

[brave lake]

i know i know. itll get smoother by the times passing

anyways, thank you for explaining, yall cleared some things up for me

[gaunt linden]

In a pinch one can probably use R^n and subspaces thereof as motivation rather than general metric spaces.

[untold lily]

yeah, I was gonna say that, but it's probably still better to go through the slight bit of extra trouble to look at metric spaces (which themselves are best motivated by R^n)

esp because metric spaces will pop up in your study of topologies anyway

[brave lake]

hey again, can you recommend me a youtube channel and/or a book about point set topology?

[rancid umbra]

munkres topology is a pretty standard book

[prime elbow]

How | |_1 : V -> R is uniformly continuous?

[ruby delta]

you're missing some text on the left

[ruby delta]

when the (euclidean norm of a) vector goes to 0, its magnitude always goes to 0 as well in any finite dimensional vector space

because you can take out the scaling factor

[prime elbow]

I don't get it

[radiant stone]

using the triangle inequality | |x|_1 - |y|_1 |<|x-y|_1

[prime elbow]

But we are taking domain with | |_2 norm

[radiant stone]

oh

yeah wait

[prime elbow]

So to prove every finite dimensional normed space is complete.

Since every two norms on finite dimensional normed space are equivalent, so it is also strongly equivalent that means they preserve completeness.

Now since V is finite dimensional normed space over R so it is isomorphic to R^n, by using the norm structure of R^n we can induce new norm structure on V, so V is complete under that norm say | |, but all other norm also equivalent to norm | |.

Hence all finite dimensional normed space is complete.

[radiant stone]

My metric topology intelligence is like having a skill issue, cannot even bring up concepts I learned

Maybe I need some more brainrots to review metric topology haha never used it after learned it tbh

[prime elbow]

Right?

[prime elbow]

I am trying to show if A \in M(n, R) and | I - A | < 1 then A is invertible.

Any hint?

First I don't know exactly which norm the author is using here

[ruby delta]

are you allowed to use the result that the set of invertible matrices are open?

[prime elbow]

Yes

[quick delta]

It doesn’t actually end up mattering
What should (1 - t)^-1 be as a power series?

(The only thing that matters is that it’s multiplicative)

[prime elbow]

\sum_{i=0}^{\infty} t^i

[quick delta]

So A = I - (I - A)
So we want to say A^-1 = sum (I - A)^n

What’s the problem?

[prime elbow]

I see

I understand this is the same as when we are working in R

But for the rigour argument what should I write here ?

[quick delta]

You need to prove that that sum converges
And then that it converges to the right thing

[prime elbow]

I mean first I have to show sum (I-A)^n is convergent, which i think can be done by using | I - A | < 1

[quick delta]

Hint: the partial sums are ||A^-1(I - (I - A)^n)||

[prime elbow]

Is that work for all norms? I don't think so

Because we can make norm on M(n,R) such that | I | < 1

[quick delta]

Basically any norm you reasonably put on the space of matrices will be multiplicative

(Or submultiplivative)
More for the reasons of “why are you doing anything else”

[prime elbow]

Can't we make such a norm on M(n, R) such that | I | < 1 ?

[quick delta]

We can
But again, why would you do that

[prime elbow]

But then this statement can't hold in that norm

[quick delta]

^ then find this out

[quick delta]

And how’s the question you’re doing written?

[prime elbow]

How partial sum is A^-1( I - (I-A)^n)? Because A^-1 is not defined yet

[quick delta]

Ok sorry
More formally, when multiplied by A, the partial sum is (I - (I - A)^n)

Because A = I - (I - A)

[prime elbow]

You mean partial sum multiply by A is I -A - (I-A)^(n+1)

No

I got it

[indigo nymph]

it doesn't matter, just take any submultiplicative norm
recall the geometric series formula
(I - B) sum B^k over k ≥ 0 = I
apply it for B = I - A
it's true since the séries of B^k converges since ||B^k|| ≤ ||B||^k

[prime elbow]

You mean norm doesn't matter?

[indigo nymph]

no, i mean you can take any submultiplicative norm

[prime elbow]

What is submiltiplicative norm?

[quick delta]

|AB| <= |A||B|

[indigo nymph]

that satisfies
||AB|| ≤ ||A||×||B||

[prime elbow]

Oh

[indigo nymph]

because without that you can't assure the convergence of sum B^k

[prime elbow]

Oh

[indigo nymph]

so that lemma is false unless we work with submultiplicative norms

[prime elbow]

So $| B^k | \leq |B|^k$ so isn't imply that $\sum | B^k |$ convergent.

Let denote $P_n$ as partial sum of series $B^k$, so we have to show $P_n$ is cauchy, so $| P_n - P_m | = | \sum_{k= n+1}^{k = m } B^k | \leq \sum_{k = n+1 }^{k = m } |B^k|$.

[prime elbow]

Is it correct?

[indigo nymph]

sorry the counterexample isn't the good one since ||I - A|| is 1

[indigo nymph]

it implies the convergence of B^k
because ||B|| < 1 and sum ||B||^k = 1/(1-||B||)

[gentle osprey]

Notknow🙇

[indigo nymph]

no you have
sum |B^k| ≤ sum |B|^k = 1/(1-|B|) so the series of B^k converges

no need to use cauchy
just triangle ineq

[prime elbow]

But I think it's not wrong to use the cauchy part?

[indigo nymph]

it's not

[prime elbow]

But its sum runs infinite time so how can we use triangle inequality

[indigo nymph]

|sum u_k over IN| ≤ sum |u_k| over IN is true when series u_n is absolutely convergent
also i said something dumb, you dont need triangle ineq
just |B^k| < |B|^k and take the sum on both sides

you can obviously do that and assure the convergence of series B^k since series |B|^k is convergent

[uneven bronze]

$x$ is an accumulation point of $A$ if $A\cap(U\setminus{x})\neq\varnothing$ for every neighborhood $U$ of $x$. Consider \

Proposition. If $X$ is a topological space, $E\subset X$, and $x\in X$, then $x$ is an accumulation point of $E$ iff there is a net in $E\setminus{x}$ that converges to $x$, and $x\in \overline{E}$ iff there is a net in $E$ that converges to $x$. \

I think I understand the first implication of the first equivalence. Consider $x_\alpha\in E\setminus{x}$ and $x_\alpha\to x$, then why does every punctured neighborhood of $x$ contain some $x_\alpha$ and hence $x$ is an accumulation point of $E$? I struggle with the second equivalence too.

[gentle osprey]

psie

[prime elbow]

Oh net, which book are you using?

[uneven bronze]

Folland 😅

[prime elbow]

Which one?

I learned about nets and filters from this

So maybe you found this one helpful

[uneven bronze]

thanks, it's his real analysis book

[prime elbow]

Okay

[prime elbow]

Yes because of the cauchy sequence

[prime elbow]

And how can I show this (I - B) sum B^k over k≥ 0 = I?

[indigo nymph]

$(I_n - B) \sum_{k=0}^{+\infty} B^k = \sum_{k=0}^{+\infty} (I_n - B)B^k = \sum_{k=0}^{+\infty} (B^k - B^{k+1}) = B^0 - 0 = I_n$

[gentle osprey]

Slomenist

[indigo nymph]

telescopic sum using the fact that when series u_n converges, u_n tends to 0

here sum B^k converges so B^k ---> 0 when k ---> +oo

[prime elbow]

I see

Thank you Slomenist, you helped me a lot

[rancid umbra]

where is this from?

[indigo nymph]

Miklós Schweitzer 1994

[uneven bronze]

I still struggle with this direction:\

$x\in \overline{A}\implies$ there's a net in $A$ converging to $x$.\

The author refers to an earlier proposition, namely that $\overline{A}=A\cup\operatorname{acc}(A)$, where $\operatorname{acc}(A)$ are the accumulation points of $A$, defined as the points $x$ such that $A\cap(U\setminus{x})\neq\varnothing$ for every neighborhood $U$ of $x$. How does the implication follow from this proposition?

[gentle osprey]

psie

psie

[tender halo]

well i mean

it follows straightforwadly?

you have found nets that converges to the accumulation points

points that are members of A are limits of trivial sequences for example

so youve won right

[uneven bronze]

well, to summarize, if x is in cl(A), then it is either in A or in acc(A). If x is in acc(A), we use what the author has previously shown, right? And if it is in A, then we choose the trivial net, or?

[tender halo]

yeah

[uneven bronze]

awesome

[uneven bronze]

What is the negation of a net converging? The definition of convergent net to x is that for every neighborhood U of x, x_a in U eventually (meaning there is some a0 such that x_a in U for a>=a0).

If the directed set is the natural numbers, i.e. a sequence, I'd say the negation is that x_a is not in U for infinitely many natural numbers a. But does infinitely many something something make sense for an arbitrary directed set?

[uneven bronze]

The reason I'm asking in the first place is because I'm reading about f being continuous at x iff for every net < x_a > converging to x, the corresponding net < f(x_a) > converges to f(x). The implied by direction proceeds by contraposition, i.e. assume f is not continuous at x. Then there is a neighborhood V of f(x) such that f^{-1}(V) is not a neighborhood of x. Ultimately we find a net such that x_a to x, but f(x_a) not to f(x) because f(x_a) not in V. I don't get the conclusion. Is it not possible for a net that is not in the same set as its limit to converge to that limit?

[uneven bronze]

Ok, I figured out the negation. It all makes sense now.

[prime elbow]

consider the metric on R d(x,y)= | arctan(x)-arctan(y) | so for sequence xn=n for all n, xn is a cauchy sequence right ? but is R complete with respect to the metric d ? but still as a ordered field R is always complete and it always satisfied lub property

Is this statement true?

[quick delta]

R is not complete wrt this metric, and it’s the standard example of “two metrics can induce the same topology, but differ on completeness”

[prime elbow]

Yes

But someone argue with me that completeness has two different point of view one is on order and second one metric

Though the concept of completeness differs slightly between Dedekind completeness and Cauchy completeness but it is not necessarily valid to assume that a shared goal implies identical methods of attainment ..... moreover the concept of orderedness in set theory is fundamentally different from the concept of a metric. The notion of the limit of a sequence and whether that sequence is Cauchy depends entirely on the definition of the metric in the given space. Therefore, completeness in the context of order and completeness in the context of a metric must be treated as distinct concepts.

[tribal palm]

i wish i were wrong in claiming this is a triangulation of the sphere (in the disk model, with all points on the boundary identified)

[quick delta]

It’s just a tetrahedron

[tribal palm]

yes, i just hate that the biggest area in the model here is a triangle

[quick delta]

Has 3 sides
Is triangle
I don’t see the problem 🙃

[tribal palm]

oh i guess this is another way to draw the same

fsr this one is much easier to read for me

[west brook]

You don’t even need a vertex on the circle

[tribal palm]

good point

[main ore]

I have here Baires category theorem. The wikipedia mentions that Baires is dependant on the axiom of choice, or equivalently Zorns lemma. Ive tried looking but i cant find out what part of this proof would demand the axiom of choice.

A stack post mentioned it is in the extension of the balls to infinity, but i dont see that exact form here.

[quick delta]

It’s where you choose the sequence x_1, x_2, …

That needs at least a weak form of choice

[main ore]

youre talking about here?

[untold lily]

yes

[main ore]

could you expand on how you saw that this was where AC was used?

[quick delta]

As soon as you’re making infinitely many choices, you always need some form of AC (unless there’s a canonical way to choose them)

[untold lily]

constructing a sequence where the n+1'st term is defined chosen based on the n'th term needs at least a weaker form of choice

yeah, excluding canonical choices

[main ore]

makes sense. Thanks so much people.

[prime elbow]

so this is true in any topological space, right? Because if O cap A is empty then O \subset A', so for x \in O, x will be limit point of A but O cap A is empty so it implies x is not limit point, contradiction.

A' is the set of all limit points of A

[prime elbow]

Any hint to show every normed space is connected?

[steep kite]

Prove that it is path connected

[prime elbow]

Okay

But author has not introduced path connected yet

I think path connected is easy to show, t -> tx + (1-t)y, right?

And by sequential criteria we can show this is continuous

[rancid umbra]

addition in a normed space is continuous

same with multiplication

[prime elbow]

Yes

Scalar multiplication

Yes

[rancid umbra]

so it’s just a composition of continuous maps

[prime elbow]

Yes

So we are done

Thank you Trivial

But now I am thinking how the author did this without using the path connected?

[steep kite]

You’re welcome, I’m still thinking if you can use your exercise to prove it directly, if I came up with something I let you know

[prime elbow]

Okay

[prime elbow]

I can share proof if you want

[steep kite]

Sure

I can't think of anything now haha

[prime elbow]

So here I don't understand how \overline{t} in (0,1) ?

[hidden abyss]

$\tilde{t}= 1$ would mean that for all $s<1$, $a+s(b-a) \in A$. Since $A$ is closed, letting $s\to 1$ yields $b\in A$, which contradicts the assumption $b \in E\setminus A$

[gentle osprey]

Jussari

[hidden abyss]

Proving that tilde{t} is not 0 is similar

[prime elbow]

I see

Thanks Jussari

[prime elbow]

But I have to there some t > 0 such that a +s(b-a) for all s < t.

[hidden abyss]

If no such t exists, then there is an infinite sequence of s_i such that s_i ->0 and a+s_i(b-a) ∉ A

Taking the limit s_i -> 0 again leads to a contradiction with the fact that E\A is closed

[prime elbow]

Oh

Got it

[uneven bronze]

A subnet of a net $\langle x_\alpha\rangle_{\alpha\in A}$ is a net $\langle y_\beta\rangle_{\beta\in B}$ together with a map $\beta\mapsto\alpha_\beta$ from $B$ to $A$ such that\

• for every $a_0\in A$ there exists $\beta_0\in B$ such that $\alpha_\beta\gtrsim\alpha_0$ whenever $\beta\gtrsim\beta_0$;\
• $y_\beta=x_{\alpha_\beta}$.\

Now, the text says that clearly a subnet converges if the net converges. Why? My attempt:\

If $(x_{\alpha_\beta}){\beta \in B}$ doesn't converge to $x$, then there is some neighborhood $U$ with $x \in U$ such that for all $\beta_0\in B$, there is $\beta$ with $\beta\gtrsim \beta_0$ such that $x{\alpha_\beta} \notin U$. How do I now show the corresponding non-convergence of $\langle x_\alpha\rangle_{\alpha\in A}$?

[gentle osprey]

psie

[queen prism]

do you know how to show the corresponding result for sequences and subsequences?

[uneven bronze]

I believe yes

[uneven bronze]

I'm thinking, by the first point of the definition, for every $\alpha_0\in A$,we can find some $\beta_0\in B$ (by the above) such that if $\beta\gtrsim\beta_0$, then $\alpha_0\lesssim\alpha_\beta\notin U$. So this says that $\langle x_\alpha\rangle_{\alpha\in A}$ does not converge, right?

[gentle osprey]

psie

[queen prism]

have you tried showing it directly?

[uneven bronze]

Is my approach not correct?

[uneven bronze]

Does anyone know if the definition of subnet I gave above is a so-called Kelley subnet, Willard subnet or something else entirely?

[uneven bronze]

It's a Kelley subnet, I'm pretty sure.

[tender halo]

its still true because then there is no O

[unreal stratus]

oh lol i didn't read properly

[plush folio]

This argument doesn't seem complete, where do you use the fact that O is open?

[unreal stratus]

This is true in any topological space. Indeed one can rephrase it as follows: let X be a topological space, A a dense subset. Then any nonempty open O has nonempty intersection with A. This is a standard fact about dense subsets

The proof I would give is ||if O \cap A is empty, then X \ O is a proper closed subset containing A, oof||

[plush folio]

Nice, proof by oof

[prime elbow]

O is open set so if x in O and O\subset A' that means x is limit point of A, then it must have intersection with A

here i used O is an open set

[indigo nymph]

any IR-vector space even, is convex (cause it's closed under linear combinations hence closed under barycentric combinations), hence star-convex, path connected and connected

[prime elbow]

IR?

[indigo nymph]

vector space over the field of real numbers

[prime elbow]

oh

[prime elbow]

then cl A \subset X\O, which contradict that O\subset cl A

[prime elbow]

the statement the diameter of open ball is twice of its radius, is wrong because in discrete metric space if i take open ball B(x,2) then diameter is 1 not 4, yes diameter of open ball B(x,r) is always <= 2r

[quick delta]

Yeah I assume that’s doing something like “working in R^n” or similar
Like B(x, 1) in the discrete metric has radius 1 but diameter 0

[prime elbow]

Yes

[hoary breach]

is E some normed vector space?

the discrete metric is not translation invariant so it does not come from a norm

[prime elbow]

Oh yeah good point

So it is valid for every normed space that diameter of an open ball is twice of its radius

[quick delta]

Yeah

WLOG it’s centred at 0

Then take some vector v =/= 0
Then (r - e)v/||v|| and (e - r)v/||v|| are in B(0, r) and are a distance of 2r - 2e apart
So the supremum of these is 2r
And the fact that’s an upper bound comes from triangle inequality

[prime elbow]

yes

[uneven bronze]

Proposition. If $(x_\alpha)$ is a net in a topological space $X$, then $x\in X$ is a cluster point (a point that the net "visits infinitely often") iff $(x_\alpha)$ has a subnet that converges to $x$.\

Consider the $\implies$ direction. Let $\mathcal{N}$ be the set of neighborhoods of $x$ and make $\mathcal{N}\times A$ into a directed set the usual way. For each $(U,\gamma)\in\mathcal{N}\times A$, there exists $\alpha_{(U,\gamma)}\in A$ such that $\alpha_{(U,\gamma)}\gtrsim\gamma$ and $x_{\alpha_{(U,\gamma)}}\in U$ (since $x$ is a cluster point). One can then show that $(x_{\alpha_{(U,\gamma)}})$ is a subnet that converges to $x$. \

I wonder, a subnet $(y_\beta){\beta\in B}$ of a net $(x\alpha){\alpha\in A}$ is a net together with a map $\beta\mapsto\alpha\beta$ such that (1) for every $a_0\in A$ there exists $\beta_0\in B$ such that $\alpha_\beta\gtrsim\alpha_0$ whenever $\beta\gtrsim\beta_0$ and (2) $y_\beta=x_{\alpha_\beta}$. What is the map $\beta\mapsto \alpha_\beta$ in the proof of the above proposition?

[gentle osprey]

psie

[indigo nymph]

is A bar the closure here ?

[prime elbow]

Yes

[indigo nymph]

then that's false, take O = A

oh i didn't see open, hmm

[uneven bronze]

I think the map is just $(U,\gamma)\mapsto\gamma$ if I'm not mistaken.

[gentle osprey]

psie

[untold lily]

you are mistaken

honestly, I don't know how this map is defined, although the proof does sort of define it

It says for each (U, gamma), alpha(U, gamma) is an element of A greater than or equal to gamma such that x_alpha(U, gamma) is in U

But you are being asked to make a choice, and idk why these choices will be consistent with one another

[uneven bronze]

yeah, (U,\gamma) should map to some element alpha_{(U,\gamma)} ... it's convoluted.

[untold lily]

I would instead consider as my directed set pairs (U, gamma) such that x_gamma is in U, and the net would be a projection as you expected

Exactly what Willard does

[uneven bronze]

would you say the above proof (of the proposition that is) is incorrect if one uses the "larger" directed set N x A?

[untold lily]

wait a sec, this is a bit tricky because there are three separare notions of subnets, and I need to be careful in my statements

I'm mostly used to the willard subnet, where the function alpha is required to be increasing

[uneven bronze]

yes indeed 😅 I'm used to the Kelley subnet definition

[untold lily]

right, kelley subnets don't need alpha to be increasing, which is what confused me

I'll have to digest the kelley definition

[uneven bronze]

sure 👍

[untold lily]

okay, taking a in A, we need a (U, b) in B such that (V, c) > (U, b) implies alpha(V,c) > a

and b can just be equal to a and this works

so I'm convinced we have a subnet

and does indeed converge to x

[uneven bronze]

yes

so the proof is correct

[untold lily]

so yeah, if you are still confused about what alpha does, it sends (U, a) to an element b of A such that b > a and x_b \in U

it's just making choices that we know we can make because it's a cluster pt

I was confused because this isn't a willard subnet

not necessarily increasing

[untold lily]

and there doesn't seem to be an easy fix to make it so, you gotta do this instead

[uneven bronze]

yeah, it can be confusing. So beta \mapsto alpha(beta) is just the map (U,gamma) \mapsto phi, where phi is in A such that phi >= gamma and x_phi in U. Ok, that works.

Thank you.

[untold lily]

[prime elbow]

Can normed vector space be compact? I know finite dimensional can't but what about infinite dimensional?

[quick delta]

There is a surjective continuous map from V to R_{\geq 0} via v -> |v|
As the image of a compact set is compact, V therefore can’t be compact

[prime elbow]

i see

[quick delta]

In fact the unit ball of an infinite dimensional NVS isn’t even compact (although this is slightly non-trivial)

[prime elbow]

yeah

if X is closed set and boundary of X is empty then X is clopen right?

[indigo nymph]

if it was compact then the closed unit ball would be compact, which contradicts Riesz theorem

[prime elbow]

oh

[quick delta]

No
Consider the unit sphere in R^3

[indigo nymph]

B is compact <=> dim < +oo

[quick delta]

You can do it in a much more low tech way 🙃

See above

[prime elbow]

are you sure?

[quick delta]

Bleh brain is too used to manifolds with boundary

Boundary of X is empty => X = interior of X, so yes

[prime elbow]

because if boundary is empty then it implies X is open

so to show in normed vector space every non empty compact set has boundary, if it has no boundary then by connectedness of normed vector space, V will be compact which is a contradiction

[quick delta]

Yes

[prime elbow]

mico, from which book did you learn these all stuffs?

[indigo nymph]

yeah nice one
since compactness is preserved under continuous maps

[indigo nymph]

.

[indigo nymph]

some books don't contain some details like this
i think you just gotta diversify at some point and meditate on theorems and definitions by yourself

[quick delta]

I didn’t tbh
I don’t know where I learned topology (it, and linear algebra are the two things I know that I don’t really have a place to point to. They just appeared in my brain at some point without an explicit learning process)

[prime elbow]

appeared in your brain

[quick delta]

I have memories of trying to use LA/topology books to learn
But I also have memories of them being fully ineffective for me
But by the time I got to the relevant courses in 2nd year, I knew them inside and out

[prime elbow]

and you are doing undergraduate, right?

[quick delta]

Yesish
I just finished 3rd year, and will be doing an integrated masters next year

[prime elbow]

can i DM you?

[quick delta]

Why?

[prime elbow]

i want to ask about integrated master program

so you will do master + PhD?

[quick delta]

Integrated master’s in UK is bachelors + masters

[prime elbow]

so that means you will do master next year

[indigo nymph]

did you do algebraic/differential topology in 3rd year ?

[quick delta]

I did algtop (homotopy + Simplicial homology) last year
I could’ve done diffgeo, but the course looked bleh and the lecturer was errrrr
One of the lecturers of all time

[quick delta]

Yes

[indigo nymph]

they let you choose which courses to take ? that's nice

[quick delta]

We get choices in 2nd and 3rd
Both years are formally “choose any number from [15, 37] courses” although practically LA and methods in 2nd year are forced, and most people do 6-10

[indigo nymph]

what would you like to talk about in your master's thesis ?

[quick delta]

It depends on what’s offered (there’s usually a choice of around 100ish) but I’ll probably do something group theoretic

[indigo nymph]

that's awesome tbh

[untold lily]

very cool idea, but one can very easily work with the direct definition also

[indigo nymph]

interesting, advanced group theory must be fun to learn

[untold lily]

B(0,n) as n varies is clearly an open cover with no finite subcover

[quick delta]

I am very it like

[quick delta]

Ah yeah
That’s essentially what you get if you unwrap the proofs of “R_\geq 0 is not compact” and “preimages of compact are compact” in my case too

[untold lily]

yes, both are about the idea that elements can have arbitrarily large norms essentially

[prime elbow]

I know this is true for any metric space but I used Tietz Extension theorem, but now the author has not introduced Tietz Extension yet

[unreal stratus]

I think you can just do casework: it's either not closed, or not bounded, and in either case you can write down a function

[prime elbow]

So for the first case when it is not closed :

So A is not closed implies there is a sequence a_n in A converging to x not in A.

So we can map a -> 1/d(x,a), which is continuous and it is unbounded because for any n in N we can choose a_n such that a_n maps to n.

Second case if A is not bounded:

Since A is not bounded, we can map v -> | v |. If this map is bounded then it implies A is bounded.

Is it correct?

[unreal stratus]

perfect, yes :)

[prime elbow]

Thank you ❤️

The hint was very helpful

[tribal palm]

this notation is so ugly 😭

$$X_0 \cup_{\sqcup \phi_\alpha} \left(\bigsqcup_\alpha D^1\right)$$

[gentle osprey]

red beet jen

[unreal stratus]

what book is this?

[rancid umbra]

eh, i guess its pretty standard

from the nlab page on cw complexes

that one is ugly because they notate the attaching maps as the actual coproduct map

the three cups make it look weird

[tribal palm]

Viro

[unreal stratus]

never heard of it lol fair

[tribal palm]

i adore it 😌

http://pdmi.ras.ru/~olegviro/topoman/index.html

it’s freely available

[cunning sleet]

what book is this?

[tribal palm]

guh

see the messages just above your question

[tender halo]

its a more or less the standard text in russia

for introductory topology

[tribal palm]

it’s a weird mix of extremely friendly but also leaving all proofs to the reader

i also love that it has plenty of more open questions which encourage reflection on things that dont necessary have an easy or clear-cut answer

[tribal palm]

this makes a lot of sense i think for the subject matter, as most of the proofs are really simple, and they break the slightly more complicated ones like urysohn’s lemma into smaller steps

also it has a number of small digressions briefly introducing various topics, like p-adics

[uneven bronze]

Definition. A (Kelley) subnet $\langle y_\beta\rangle_{\beta\in B}$ of a net $\langle x_\alpha\rangle_{\alpha\in A}$ is a net together with a map $\beta\mapsto\alpha_\beta$ such that (1) for every $a_0\in A$ there exists $\beta_0\in B$ such that $\alpha_\beta\gtrsim\alpha_0$ whenever $\beta\gtrsim\beta_0$ and (2) $y_\beta=x_{\alpha_\beta}$. \

Exercise. Let $\langle x_n\rangle_{n\in\mathbb{N}}$ be a sequence. If $k\to n_k$ is a map from $\mathbb{N}$ to itself, then $\langle x_{n_k}\rangle_{k\in\mathbb{N}}$ is a subnet of $\langle x_n\rangle $ iff $n_k\to\infty$ as $k\to\infty$, and it is a subsequence iff $n_k$ is strictly increasing in $k$.\

Attempt. The second equivalence is obvious by definition; a subsequence of a sequence $f:\mathbb{N}\to X$ is the composition $f\circ g$, where $g:\mathbb{N}\to\mathbb{N}$ is strictly increasing. I'm curious about the first equivalence; how does the condition $n_k\to\infty$ as $k\to\infty$ ensure $\langle x_{n_k}\rangle$ is a subnet and vice versa?

[gentle osprey]

psie

[uneven bronze]

Actually, it's kind of obvious too. So disregard.

[uneven bronze]

Here's a more advanced question. Consider the exercise above.\

(continued) There is a natural one-to-one correspondence between the subsequences of $\langle x_n\rangle$ and the subnets $\langle x_n\rangle$ defined by cofinal sets.\

What is this one-to-one correspondence?

[gentle osprey]

psie

[uneven bronze]

Anyone who's got any idea? 😔

[untold lily]

what is a subnet defined by a cofinal set?

I see, just looked it up

I mean, isn't it clear what you do here? there seems to be a clear way to define a subsequence using a cofinal subset and vice versa

[uneven bronze]

indeed, but they seem to be asking for more; an invertible map that sends a subsequence to a unique subnet. By the way, since you asked, a cofinal set A in the naturals N gives rise to the inclusion map which makes (x_a)_{a in A} a subnet of (x_n)_{n in N}.

[untold lily]

what exactly are you struggling with?

a cofinal set uniquely defines both a subnet and a subsequence

all you're doing is shuffling this one set around

[uneven bronze]

ok, so it is this subnet and subsequences that are in one-to-one correspondence?

[untold lily]

subnets defined by cofinal sets on one side

subsequences on the other

these two sides are in one to one correspondence

[tender halo]

(note that a subnet of a sequence need not be a sequence itself)

[uneven bronze]

but you said that a single cofinal set defines both a subnet and a subsequence, I was wondering if they are in one-to-one correspondence?

[untold lily]

what do you mean by "they"

[uneven bronze]

I mean the subnet and subsequence defined by the cofinal set

[untold lily]

the one subnet and the one subsequence is in one-to-one correspondence? I mean yeah, 1 element on both sides

you got me confused now

I'm saying that a subsequence* is uniquely defined by a cofinal set

and so is a subnet defined by a cofinal set

so, both of these sets are in one to one correspondence with the set of all cofinal subsets of N

[uneven bronze]

hmm, I still don't see the one-to-one correspondence. Here you said a cofinal set uniquely defines both a subnet and a subsequence. So I think I misunderstood you to mean that a single cofinal set both defines a subnet and a subsequence, but I guess you meant that there is one cofinal set that defines a subnet and another cofinal set that defines a subsequence, correct?

[untold lily]

no, I do mean the first thing

the same damn set

[uneven bronze]

haha, ok

[untold lily]

say A is cofinal in N

do you understand how to construct a subsequence with A

[uneven bronze]

the inclusion map?

[untold lily]

you obviously understand how to construct a cofinal subnet with A

[uneven bronze]

yes, actually that requires the inclusion map

[untold lily]

no, a subsequence requires an increasing function from N to N

[uneven bronze]

yes, correct

[untold lily]

but think conceptually

you have an infinite subset of natural numbers

how would this define a subsequence

[uneven bronze]

we just take this infinite subset to be our index set

[untold lily]

ok, let me put it more blatantly, we need a function from N to N

(assuming N starts at 1) where does 1 map to

[uneven bronze]

ah ok, so we enumerate the infinite subset of natural numbers. This will be increasing I think.

[untold lily]

idk if I'd call it enumerate, you simply order the subset from smallest to largest

1 maps to least element of A, then 2 maps to least element after that, etc

[uneven bronze]

yes, that's a better way of putting it. Definitely increasing then.

[untold lily]

yes, you make it so

the point is, a subsequence is uniquely defined by the image of that increasing map from N to N

which is the cofinal set

[uneven bronze]

aha, makes sense

[untold lily]

try to think about that, two of these maps cant have the same image and be different subsequences

[uneven bronze]

indeed, so you are saying if g and h are both strictly increasing from N to N, g(N) = h(N) implies g = h

[untold lily]

yes

exactly

and conversely, given a cofinal set, you can construct such a strictly increasing function

so, you now have injectivity and surjectivity of this "one-to-one correspondence"

[main ore]

Does the fact that we can calculate trigonometric identities depend on the metric of the reals or could we do it without an explicit metric?

[timid helm]

HOWDY

[untold lily]

This feels a bit vague to me, how are you defining the trigonometric functions?

[rancid umbra]

continuity and completeness will be the only things affected by the metric. the algebraic identities should still hold tho…

obviously anything with like, an absolute value may change

[main ore]

cosine sine and such

[untold lily]

how are you defining them

[main ore]

what do you mean by that

[warped shore]

If you are defined trig functions geometrically then ofc you need metric, that's like the definition of geometry

if you are defining trig functions just the normal way then nah

[rancid umbra]

metrics do not affect algebraic identies

[untold lily]

they'll generally be given by power series, for power series to even make sense you need a norm on the function space, which will then depend on a norm of R, but all norms on R are equivalent

[rancid umbra]

the norm doesn’t really matter

[untold lily]

so yes, it depends on the norm, which is a length, but there is a unique norm on R up to equivalence

[rancid umbra]

just work with formal power series

[untold lily]

well yeah, you could do that, in which case everything is algebraic

[warped shore]

geometric properties of trig functions are a side effect 😎

[rancid umbra]

right, thats what they were asking

like, any identities you prove about the usual trigonometric functions with formal power series will be true once you add a metric

[untold lily]

yes, you're right

[main ore]

wouldent that require that you can show the power series converges to the identity?

[warped shore]

Yeah

[main ore]

how do you do that without the usual metric of some kind on R^{n}

[rancid umbra]

sorry, what do you mean by this?

[main ore]

you said you wouldent need a metric if you defined it by a power series

right?

[warped shore]

wait

trig identity right?

[main ore]

right

[warped shore]

like sin^2+cos^2

yeah im curious what the algebraic proof is

[rancid umbra]

pretty sure you can do this with de moivre’s over C?

[rancid umbra]

yes

[main ore]

how so?

[warped shore]

How to introduce i

[rancid umbra]

you work over C

[untold lily]

https://math.stackexchange.com/a/2757088

messy like all computations with formal power series that involve multiplication, but it werks

[warped shore]

ahhh

this bit is clever

so thats how it works

[rancid umbra]

that’s what i had in mind

[warped shore]

im so cooked if i ever have to come up with things like this in undergrad

[rancid umbra]

but formal power series can just be defined as sequences of real or complex numbers

with addition and multiplication defined correctly

[warped shore]

bad crop

there

[rancid umbra]

showing that you have convergence in the metric is what’s required of you, once you place a metric on the space. so if you prove some step in the algebraic proof actually converges in your metric, the identities may be able to tell you exactly what it converges to.

but formally, the algebraic identities will hold

since we are just manipulating sequences of real or complex numbers

[prime elbow]

We can do this by using the fact any normed vector space ( not {0} ) can't be bounded so it cannot be compact even

[rancid umbra]

for any non-zero unit vector v in V, Fv = {av : a in F} is isomorphic to F as normed spaces whenever F = R or C. since its also closed, the V can't be compact, otherwise F would be

sorry worded this kinda poorly

[prime elbow]

And Fv is finite dimensional so it cannot be compact

[rancid umbra]

when F is R or C, yea

[prime elbow]

But I think it is a pretty good argument that if we use the fact that normed vector space ( not {0} ) is always unbounded so it cannot be compact

[rancid umbra]

oh oh, ur using heine borel

[prime elbow]

Not Heine Borel

I mean compact space always bounded in metric space

[rancid umbra]

oh my bad

nice

[alpine nest]

Totally bounded, in fact.

[prime elbow]

yes

any hint to show bounded set is totally bounded in nvs

i know in R it is true

[alpine nest]

That's quite difficult to show ||because it's not true in an arbitrary normed vector space||

[prime elbow]

so hint?

[alpine nest]

How can I give you a hint for showing something that's not true?

A totally bounded and complete set is compact (in fact total boundedness + completeness is equivalent to compactness).

[prime elbow]

yes

[alpine nest]

So for example the closed unit ball in a Banach space (which is bounded) can never be totally bounded unless the dimension is finite

[prime elbow]

sorry i forget to mention it is finite dimensional

[alpine nest]

Then just copy the proof from R; every finite-dimensional normed vector space is isomorphic to R^n

[prime elbow]

yes

[alpine nest]

In fact you could probably just use the isomorphism and invoke the fact from R^n rather than copying the argument

[prime elbow]

yes

[prime elbow]

yes in infinite dimensional space sphere is not compact so closed unit ball cannot be compact

[unreal stratus]

Banach-Alaoglu my beloved

[alpine nest]

I'm not sure it's that; Banach-Alaoglu is about compactness of the closed unit ball in the weak* topology.

Oh, but apparently the proof of the theorem "the closed unit ball of a normed vector space is compact if and only if the space is finitely dimensional" does draw on the dual space somehow?

[quartz horizon]

Yes iirc

[unreal stratus]

Indeed lol I was joking that like

This is a fix ig

Like this is why Banach-Alaoglu is cool

[alpine nest]

Also "Alaoglu" is just a great name

[prime elbow]

I don't get it, dual space reference here

[uneven bronze]

\textbf{Fact 1.} If $\langle x_\alpha\rangle {\alpha\in A}$ is a net in a topological space, then $\langle x\alpha\rangle $ converges to $x$ iff for every cofinal $B\subset A$ there is cofinal $C\subset B$ such that $\langle x_\gamma\rangle_{\gamma\in C}$ converges to $x$. \

\textbf{Fact 2.} There is a one-to-one correspondence between the subsequences of $\langle x_n\rangle$ and the subnets of $\langle x_n\rangle$ defined by cofinal sets. \

\textbf{Corollary.} If $f_n\to f$ in $L^1$, there is a subsequence ${f_{n_j}}$ such that $f_{n_j}\to f$ a.e. \

\textbf{Exercise.} Let $X$ be the set of Lebesgue measurable complex-valued functions on $[0,1]$. There is no topology $\mathcal{T}$ of $X$ such that a sequence $\langle f_n\rangle$ converges to $f$ with respect to $\mathcal{T}$ iff $f_n\to f$ a.e. (Use the Corollary and Fact 1 and 2.)

[gentle osprey]

psie

[uneven bronze]

I'm overwhelmed by the information provided in the exercise above. What exactly is it that I need to show?

[prime elbow]

so equivalent is used under same underlying set and isomorphism doesn't depend on the underlying set.

And if a space V is isomorphic to R^n(vector isomorphic) then by using that isomorphism we can defined the new norm on R^n and then V is isomorphic to R^n(with new norm).

Same for W( dim W = n ), so W is isomorphic to R^n ( not necessarily with above norm ), but all norms are equivalent(strongly equivalent ) thus they are homeomorphic or say isomorphic.
By transtivity V and W are isomoprhic.
is there any mistake in my argument?

[alpine nest]

Facts 1 and 2 together tell you that a sequence (x_n) is convergent to x in some topology if and only if its every subsequence has a subsequence converging to x. You can use that to show that a.e. convergence is not topologizable by showing that a.e. convergence doesn't have this subsequential property.

[uneven bronze]

Ok, I will try. Thank you.

[uneven bronze]

So let $f_n\to f$ in $L^1$, but not $f_n\to f$ a.e. (what would be an example of such a function on $[0,1]$?). Then every subsequence of ${f_n}$ also converges to $f$ in $L^1$, but, by the corollary above, each of these subsequences contains a subsequence converging to $f$ a.e. Does this go against the property you stated? After all, we didn't assume ${f_n}$ converged almost everywhere from the beginning?

[gentle osprey]

psie

[unreal stratus]

Pretty sure an example of such a sequence which converges in L1 but not a.e. is given by considering indicators of the following intervals: [1/2,1], [0,1/2], [3/4,1], [1/2, 3/4], [1/4, 1/2], [0,1/4], [7/8,1],...

For each x in [0,1] the sequence f_n(x) diverges but clearly f_n -> 0 in L1

[uneven bronze]

ah yes, typewriter sequence 🙂 I forgot

[alpine nest]

Yep, that's the one. Every subsequence of this has a further subsequence which converges a.e. to 0, but the sequence itself doesn't converge a.e. to 0 (and if a.e. convergence was topologizable, then it would)

[untold lily]

what book is this?

[untold lily]

Looks good

to say the norms are equivalent is the statement that I : (R^n, new norm) -> (R^n, usual norm) is a topological linear isomorphism (i.e. usual notion of isomorphism of normed spaces)

and so you are composing two of these isomorphisms essentially to say every normed space V of dim n is isomorphic as normed spaces to R^n with usual norm

I'm probably just saying stuff you know

[prime elbow]

Yes

[heady skiff]

Are closed subsets of compact Hausdorff spaces compact

I think yes but I lowkey forgot everything from my point set class

Something like take the complement and use compactness

[quartz horizon]

yes, in general a closed subspace of a compact space is compact

[heady skiff]

Thanks

Are compact subspaces of compact Hausdorff spaces closed 🤔

[quartz horizon]

yes, in general a compact subspace of a hausdorff space is closed

[untold lily]

compact implies closed for hausdorff spaces

[heady skiff]

Okay thanks 😂

I see where we used the compactness and Hausdorffness now

[quartz horizon]

yes compactness and hausdorffness play quite nicely with each other

here's a lemma you might like to try proving

let $X$ be a compact hausdorff space. then:

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

a) any finer topology on $X$ cannot be compact

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

b) any coarser topology on $X$ cannot be hausdorff

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

in this sense they're at a very nice "sweet spot"

it's true generally that coarser topologies than compact ones are compact, and finer topologies than hausdorff ones are hausdorff

another way of saying this is - if $X$ is a set, then there is at most $\textit{one}$ topology on $X$ which makes it a compact hausdorff space

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

(and actually since compact hausdorff spaces of arbitrary cardinality exist, there's exactly one)

[quartz horizon]

actually nvm this is false lol

i assumed coarseness was a total order on the collection of topologies on a given set

it's only a partial one

the previous statement i said is true, though

[tender halo]

this just in S1 is homeomorphic to S2 and S3

[uneven bronze]

It is Folland’s real analysis, chapter 4 🙂

[untold lily]

oh, I never looked at his material on nets

that's some cool stuff

[unreal stratus]

is this just cause of considering ordinals?

[quartz horizon]

This statement is false lol

[quartz horizon]

This one is right

[quartz horizon]

.

[unreal stratus]

Well this is false but like

[quartz horizon]

.

[unreal stratus]

I thought it is true that there are compact hausdorffs of arbitrary cardinality though

Yeah you can just take a set with discrete topology and one point compactify i guess

lol

[quartz horizon]

Yes

[unreal stratus]

but that's what i asked about lol oop

but cool

[uneven bronze]

I have a quite elementary question. The definition of compact of a topological space $X$ is that for every open cover $\bigcup_{\alpha\in A}U_\alpha=X$ there is a finite subset $B$ of $A$ such that $\bigcup_{\alpha\in B}U_\alpha=X$. Now the definition of compact of a subset $Y\subset X$ is defined as compact in the relative topology. I take this to mean that for every $Y=\bigcup_{\alpha\in A}(U_\alpha\cap Y)$ where $U_\alpha$ is open in $X$, there exists finite $B\subset A$ such that $Y=\bigcup_{\alpha\in B}(U_\alpha\cap Y)$. My book expresses compact of a subset a bit differently, just want to check if my formulation is correct.

[gentle osprey]

psie

[rancid umbra]

sorry

your formulation of compact in the relative topology is correct

[uneven bronze]

ok, thanks 👍

[rancid umbra]

does your book express it as a Y being a subset of the union U_a ∩ Y over a?

[uneven bronze]

it expresses it as Y being a subset of the union U_a simply, where U_a is open in X

I guess this is an equivalent formulation

[rancid umbra]

yea, these should be equivalent

that’s what i meant earlier by compactness not being a relative property

[uneven bronze]

ah ok, makes sense

[uneven bronze]

I guess the equivalence simply follows from $A\cap B\subset A\cap C$ if $B\subset C$.

[gentle osprey]

psie

[rancid umbra]

yup

[pallid comet]

We know that there is a countable LOTS (linearly ordered topological space, aka space generated by the order topology) that any countable LOTS embeds into - namely, the rational numbers Q.

Is there an analog for higher cardinalities - that is, for instance, is there a LOTS X of cardinality aleph 1 that every LOTS Y of size aleph 1 has an embedding into X?

[rancid umbra]

i think this paper answers in the negative

@pallid comet

[pallid comet]

thank you!

[prime elbow]

so E is normed algebra so by definition, | xy | <= |x||y|, but i don't know which norm are they using on E^2, if it is max norm then we are done, any idea?

[prime elbow]

so when we are working in normed space we are interested in continuous linear tranformation between two space because then we define norm on the linear transformation, right?

[uneven bronze]

Consider the definition of a directed set A; it is a set with a binary relation > which must satisfy reflexivity, transitivity and directedness (the last property is that for each a,b in A, there is a c such that c > a and c > b).

Just want to double check a simple phrase; we say that a set A is directed by reverse inclusion when U > V iff U \subset V for U,V in A. Can we also direct by simply inclusion? I think I've seen the author having used this.

[tender halo]

sure, any ideal is directed by inclusion for example

[uneven bronze]

Consider $X = (-1,1)\cup\lbrace 0^{\prime} \rbrace$, where $0^{\prime}$ denotes a point that is not an element of $(-1, 1)$. And $X$ is equipped with the topology generated by $(-1,a), (a,1), (-1,b)\setminus \lbrace 0 \rbrace \cup \lbrace 0^{\prime} \rbrace, (c,1) \setminus \lbrace 0\rbrace \cup \lbrace 0^{\prime} \rbrace$, where $-1<a<1, 0<b<1, -1<c<0$. It says one should think of $X$ as $(-1,1)$ with $0$ split in two. I don't really understand how to picture that, but...\

Define $f,g:(-1,1)\to X$ by $f(x)=x$ for all $x$, $g(x)=x$ for $x\neq 0$, and $g(0)=0'$. Then in regards to the above space $X$, $f$ is clearly a homeomorphism onto its range (it is the inclusion map of $(-1,1)$), but what about $g$?

[gentle osprey]

psie

[hidden abyss]

You should probably prove that the map X->X which swaps 0 and 0' is a homeomorphism

Then you can relate g to f

[uneven bronze]

what map is this?

[hidden abyss]

Also I'd be careful with saying that f is "obviously" inclusion. To me its not clear a priori that the subspace (-1,1) of X has the standard topology

[uneven bronze]

yes, on second thought, it is not clear to me either that X has the subspace topology of R.

[hidden abyss]

The map h:X->X with h(0)=0', h(0')=0 and h(x) = x otherwise

[tender halo]

they are both embeddings

bug eyed line is "almost" R

its not just an embedding, its an injective open map, for both f and g

[unreal stratus]

Ye bug eyed line minus either origin is R

[tender halo]

no

[hidden abyss]

Embeddings need not be open

[tender halo]

R -> R^2: x -> (x, 0) is an embedding but not open

[unreal stratus]

Sure but that's equivalent to being an embedding onto an open subset

Ig could say okay open embedding fair

[tender halo]

oh trve

[unreal stratus]

Sorry I deleted too much lol

[warped garnet]

Does anybody have an idea on how to construct an uncountable Hausdorff space in which the complement of the closure of any nonempty open set is countable.

[uneven bronze]

We know a subbase for the topology on X is (-1,a), (a,1), (-1,b) {0} U {0'}, (c,1) \ {0} U {0'}. Now these sets in the subbase intersected with (-1,1) form a subbase for (-1,1). These sets are of the form (-1,a), (a,1), (-1,b) \ {0}, (c,1) \ {0}. Do they generate the subspace topology of (-1,1) as a subset of R? I'm confused about the sets (-1,b) \ {0}, (c,1) \ {0}?

[uneven bronze]

Although I don't understand your concern with the subspace (-1,1) of X having the standard topology. Why is that a concern?

[hidden abyss]

Let A = (0,1) with the discrete topology, and B = ℝ with the standard topology. Then the (set theoretic) inclusion A->B is not a homeomorphism onto its image

You need to check that something like that doesnt gappen with (-1,1) and X

One way is to show that the subspace topology of the image is the same as the topology on (-1,1)

Or you could show that f is an open map

[uneven bronze]

Ok, thank you. There are still many pieces I'm confused about. Perhaps showing f is an open map is easier, though f maps from (-1,1) to X. Is (-1,1) assumed to have the subspace topology from R in the definition of f?

[hidden abyss]

Yeah, (-1,1) has the standard topology

[uneven bronze]

For instance, since f maps (-1,1) to (-1,1) in X, and X has subbase sets (-1,a), (a,1), (-1,b) {0} U {0'}, (c,1) \ {0} U {0'}, the subbase sets of the range of f are (-1,a), (a,1), (-1,b) \ {0}, (c,1) \ {0}. But these aren't subbase sets in the standard topology in (-1,1) if I'm not mistaken.

Declaring e.g. (-1,b) \ {0} and (c,1) \ {0} to be open is not something I would expect in the standard topology of (-1,1), or perhaps I'm wrong?

[hidden abyss]

(-1,b)\{0} is open in the standard topology of (-1,1) though

So it works out

Assuming that youre proving continuity of f right now

[uneven bronze]

There seem to be a lot of side projects in this exercise. I just wanted to show f,g: (-1,1) -> X are homeomorphism onto their ranges. I get that for f, if I want to say that f is a homeomorphism because it is the inclusion map, I have to show the range inherits the subspace topology of X. I feel like I've been showing something slightly different, namely that the range inherits the subspace topology from R.

[hidden abyss]

Maybe it would be beneficial to work it out from the definitions

[uneven bronze]

that would be great 😅

[hidden abyss]

So first to show that f is an open map, we need to show that any open U ⊆ (-1,1) gets mapped to an open set in X

(-1,1) has a basis consisting of intervals of the form (a,b) with -1<a<b<1, so we only need to check that f((a,b)) is open in X

[uneven bronze]

makes sense

[unreal stratus]

Yeah this is kinda funny cause imo the line is two origins is best defined by gluing and then thjs is kinda clear

[hidden abyss]

There's two cases to check, either (a,b) contains 0 or it doesn't

In the first case, (a,b) just gets sent to (a,b) ⊆ X, and thats open in X

Since (a,b) = (-1,b) ∩ (a,1)

In the second case, (a,b) gets sent to (a,b) \ 0 ∪ {0'}, and thats again open in X

[uneven bronze]

ah ok, I understand everything so far

[hidden abyss]

Then you need to check that f is continuous

And you can reduce this to checking it on the subbasis because intersections and unions play nice with preimages

[uneven bronze]

So if it is continuous and open, it is a homeomorphism?

[hidden abyss]

Maybe the idea is to obfuscate the fact that 0 and 0' are indistinguishable from each other?

[hidden abyss]

It's also injective so it's a homeomorphism onto its image

i.e. an embedding

[uneven bronze]

Ok, and for g, I just do a similar analysis basically?

[hidden abyss]

Yeah you can do that

[hidden abyss]

You could also prove that this map is a homeomorphism

Because g = h ∘ f

[hidden abyss]

Since h is its own inverse, you only need to check that h is continuous

[tender halo]

is that possible?

it means uhh that every non-dense open set is countable right

hmm

[uneven bronze]

just a tiny little remark (I have bothered you for some time now and am very grateful for your help 😅 ). But under f (i.e. the inclusion from (-1,1) to X), if (a,b) contains 0, isn't it also being sent (a,b) in X and not (a,b) \ 0 ∪ {0'}?

[hidden abyss]

Yeah you're right

So there's no need to need to do case distinction there

[prime elbow]

i proved cl I is left sided ideal but what if cl I is A, so how can I say maximal ideal is always closed?

[untold lily]

is your algebra unital?

also, I feel like this is better suited to #advanced-analysis

[prime elbow]

Yes

[prime elbow]

Okay i will ask there

[uneven bronze]

Consider $X = (-1,1)\cup{0^{\prime} }$, where $0^{\prime}$ denotes a point that is not an element of $(-1, 1)$. And $X$ is equipped with the topology generated by $(-1,a), (a,1), (-1,b)\setminus {0} \cup {0^{\prime}}, (c,1) \setminus {0} \cup {0^{\prime}}$, where $-1<a<1, 0<b<1, -1<c<0$. \

Now, previously I have shown that $X\setminus{0'}$ and $X\setminus {0}$ are both homeomorphic to $(-1,1)$. Now I'm asked to show that $X$ is $T_1$ but not Hausdorff, although each point of $X$ has a neighborhood that is homeomorphic to $(-1,1)$ (and hence is Hausdorff).\

For $T_1$, I consider cases. Suppose $x\neq 0,0'$. Then ${x}^c=(-1,x)\cup (x,1)\cup{0'}$. Is this an open set?

[gentle osprey]

psie

[tender halo]

yes?

i mean its plainly is an open set

[uneven bronze]

the thing is, the {0'} confuses me. Why is this also open? For only then, the union of the three sets is open, or?

[tender halo]

zero singleton is not open, but it is closed

the union is open

because it has a nbhd of zero

[uneven bronze]

why is a set open in this topology if it is a nbhd of zero?

[tender halo]

because thats how nbhds of 0 and 0' look

you have it in your definition

[untold lily]

wlog, x > 0

then (x,1) \ {0} cup {0'} is open and this gets you 0'

[uneven bronze]

exactly, suppose without loss of generality that x>0. Then we can write {x}^ c as the set (-1, x) \ {0} U {0'} (note that the subbase sets are defined for x in a specific interval), which is open, unioned together with {0} U (x,1). I have no clue why {0} U (x,1) would be open.

[tender halo]

what no

[untold lily]

no no no, (x,1) \ {0} union {0'}

[tender halo]

(x, 1) does not have zero when x is bigger than 0

[untold lily]

what's left to add?

[tender halo]

x is less than zeo

[uneven bronze]

but that is not an element in the subbase, or?

[untold lily]

oh, right I saw it wrong, sec

so, you add (-1, x) \ {0} union {0'}

and you add (x,1)

you can't figure out the rest?

[uneven bronze]

but then {0} is missing? 😄

[untold lily]

theres a very easy way

to add 0

figure it out

[tender halo]

it is not open, when x is bigger than 0

it is open when x is less than 0

[uneven bronze]

I have to meditate on this one

[untold lily]

here's the answer, ||add (-1, x)||

[tender halo]

nbhds of 0 and 0' are really similar

they only differ by one point

you have the regular R and then this extra point that has as nbhds open segments around 0 (excluding 0)

[uneven bronze]

ok, this one scares me a bit

[tender halo]

this is how it looks

[uneven bronze]

nice with an illustration, thanks 👍

[untold lily]

I've seen this example before, but never investigated it

wouldn't expect it to be T1

[tender halo]

its locally hausdorff

so it is T1

[untold lily]

first time I'm seeing locally hausdorff, so this just confuses me further lol

[tender halo]

its what it says on the tin

[untold lily]

interesting

[uneven bronze]

Just one final question about this T1-ness. Consider {0'}^c = (-1,1). How do I write (-1,1) as elements from the subbase? Recall these are (-1,a), (a,1) where -1 < a < 1, and further sets which I think are not relevant, as they include {0'}.

I have already showed that X \ {0'} is homeomorphic to (-1,1), so it should be open, right?

[untold lily]

uhh

I think you should take a break

and come back to this once you've rested your mind

[uneven bronze]

good idea 🙂

[untold lily]

you will see it instantly

[pure shale]

Hi, basic question about Zorn's lemma here, I feel I'm making some categorical misunderstanding. Below, use the convention that the naturals start at 1.

Suppose we take a set $X = { 1 - \frac{1}{n} : n \in \mathbb{N} }$. Then this set clearly has an upper bound (it is 1) and yet there is no maximal element.

Indeed if we take the subset $S \subset X$, defined $S = { 1 - \frac{1}{2n}: n \in \mathbb{N} }$, then $S$ has the upper bound of 1. However as we have seen, there is no maximal element of $X$.

This, to me, looks like a direct contradiction of Zorn's lemma. But this is also such a common set that I am certain I've misunderstood something. Any assistance would be appreciated.

[gentle osprey]

Attelis

[untold lily]

For Zorn's lemma, the upper bound of a chain needs to belong to the set

neither X nor S has an upper bound for the chain of the entire set

[pure shale]

Oh!

Yep that'll do it. Thank you kindly @untold lily :)

[prime elbow]

I am thinking of an example such that in metric space M, cl A is dense but not A. Any hint?

Oh no

A has to be dense

Because cl A dense implies cl A = M

So A is dense

[prime elbow]

I am thinking that is that true there is an open dense set A of R such that R\A is uncountable, R with usual metric space?

[uneven bronze]

Let $0'$ denote a point that is not an element of $(-1,1)$, and let $X = (-1,1)\cup{0^{\prime}}$. Let $\mathcal{T}$ be the topology on $X$ generated by the sets $(-1,a), (a,1), [(-1,b)\setminus {0}] \cup {0^{\prime}}$ and $[(c,1) \setminus {0}] \cup {0^{\prime}}$, where $-1<a<1, 0<b<1$, and $-1<c<0$.\

Exercise: The sets $[-\frac12,\frac12]$ and $([-\frac12,\frac12]\setminus {0})\cup{0'}$ are compact but not closed in $X$, and their intersection is not compact. \

I think I know how to show these sets are closed, but how do I show they are compact and their intersection isn't?

[gentle osprey]

psie

[rancid umbra]

hint: ||the cantor set doesn't contain any intervals||

[uneven bronze]

Take for instance [-1/2,1/2]. We need to consider for every open cover (of open sets in X??) that it has a finite subcover. I'm not really sure how to do this. 😔

[tender halo]

A is the complement of the cantor set

[rancid umbra]

there are two maps (-1,1) -> X which may be helpful

[uneven bronze]

yes, the inclusion map of (-1,1) to X, and g(0)=0' and g(x)=x otherwise on (-1,1), right?