point-set-topology

[rancid umbra]

yea

[uneven bronze]

Indeed, I have shown they are homeomorphism onto their ranges.

[rancid umbra]

yup

continuous functions take compact sets to compact sets

so you have shown a bit more than what you need to, actually

but its good that you did

[uneven bronze]

thank you! the intersection of these two sets is [-1/2,1/2] \ {0}. How can I convince myself that this is not compact in X?

[rancid umbra]

use the first map

[uneven bronze]

Ok 👍

[rancid umbra]

you may need this for it

[lucid ocean]

hello

[uneven bronze]

Just want to check a basic thingy about homeomorphisms. If I have two spaces (X, T) and (X,T'), and I'm able to construct a homeomorphism between these spaces, then T = T', right?

[quartz horizon]

not necessarily, i think?

[jolly umbra]

Consider
X = {x, y}
T = {X, ∅, {x}}
T' = {X, ∅, {y}}

f: (X,T) -> (X,T') given by f(x)=y and f(y)=x is a homeomorphism.

[uneven bronze]

Ok. 👍

[lucid ocean]

makes me wonder what are the spaces where it's exactly true, that if you have a homeomorphism f: X -> X (from (X, tau) to (X, tau')), then tau = tau'

a kind of homogenity property

obvious candidates are like (X, P(X)) or (X, {emptyset, X}), and probably (R, tau_R), though I'll have to double check

[hidden abyss]

Something like iff for every x, y there is an isomorphism of posets between the open neighbourhoods of x and the open neighbourhoods of y?

[tender halo]

its not true for R

[hidden abyss]

Huh really

Oh right

[lucid ocean]

crap am I missing something obvious

[hidden abyss]

You could switch the labels on 0 and 1 for example

[lucid ocean]

oh right

are all such spaces "trivial" in a way

by simple renaming

[tender halo]

take any bijection from X to itself that takes an open set to a non-open set and lift the topology through it

[hidden abyss]

In a sense yes since a homeomorphism R->R' will induce a renaming

[tender halo]

so we get that all subsets of certain cardinalities are open

[lucid ocean]

right, okay

yeah nevermind, wasn't as interesting as I thought 😔

[prime elbow]

Thank you

Yes

[hidden abyss]

Oh this isnt even true if you assume T ⊆ T'

[lucid ocean]

I'm getting flashbacks to asking a problem I was curious about and getting the correct but very unsatisfying answer of yes because the empty set exists

and similar stuff like that

[prime elbow]

And complement of first category is not necessarily to be second category, right?

We can take metric space Q and take set A = { x ≥0 | x in Q } then both A and Q\A is first category.

And complement of second category is not necessary to be second category because we can take discrete metric space, say R and A = { x≥0 | x in R}, right?

[tender halo]

so yeah in the end we get that the discrete and indiscrete topologies are the only answers

i think the matter can be different in ZF instead of ZFC?

[lucid ocean]

hmm would it not work with say cofinite even?

like you said, just topologies defined by specific cardinality of their open or closed sets

[hidden abyss]

You could have something $\bigsqcup_{\bZ} \bR$, where in T, the negative indexed copies of $\bR$ are discrete and the non-negatives are euclidean, and in T' the non-positives are discrete and positives are euclidean

[gentle osprey]

Jussari

[tender halo]

huh i guess cofinite does work hmm

[hidden abyss]

Shifting indices by one is a homeomorphism and T is a proper subset of T'

[lucid ocean]

co-K (all subsets with < K elements removed), for any cardinal K should work, because under bijection, f(X\S) = X \ f(S) where of course f(S) has the same cardinality

but that's probably it, you could cook up some map that swapped labels for anything else

[uneven bronze]

If we assume the homeomorphism is the identity map, then it is true, right? This gives us then the other inclusion.

[quartz horizon]

yes

[lucid ocean]

my current favourite basic topological fact -- continuous surjections map to "smaller", less-separated spaces necessarily, e.g. continuous surjection f: X -> X implies the target topology is coarser than initial

[tender halo]

coarser spaces can be more separated

wrt T3 and T4

[lucid ocean]

that's true

tho I don't really think of those as necessarily separation even, more regularity type properties

but that's just my personal feeling (the "pure" separation properties are the ones preserved under preimage of continuous injection to me)

[tender halo]

huh why?

do you mean image?

you can inject anything into the antidiscrete space

[lucid ocean]

as in if (Y, tau_Y) is P, and there exists continuous injection from X to Y, then one can deduce that (X, tau_X) is P

where P can stand for T2, T1, etc.

[tender halo]

almost nothing is preserved that way

except for density

wait uhh preimage

[lucid ocean]

T2, T1, T0, T2_1/2 and complete hausdorffness are preserved that way

as well as total disconnectedness

[tender halo]

ah right hmm

to me a more natural candidate are proper maps

[lucid ocean]

all such properties are preserved under subspace, homeo and expansion of topology

[unreal stratus]

the only valid separation axioms are T0, T1, T2 and then being a metric space

[lucid ocean]

and it's equivalent to all three in conjunction

[tender halo]

T3.5 is the most natural separation axiom

[lucid ocean]

ironically, metrizability would not fit into my system above

because K-topology on R fails to be T3, T4, metrisable, etc. but euclidean R is, and K-topology is finer

[unreal stratus]

is K-topology the weird one with {1/n}

[lucid ocean]

yes, or I guess deleted sequence topology according to the book

[unreal stratus]

I ask this because lol there's uh

similar notation used for compactly generated spaces if you know those

[tender halo]

its the most popular example for a hasudorff non regular space

[lucid ocean]

I don't but that's good to know I guess

[lucid ocean]

yeah it's a bit weird, I crafted my definitions so that the lower T axioms would be preserved

and then got a few other metaproperties as a consequence

analogously, antiseparation properties according to (a shower thought I had a year ago) are those preserved by continuous surjection, which is more straightforward

compactness, connectedness, path connectedness, separability, sequential compactness etc. satisfy antiseparation

they are equivalent to preservation under quotients, coarsening and homeomorphism (or just quotient maps and coarsening)

also antisep + sep => "trivial" in a sense as well, in that a property satisfying both cannot distinguish topology, e.g. the property of being a countable space

[lucid ocean]

a urysohn stamp of approval

or was it tychonoff

okay so it's named after tychonoff but urysohn worked with it before

[radiant stone]

me when finding out separation axioms were useless when dealing with schemes:

[lucid ocean]

I have lots to look forward to then

[lucid ocean]

what's a proper map? Is it the definition wikipedia supplies me with

[prime elbow]

.

[tender halo]

yea, closed with compact fibers

it preserves a lot

[lucid ocean]

interesting

so like preimage under proper maps preserves P or image under proper maps preserves P?

where "preimage" is modified slightly like above

(I don't have an intuition for proper maps yet, it's also the first time I'm learning of it)

[tender halo]

it preserves all axioms of separation

under continuous surjections

[lucid ocean]

oh damn

[tender halo]

also local compactness

local compactness, compactness and T3 are preserved under preimages

[lucid ocean]

yeah but then it's a bit too permissive to define the separation metaproperty

[tender halo]

also metrizability to boot

[lucid ocean]

compactness would be considered a separation property which just sounds wrong

it's a cool property of maps tho, I'll mess around with it more

[tender halo]

proper maps guarantee locality

in a certain sense

they tell us that preimages of compact sets are compact

so preimages of small sets are small

[lucid ocean]

hmm would it be reasonable to say that they don't "squish" too much space together

yeah, if you "squish" space together, then your preimages are big (in a way)

[lucid ocean]

well I'll figure out the intuition through this I'm sure

[lucid ocean]

(but also at the same time, if the domain and codomain are compact, then preimage of closed sets are closed so these maps are also automatically continuous)

[gritty widget]

hello guys what are you working on here?

[lucid ocean]

just playing around in the topological sandbox, nothing much

[gritty widget]

cool

[lucid ocean]

there this old paper on making "opposite" versions of topological properties

where connectedness (antisep) |-> total disconnectedness (sep)

compactness (antisep) |-> anticompactness (sep)

I wonder if P is antiseparation property => anti-P is separation property (with cont. inj)

[lucid ocean]

re bump so someone else with more knowledge can answer

[gritty widget]

Your example is correct in the metric space 'Q' (rationals with usual metric):

A = {x ≥ 0 | x ∈ Q} (non-negative rationals)
Q\A = {x < 0 | x ∈ Q} (negative rationals)

Both are first category because:

Each can be written as a countable union of singletons
Each singleton {r} is nowhere dense in Q (its closure is {r}, which has empty interior in Q)
Therefore both A and Q\A are first category

This shows: First category + First category = Entire space
Part 2: Issue with Your Discrete Metric Example
Your discrete metric example doesn't work as intended. In discrete metric space R:

A = {x ≥ 0 | x ∈ R} is second category
R\A = {x < 0 | x ∈ R} is also second category

Why? In discrete metric:

Every point is isolated (open ball of radius 1/2 around any point contains only that point)
Every infinite set has non-empty interior (every point is an interior point)
Since A and R\A are both infinite, both have non-empty interior
Therefore neither is nowhere dense, so both are second category

for the part 2
To show "complement of second category ≠ necessarily second category":
Take R with usual metric:

A = [0,1] is second category (by Baire's theorem - complete metric space)
R\A = (-∞,0) ∪ (1,∞) is also second category

But if we take A = R{0} (all reals except 0):

A is second category
R\A = {0} is first category (single point)

So yes, complement of second category can be either first or second category, but your discrete example showed the opposite of what you intended.

[prime elbow]

Sorry it is a typo, second part is complement of second category not necessary to be first category, so discrete metric space works

[prime elbow]

And if I am right then the first category has an empty interior

Thank you, you wrote a lot ❤️

[uneven bronze]

Statement: If $X$ is countably compact, then every sequence in $X$ has a cluster point (i.e. every neighborhood of this point contains infinitely many elements of the sequence). \

Proof: Let $\langle x_n\rangle_{n=1}^\infty$ be a sequence in $X$, and for each $n\in\mathbb{N}$, define $E_n={x_k}{k=n}^\infty$. Suppose $\bigcap_1^\infty \overline{E}n=\varnothing$. Then ${(\overline{E}n)^c}1^\infty$ is a countable open cover of $X$, which has a finite subcover ${(\overline{E}{n_1})^c,(\overline{E}{n_2})^c,\ldots,(\overline{E}{n_m})^c}$. It follows that $$\bigcap_1^m E{n_k}\subset \bigcap_1^m \overline{E}{n_k}=\left(\bigcup_1^m(\overline{E}{n_k})^c\right)^c=X^c=\varnothing,$$a contradiction because $x_N\in \bigcap_1^m E_{n_k}$ for $N=\max{n_k}1^m$. Hence there exists $x_0\in \bigcap_1^\infty \overline{E}{n}$. Now let $U$ be a neighborhood of $x_0$ and $n\in\mathbb{N}$. Then $U$ meets $E_n$, and so there exists $k\geq n$ such that $x_k\in U$.

[gentle osprey]

psie

[uneven bronze]

I was wondering about a tiny little detail. Do we need to be concerned about whether U meets En in only x0?

[gritty widget]

no prob

You're absolutely right about the correction your discrete metric example perfectly demonstrates that the complement of a second category set is not necessarily first category.
Part 1 (Q with usual metric): Correct

Both A = {x ≥ 0 | x ∈ Q} and Q\A are first category
Shows: complement of first category ≠ necessarily second category

Part 2 (R with discrete metric): Correct after your clarification

Both A = {x ≥ 0 | x ∈ R} and R\A are second category
Shows: complement of second category ≠ necessarily first category

Regarding your final question about "first category has empty interior" - this is not true in general.
A first category set is defined as a countable union of nowhere dense sets. While each nowhere dense set has empty interior, their countable union can still have non-empty interior.
Counterexample: In the discrete metric space, consider the set of all even integers E = {..., -2, 0, 2, 4, ...}. We can write E as a countable union of singletons E = ∪{2n}, where each singleton {2n} is nowhere dense. So E is first category, but E has non-empty interior (every point in E is an interior point in the discrete metric).
However, in complete metric spaces, Baire's theorem tells us that first category sets have empty interior, which might be what you're thinking of. But this is a special property of complete spaces, not a general property of first category sets.

[gritty widget]

What is blessing my eyes

[uneven bronze]

what do you mean? 🙂

[gritty widget]

it's a meme

i would like to help you but i need to buy grocieries

maybe later

[uneven bronze]

ah ok, no worries, no hurries! 🙂

[prime elbow]

How do you write it so fast? isn't chatgpt?

[gritty widget]

bro i just copied and pasted the old parts of the message i've sent and then did some changes

[prime elbow]

Don't mind

Ah okay thanks man

[gritty widget]

if you have other questions ask them

[prime elbow]

We cannot write R^2 as a countable union of lines. Since line { (x,ax+b), x in R} is closed set in R^2 and interior set is empty so by Baire category theorem it cannot be possible that R^2 is the first category.

Actually the set (x,ax+b) is closed easily but I know its interior set is empty but I have no rigorous argument for it

[gritty widget]

Mh

Give me some time

I’ll respond in 30/40min

[hidden abyss]

Suppose the interior of a line L was nonempty. Then there's some (x,y) ∈ L and an ε>0 such that all points within an ε-distance of (x,y) lie in L too

but for example at least one of the points (x+ε, y) and (x,y+ε) will lie outside of L, a contradiction

[prime elbow]

Yeah, can I use the fact proper subspace has an empty interior?, I know it is not subspace but it is affine subspace

Oh no it is not

[prime elbow]

Is it independent of norm?

[hidden abyss]

Thats not true for general subspaces. Or do you mean linear subspaces?

[prime elbow]

I mean yes I am talking about normed vector space, where subspace is linear subspace

Oh every norm on finite dimensional is equivalent

So it doesn't change

[hidden abyss]

I was assuming the euclidean norm

I dont think R² has norms that induce a different topology though

[prime elbow]

Yes because every norm on R^2 is equivalent so they induce the same topology

[hidden abyss]

Uhh yeah I guess so then

I guess the easiest proof is something like wlog assume the point is 0, then we can scale a basis (of the whole space) down so that they lie in an epsilon ball and since the subspace has smaller dimension, one of those is not contained in it

[gaunt linden]

Does it need to be a topological proof? It would seem to be easier just to appeal to cardinality:
There are only countably many slopes of the lines, and so R is uncountable we can choose a0 such that y=a0·x is not parallel to any of the lines.
Then each line has exactly one intersection with y=a0·x, and these at most countably many intersections cannot be all of the points on y=a0·x.
So there are points in the plane that are not in any of the lines.

[lucid ocean]

my old (and now redoing new) notes on the topic in mini-article form

[prime elbow]

Actually I have to use Baire category therefore I used it

[gaunt linden]

Okay.

[prime elbow]

How countably many slopes?

[gaunt linden]

Only countably many lines, and each line has one slope.

[prime elbow]

Oh yeah

[gaunt linden]

I should perhaps have said "at most countably many slopes"

[prime elbow]

I got it, its good, thanks

[unreal stratus]

If R^2 is a countable union of lines then by measure theory it has measure 0

Jk

[main ore]

does anyone have a paper or article or something with the proof that the felix Hausdorff and our more modern day one are equivalent?

[gaunt linden]

Hmm, that could probably be an exercise in a first introduction to point-set.
Which means finding it actually written down could be a bit hit-and-miss.

[untold lily]

ronald brown's topology and groupoids has it

(I just followed the citation given in the wiki)

[rancid umbra]

of course it does

i swear that book has everything in it

[main ore]

oh thanks i thought it was mostly something else

[untold lily]

I remember trying to prove this long ago and got stuck in the last axiom

I think I resolved that by finding it in another book

wasnt browns book I think, very old text

[hidden abyss]

I think Armstrongs Basic topology also proves it

[main ore]

it looks like a pretty nice proof actually

[gaunt linden]

I think that last axiom just ought to be definition-chasing, though? M is the open set sitting in between x and N which certifies that N is in fact a neighborhood of x.

[untold lily]

I think what I had suffered was: I thought I had proven the equivalence without using axiom 4, so I didn't understand why it was necessary

going from nbhd definition to open set definition

[main ore]

it made me think of another question though. If all topological spaces were metrizable, would we say topological spaces and metric spaces are the same definition?

[untold lily]

why you need it turns out to be subtle iirc

[untold lily]

sort of but not really: if a space is metrizable, there'll be many different metrics inducing the same topology

"metric space" implies you've fixed a metric

but topologically they'd behave the same, so when proving stuff in the theory the distinction wouldn't matter in a fundamental way

[hidden abyss]

Otherwise we could force every neighborhood to contain some fixed point

[main ore]

sounds fair to me, but let me pivot back to this proof then. The proof aims to show that N induces the same kind of open sets in the modern definition, aswell as those open sets have the traits we have in the modern definition. Now this would make the definitions the same, only if each N imposes a unique topology on X, by your example. Correct?

[hidden abyss]

Like taking R and defining the neighbourhoods of x to be precisely sets A ∪ {0} where A is a nbhd of x in the standard topology on R

[untold lily]

yeah ik it had to do with how you get weird nbhds if you don't add 4

tho you get same open sets

[hidden abyss]

Then the only open sets we get are the empty set and those opens of R which contain 0

[untold lily]

uh, yes

[hidden abyss]

Though this is a valid topologh isnt it? I guess the problem is that nonzero points won't have any open neighbourhoods

[untold lily]

basically, given a topology defined by open sets, you get a collection of nbhds that satisfy the nbhd axioms

and if you forget where the nbhds came from, just look at the nbhds, and generate open sets based on the nbhd axioms, you get back what you started with

but this is a bit difficult to phrase

and conversely ofc, if you start with a collection of nbhds obeying the axioms, it gives you a collection of open sets, and if you then forget the nbhds you started with and look at the nbhds generated by the open sets, you get the nbhds you started with

[gaunt linden]

Ah, I see.
My gut feeling (without actually working out the proofs) is that the fourth axiom might be necessary for making sure you get the same neighborhood system back when you convert it to a topology and back again.

[untold lily]

that is exactly it

the open sets turn out to be the same, but the open sets don't generate the same nbhds you started with

[gaunt linden]

Yeah -- it's easy to see that a neighborhood system derived from a topology will always satisfy that axiom, if we start from a neighborhood system that doesn't, it can't possibly round-trip correctly.
We then just need an example showing the first three axioms don't imply the fourth.

[gaunt linden]

(But it looks like Jussari's example will work for that, actually).

[median sand]

If X is completely regular (points may be separated from closed sets by functions), how do I show X embeds into a product of pseudometric spaces? This is stated w/o proof in Kelley. I know how it's done for Tychonoff spaces, but this uses the T_1 condition. I tried doing something like defining for each A<=X closed a metric by |f(x)-f(y)|, where f is a separating function, and bundling these into a product, but this didn't seem to work.

[gaunt linden]

If you let each factor be |X| with the pseudometric induced by a separating function, then the map from X to the diagonal of the product is automatically a bijection to its image, and also continuous since the composition with each projection is.
So all you really need to show is that it this map is open ...

[tender halo]

why is it a bijection?

[gaunt linden]

It is injective because the underlying set of the product is just a Cartesian product of a lot of copies of |X|, and surjective because of "to its image".

[tender halo]

i dont follow the injectivity

[gaunt linden]

f(x) = (x,x,x,x,x,x,x,...)

So if f(x)=f(y) it can only be because x=y.

[tender halo]

ah i guess, that makes sense actually

it is an embedding then

because its a diagonal of a set of functions that separate points and separate points and closed sets

[gaunt linden]

The natural idea that doesn't work (I think) is to take the product of a lot of copies of R instead -- then the natural map from X might not be injective.

[tender halo]

thats called diagonal lemma or something

[uneven bronze]

Exercise: For $x\in[0,1)$, let $\sum_1^\infty a_n(x)2^{-n}$ (where $a_n(x)=0$ or $1$) be the base-$2$ decimal expansion of $x$. (If $x$ is a dyadic rational , choose the expansion such that $a_n(x)=0$ for $n$ large.) Then the sequence $\langle a_n\rangle$ in ${0,1}^{[0,1)}$ has no pointwise convergent subsequence. (Hence ${0,1}^{[0,1)}$, with the product topology arising form the discrete topology on ${0,1}$, is not sequentially compact. It is, however, compact.) \

Couple of questions; why can we and why shall we choose the expansion such that for dyadic rationals, the sequence $\langle a_n\rangle$ is eventually $0$? Also, what does it mean for the product topology to arise from the discrete topology? I know e.g. that $f_n\to f$ in the product topology iff $f_n\to f$ pointwise.

[gentle osprey]

psie

[pallid comet]

sorry - I don't really know where to ask this, but

if (C, p) is a covering space of X (that is, for any point x in X, there is an open U containing x such that the map p acts like a homeomorphism on connected components of p^-1(U)) (also, C and X are assumed to be path connected and locally path connected metric spaces)

then for any point X, must there be an open set U containing X for which the set of connected components of p^-1(U) is locally finite?

(furthermore, must the set of connected components of any open set be locally finite?)

[uneven bronze]

I think I know how to answer these questions now.

[gaunt linden]

For a counterexample to the question I think you're trying to ask, consider R as a covering space of the unit circle by p(t)=(cos(t),sin(t))

[pallid comet]

sorry if i'm missing something, but wouldn't the preimage of any open set on the circle consist of various disjoint open sets in R spaced integer amounts apart?

so the set of components will be locally finite

[gaunt linden]

Sorry, seems I missed "locally" finite.

But what exactly does that mean in this context?

[pallid comet]

for any point c in C, there exists a neighborhood V of c that only intersects finitely many connected components of p^-1(U)

[gaunt linden]

Can't you just choose one of those connected components as V?

[pallid comet]

you could, but c need not be in any component*

so for instance you could have something like...

[gaunt linden]

Oh.

[pallid comet]

(or something)

[gaunt linden]

Good point.

Might be tricky for there to be a point in X that this c can map to continuously, though.

[pallid comet]

let's perhaps clear up the names a bit here - say we want to prove/disprove the following statement:

let (C, p) be a covering space of X. for any point x in X, there exists a neighborhood U of x such that: for any point c in C, there exists a neighborhood V of c that only intersects finitely many connected components of p^-1(U).

then c definitely can't be a limit point of the set p^-1(x) due to the definition of covering space -

if c was a limit point of p^-1(x), we'd have p(c) = x and thus c must lie in some component H of p^-1(U), and H must actually be open since C is locally path connected, so H must contain at least another point mapping to x, but then the mapping p restricted to H wouldn't be a homeomorphism as 2 points map to x

i don't see any problems with the set of components of the preimage of U being not locally finite though -

[red yoke]

LCH is sufficient

[pallid comet]

LCH?

[red yoke]

Locally compact hausdorff

(wait am I tripping)

No I am not

[pallid comet]

sorry - i don't really see how that can be used -

[red yoke]

Recall LCH spaces are T3

Pick U ⊃ x satisfying the covering condition, pick U' ⊃ x such that cl(U') is compact and contained in U

If p(c) is in U, pick V homeo to U

Otherwise apply T3 to pick V disjoint from p^-1(U)

[pallid comet]

isn't p^-1(U) open though?

[red yoke]

Sorry, disjoint from p^-1(U')

[pallid comet]

oh yeah how didn't i think of this :/

[red yoke]

Perhaps unexpected usage of compactness

[pallid comet]

actually i was working with metric spaces already, so t3 was already a given :3

but anyway thanks!

[alpine nest]

Lesser continuum hypothesis

[pallid comet]

...i just realized that p^-1(cl(U')) needs not be closed

wait - nevermind, it does. my brain just malfunctioned. sorry for the ping!

[red yoke]

That was what I used compactness for at fIrst, I'm now realizing it's redundant

You just need T3 and cl(U') ⊂ U

[pallid comet]

yup, you just need t3

[uneven bronze]

Exercise. If $X$ is normal, then $X$ is countably compact iff $C(X)=BC(X)$. \

Attempt. $\implies$: If $f\in C(X)$, then $f(X)$ is countably compact. We know that $\mathbb{C}$ is first countable. Does it follow that $f(X)$ is also first countable? If so, then I know that $f(X)$ is sequentially compact, and hence bounded by Heine-Borel.\

$\impliedby$: This direction is trickier I think. Any help is really appreciated.

[gentle osprey]

psie

[uneven bronze]

C(X) are the continuous, complex-valued functions on X, and BC(X) the bounded, continuous, complex-valued functions on X.

Ok, I think it is obvious that f(X) is first countable if it is a subset of a first countable space. If you have some idea on how to approach the converse, let me know. 🙂

[alpine nest]

My first thought is to do a contrapositive;; so assume X is not countably compact, and use the countable cover with no finite subcover to construct a continuous nonbounded function.

Normality will probably help to get the function in question.

[lucid ocean]

revised my notes

[tender halo]

subspaces of first and second countable spaces are first and second countable yea

your question is as follows: prove that in normal spaces pseudocompactness implies countable compactness (the reverse is true everywhere, because the image of a countably compact space is countably compact and therefore compact in R and therefore bounded)

for the reverse, you need to use tietze-uryhson and the fact that a space is countably compact iff there are no countably infinite discrete subsets

[lucid ocean]

is hereditary separability preserved under continuous surjection?

[tender halo]

well separability is

so hereditary separability too

[lucid ocean]

phew, I thought I was going insane

but is that really true

I guess suppose for a contradiction it were false for some continuous surjection f: X -> Y, then there is some subset S subset Y not separable

then (edit) f^(-1)(S) subset X and is separable as a subspace

then we have a contradiction

[tender halo]

you dont need contradiction

also why choice, just take the preimage

also you dont need axiom of choice to choose one thing

[lucid ocean]

wait yeah true

what's a direct proof

hmm wait I guess

if S subset Y, then f(f^(-1) (S)) = S and we just have cont image of separable as per usual

yeah nevermind, thanks

[lucid ocean]

so at least I haven't accidentally proven something known to be false, reason why I was so paranoid

[uneven bronze]

here's what I know and understand so far. If X is not countably compact, it is not limit-point compact. This means there is an infinite subset with no limit/accumulation point; we can take a countable subset of this infinite set and it will still have no limit/accumulation points, so it is closed. Now, is it possible to go directly from here to concluding that X is not pseudocompact, or do I have to construct some function on this closed subset somehow (I don't know how though...) and extend it via normality + Tietze?

[tender halo]

the latter

ask yourself this, what is the topology on that subspace

[uneven bronze]

the subspace topology from X?

[tender halo]

yea

what is it exactly

having no limit points implies something about what the topology is

[uneven bronze]

Ok, I'll just accept for now the discrete one.

and so every subset is closed and open at the same time

[tender halo]

yea, its the discrete topology

its pretty easy to make an unbounded function on N

[uneven bronze]

if x1, x2, ... is the enumeration of the set, we just take x_n -> n, or? Then it's continuous (and unbounded), and voila, extension via Tiezte (possible through normality and the fact that the set is closed) furnishes a countinuous unbounded function on C(X) \neq BC(X). Baam!

[tender halo]

yep

in fact, a space is pseudocompact iff it has no C-embedded copy of N as a subspace

[median sand]

Where can I find a thorough exposition of function space topologies (uniform structures, compact-open topology, Arzela-Ascoli, etc.) besides Kelley or Willard?

[craggy breach]

im
confused on how if a set is open they why is the complement of that set closed ?

[fringe thorn]

that's the definition of a closed set

a set whose complement is open

if you're seeking intuition, a open set is one that contains none of its boundary points

and a closed set is one that contains all of its boundary points

a set and its complement share the same boundary, so if a set is open, its boundary is entirely contained by its complement, making it closed

[craggy breach]

i see thanks for help

[gritty widget]

I mean you don't really need that much topology to understand function spaces

[tender halo]

do you not?

even C_p theory (the theory of topology of pointwise convergence) is pretty broad

[untold lily]

Dugundji

[tardy carbon]

to understand the metric space version of Arzela Ascoli, you do not. It's covered in intro analysis books such as https://mtaylor.web.unc.edu/wp-content/uploads/sites/16915/2018/04/anal1v.pdf

[untold lily]

feel like you're overdoing the gimmick here

[tender halo]

ok? thats like saying that you dont need to know topology to prove that a segment is compact because its covered in first year real analysis book

i mean its true but also irrelevant

[tardy carbon]

No, I mean the metric space version of Arzela Ascoli has been enough for all my applications. It's pretty general

[tawny tundra]

Does anybody know any papers that go through properties of nets and filters in topological spaces?

[untold lily]

does it need to be papers or can it be books?

though there are peter clark's lecture notes on convergence that delve into this, seems like the notes might have been deleted though

[tardy carbon]

Folland's real analysis book covers nets

[untold lily]

a bunch of books cover this: willard, kelley, first chapter of Pedersen's Analysis NOW

[pure shale]

TVS by Schaefer covers it iirc

[untold lily]

yeah there's a lot more from the analysis side of things: infinite dimensional analysis: a hitchhiker's guide and handbook of analysis and its foundations

those are two books meant to be general analysis references, they cover it

other books on TVS will cover it as well, like treves

[tawny tundra]

books or papers

I just want a list of results on nets that are not trivial

I've read some books on topology but the results are a bit trivial though

Things such characterizing being T2 through nets and such

But I'm looking for theorems on nets specific to spaces with additional structure such as monotone convergence and such

[tender halo]

nets are not particularly interesting, i am yet to find any interesting theorems where nets are somehow more natural as a method of bookkeeping

maybe its more natural in stuff that is not in general topology

[tawny tundra]

hmm

[tender halo]

like complete partial orders

[tawny tundra]

I'm interested in having a formal way of manipulating limits in topological spaces, and have heard that there are some monotone convergence theorems on nets, but haven't found many theorems on nets

pretty much all I've heard on nets are characterizations of separation axioms, continuity, etc.

I haven't found much use on them aside from that

[tender halo]

ultrafilters are everywhere, i would look at those first if you are interested in juggling limits

[tawny tundra]

Alrighty

Do you know of any nice resources on them?

I mean, the books on topology I'm looking at pretty much just give out definitions and work through some properties but don't delve too deep

[tender halo]

rings of continuous functions by gillman, ultrafilters by comfort and negrepontis, these are two resources that are mentioned everywhere basically

[tawny tundra]

why do ultrafilters come up when juggling limits?

[tender halo]

an ultrafilter is what a limit is, from a certain point of view

[untold lily]

as I mentioned above, look at willard and kelley

or, you could look at the book "convergence foundations of topology" which I haven't read but attemps to construct all of topology starting from convergence structures or something

so surely, whatever you are looking for would be in that book

[tender halo]

a topological filter is a map telling you which subsets do and dont contain your limit

a space is compact iff all filters converge, i.e. all limit points are inside the space

[tender halo]

i would really recommend working through a couple chapters of gillman, comfort is pretty dry and is not particularly motivated if you dont already know why you are there, but gillman is beautiful and straightforward, the wealth of results self-evident

[tawny tundra]

Thanks!

I've looked at all the books recommended and I've got enough material to look at (and the books have got what I'm looking for)

https://tenor.com/view/space-cat-gif-20879517

[uneven bronze]

Proposition 1. If $F$ is a compact subset of a Hausdorff space $X$ and $x\notin F$, there are disjoint open sets $U,V$ such that $x\in U$ and $F\subset V$.\

Proposition 2. If $X$ is an LCH space, $U\subset X$ is open, and $x\in U$, there is a compact neighborhood $N$ of $x$ such that $N\subset U$.\

Proof. We may assume $\overline{U}$ is compact; otherwise, replace $U$ by $U\cap F^\circ$ where $F$ is a compact neighborhood of $x$. By Proposition 1, there are disjoint relatively open sets $V,W$ in $\overline{U}$ with $x\in V$ and $\partial U\subset W$. And the proof continues ...\

Questions. There are several question marks. I don't get why we can assume $\overline{U}$ is compact. $U\cap F^\circ$ is not compact, right? Also, I don't understand how proposition 1 is used here; it says that $x$ should not be in $\overline{U}$, yet it is. Neither do I understand why $V,W$ are relatively open in $\overline{U}$ and why $\partial U\subset W$?

[gentle osprey]

psie

[uneven bronze]

Ok, I think I understand the first question; we only want U to be a subset of some compact set, so we can replace it with U \cap F^o if it isn't.

[uneven bronze]

For the second question, I think what is going on is that Proposition 1 is applied to the Hausdorff space cl(U) (as a subspace of X it is Hausdorff) with compact subset bdry(U).

[uneven bronze]

I have a follow up question to Proposition 2 above. \

Proof (continued). Then $V$ is open in $X$ since $V\subset U$, and $\overline{V}$ is closed and hence compact subset of $U\setminus W$. Thus we may take $N=\overline{V}$.\

Why is $\overline{V}$ a compact subset of $U\setminus W$?

[gentle osprey]

psie

[uneven bronze]

Still thinking on this one. 😔

[indigo nymph]

in which book are these from, along with the normal countably compact space problem that you shared yesterday ?

[uneven bronze]

It's Folland's real analysis book. 🙂

[indigo nymph]

thanks

[uneven bronze]

Urysohn's lemma for LCH. If $X$ is an LCH space and $K\subset U\subset X$ where $K$ is compact and $U$ is open, there exists $f\in C(X,[0,1])$ such that $f=1$ on $K$ and $f=0$ outside a compact subset of $U$.\

Corollary. Every LCH space is completely regular (meaning it is $T_1$ and for each closed subset $A$ and each $x\notin A$, there exists $f\in C(X,[0,1])$ such that $f(x)=1$ and $f=0$ on $A$).\

Attempt at proving the corollary: so let $A$ be a (proper) closed subset of an LCH space. Then $A^c$ is open. Thus there exists some compact neighborhood $N_x$ of each point in $x\in A^c$ such that $N_x\subset A^c$. By Urysohn's, we have that $f=1$ on $N_x$ and $f=0$ outside a compact subset of $A^c$. In particular, $f(x)=1$ and $f=0$ on $A$.

[gentle osprey]

psie

[uneven bronze]

Is this a correct proof of the corollary? I think the completely regular condition should mention that A needs to be a proper closed subset, for if it is the whole space, there can not be any x not in A.

Appreciate it if you also think otherwise or can not follow the proof.

[tender halo]

i mean it works

strangely formatted though, you are given x, not choosing it

also the corollary is strictly weaker than what you call Uryhson lemma? you dont even need proof reall,y just substituting a singleton for K gives you that LCH are Tyhcnoff

i am curious how you proved the first statement without referring to tychonoffness

because in my mind you need to first establish complete regularity to prove it

[uneven bronze]

Thanks. Yes, it has a bit of a weird format. I think what needs to be fixed is that for each x in A^c, there is a function f_x, i.e. the function depends on x.

[uneven bronze]

If we take K = {x} in Urysohn's lemma, what do we take U as in that lemma?

Also, I'm not sure the corollary is strictly weaker than the Urysohn's lemma for LCH. Are you sure you read the part "...f = 0 outside a compact subset of U."? 🙂

[tender halo]

K is the singleton of x, U is the complement of A

[tender halo]

im not sure how thats relevant, thats just an extra consequence

[heady skiff]

Is $\mathbb{R} P^1 \approx S^1$ just by $[x_1, x_2] \mapsto \frac{(x_1, x_2)}{\norm{(x_1, x_2)}}$?

[gentle osprey]

okeyokay

[gaunt linden]

No, that is not well defined -- if you start by viewing the same element of RP1 as [-x1, -x2] instead, you end at a different point in S1.

[heady skiff]

Damn it

I'm trying to come up with something that is well-defined and the canonical choice seems to normalize it but that's a good point

[unreal stratus]

It's easiest if you think about RP^1 as a quotient of S^1

[gaunt linden]

If you view $S^1$ as the complex unit circle, you can use
$$[x_1, x_2] \mapsto \frac{(x_1+x_2i)^2}{x_1^2+x_2^2}$$
and then you can actually prove that $[x_1,x_2]$ and $[tx_1,tx_2]$ go to the same place for any real $t\ne 0$.

[gentle osprey]

Troposphere

[gaunt linden]

(The squaring gets rid of the distinction between antipodal points).

[plush folio]

So if you represent a line in RP^1 by its smallest non-negative angle with the x-axis, call it theta, then is this map basically just theta |-> exp(2*pi*i*(2*theta))? Like, rotating a line in RP^1 gives a point rotating at double speed in S^1 under this map?

[gaunt linden]

Yeah.

In general S^n is a double cover of RP^n; and it's only accidentally that there's a way to step around that in the n=1 case by "rotating at double speed".

[uneven bronze]

Tiezte (for LCH). Suppose that $X$ is an LCH space and $K\subset X$ is compact. If $f\in C(K)$, there exists $F\in C(X)$ such that $F|K=f$. Moreover $F$ may be taken to vanish outside a compact set. \

So I need to prove this version of Tietze. I have couple of tools to my disposal, but I'm unsure where to start. To apply regular Tietze, we need a normal space and a closed subset $A$ of this normal space and $g\in C(A,[a,b])$. How do I get these from only knowing $f\in C(K)$, where $K$ is compact?

[gentle osprey]

psie

[uneven bronze]

I think I know how to construct a function F as in the theorem/proposition that vanishes outside a compact set. But the theorem/proposition says it "...[it] may be taken to vanish outside a compact set." So I interpret this to mean that we don't have to make it vanish outside a compact set. Is this a correct interpretation and if so, how do we not make it vanish outside a compact set?

[tender halo]

you can just repeat the proof of tietze

the essential part of tietze is that any two completely separated subsets of A (or K, here) are completely separated in X

the theorem has little to do with LCH necessarily, the essential part is that the space is completely regular

[uneven bronze]

here's my proof of the theorem. \

Let $f\in C(K,[a,b])$. Then there exists precompact open $V$ such that $K\subset V\subset\overline{V}$. Since $\overline{V}$ is compact Hausdorff, it is normal. Thus apply Tietze to obtain an extension $g\in C(\overline{V},[a,b])$. Again since $\overline{V}$ is normal, apply Urysohn's to the disjoint closed sets $K\subset \overline{V}$ and $\partial V$ to obtain a function $h:\overline{V}\to[0,1]$ such that $h=1$ on $K$ and $h=0$ on $\partial V$. Finally, consider $gh$, which vanishes on $\partial V$. Then by extending $gh$ to $F$ such that $F=gh$ on $\overline{V}$ and $F=0$ otherwise (i.e. on $\partial V\cup L^c$), we obtain a continuous function on $X$ by the pasting lemma that vanishes outside a compact set. \

The big question remains; how do I not make it vanish outside a compact set?

[gentle osprey]

psie

[tardy carbon]

you want it to vanish outside a compact set right?

why not vanish?

[uneven bronze]

Well, the theorem says “Moreover, F may be taken to vanish outside a compact set.” I interpret this to mean we can actually choose if it does or not. Perhaps I’m reading too much into the language of how it’s phrased, but how would one not choose it vanishes outside a compact set?

[uneven bronze]

The whole statement of the theorem is given here, word for word. 😅

[tardy carbon]

The phrase "Moreover, F may be taken to vanish outside a compact set.", means we can find an extension F that vanishes outside a compact set - that this is an option for us.

[tender halo]

i mean yeah this works

[uneven bronze]

Exactly; so a proof of the theorem needs to highlight this I think, which my proof doesn't. It only constructs a function that vanishes outside a compact set.

[tender halo]

no, it means that a function that vanishes outside a compact set exists, it says nothing about functions that don't

[uneven bronze]

Ok, that would make sense actually.

[tender halo]

the "moreover, X may be taken to be Y" is standard text to indicate that you can add additional constraints on the results of the theorem

but also like uhh

you can fiddle the function around and force it to be any constant outside of V really

also say f doesnt vanish everywhere

take any point where it doesnt vanish, and take some point other than that one

f doesnt vanish outside the second point

which is a compact subset

[lucid ocean]

been messing with a problem which for some reason got me to make this figure

enjoy...?

[safe torrent]

[rancid umbra]

is this AI?

[tender halo]

why would it be ai

no emdashes in sight

[safe torrent]

Valid

[gritty widget]

not only are there no emdashes, but the figure seems quite structured overall

[iron bolt]

lexicographical ordering on the ω₁-fold product of the interval with itself my beloved

[lucid ocean]

would be impressive if it were

[lucid ocean]

||also idk of I've prefaced that I'm actually a highschool student before, I guess I blended in sufficiently well||

[queen prism]

<@&268886789983436800> troll, see message history

[lucid ocean]

I wonder what happened

[lucid ocean]

no, just the result of coming up with a topological space weird enough for my desires

I was looking for a topological space such that for any two distinct points x and y, one can find a continuous function f: X -> [0, omega1] such that f(x) = 0 and f(y) = omega1

and one that's connected with more than one point, because it's trivial otherwise

I (believe) found an example, the space [0, 1]^omega1 lexicographically ordered

between any two points, there's an uncountable number of "digits" or "levels of zoom" that one can use to map to [0, omega1] continuously

[red yoke]

Is [0, ω1] the long ray with both endpoints?

[prime elbow]

Hi Arki, your website not working

[red yoke]

Indeed, I have been procrastinating making webpages

[lucid ocean]

yes

[prime elbow]

[red yoke]

Hm what happens if some component of x is larger than that of y and some component of y is larger than that of x

[lucid ocean]

as in, if like x(3) > y(3) but x(4) < y(4)?

[red yoke]

Yea

[lucid ocean]

and x(a) = y(a) for 0 <= a < 3 for example

then y < x

by the lexicographic order

[red yoke]

Let's say x(a) = y(a) for a > 3 too

How does your construction work in this case

[lucid ocean]

I think I understand what you're trying to say

I've tried to work around that

notation if it doesn't make sense

[lucid ocean]

hence in my diagram, I've chosen a very particular point z, where the issues don't arise

to take a finite analogy, suppose x = .1478(1)17762... and y = .1478(9)17283...

then we have alpha = 4

so our map is to [4, omega] in this more finite case

z = .1478(5)9999999...

then if the alpha'th digit lies between 1 and 5, the output lies between 4 and 5

if the alpha'th digit equals that for z, then you look at the alpha+1'st and see where it lies between 0 and 9 and output between 5 and 6

etc. just take this but replace digits by elements in [0, 1]

and have aleph_1 digits

if your number lies before x, then it gets 0 and if it lies after z, you get omega (or omega1) of course

I probably could've saved myself some effort by showing that between two points one can find a continuous map to itself sending x to the 0 map, and y to the 1 map

then use that simpler scenario

[red yoke]

Hm so this gives you a map separating x and z

[lucid ocean]

yeah, but since y comes after z, you can just set g(y) = g(z) = omega1

[red yoke]

Why does that extend to a continuous map?

[lucid ocean]

as in setting it to be constant there?

[red yoke]

Oh I see

That checks out

[lucid ocean]

there's like an absurd amount of room in this space, so you can get really inefficient

I wonder if a smaller example existed, one with cardinality of the continuum instead

[lucid ocean]

etc. just take this but replace digits by elements in [0, 1]
and have aleph_1 digits

[median sand]

If F is a filter on X and A<=X is a subset that intersects every member of F, then the trace of F on A is the set of intersections of its members with A. This is kinda dumb, but why is this a filter (that it is a prefilter is clear)? If B<=A contains some A\cap C (for C\in F), why should it belong to the trace?

[cerulean oriole]

If G is a locally compact Hausdorff topological group and H is a compact subgroup, is the action of G on G/H (with the quotient topology) necessarily proper?

[cerulean oriole]

Let D = B \ (A ∩ C); then B = A ∩ (C ∪ D) and C ∪ D ∈ F.

[hidden abyss]

Does the existence of continuous bijections X->Y and Y->X imply that X and Y are homeomorphic

[quartz horizon]

i think no but idk what the counterexample is

[tiny obsidian]

It's probably something where you take two non homeomorphic spaces X and Y that have continuous injections into each other, disjoint union each with a sufficiently large discrete space, and then use points from the discrete spaces to plug in gaps to also get surjectivity without breaking continuity

[hidden abyss]

Hmm

Something like $X =$ countable discrete space $\sqcup \bigsqcup_{i\geq 0} X_i$ where $X_i$ is a finite indiscrete space with $i$ points

[gentle osprey]

Jussari

[hidden abyss]

And Y the same but omitting X_2

Then for X->Y we can send X_i to X_{i+1} and patch the last point from the discrete space

[tiny obsidian]

Yes ig

I was thinking like

An interval and two intervals

But this might be easier to actually write down the bijections for

[hidden abyss]

Oh lol

Thats conceptually a lot easier

[uneven bronze]

Basic question probably, but let X be locally compact Hausdorff. How do I show its one-point compactification is also Hausdorff? I've managed to show it is compact, yet this part I don't see immediately.

If we are given x and y and they do not equal, then either they both belong to X and can be separated, or say y = infty and x in X. How do I separate these two given that the open sets are either open in X or, if they contain infty, its complement in X is compact?

[tender halo]

the question is equivalent to whether there are two topologies on a set, one coarser than the other

to which the answer is obviously yes

uhh

three topologies

[hidden abyss]

x has a compact neighbourhood K in X. Use that to form the required neighbourhoods in the compactification

[tender halo]

X and Y and Z, such that X is coarser than Y and Y is coarser than Z but X and Z are homeomorphic

the answer is also yes

[hidden abyss]

Makes sense

[tender halo]

consider like

[rugged escarp]

you've to basically make cases

essentially you can take the open sets to be of two categories. One which are in the old topology, and the ones which contain the compactified points. The first set which comes out of the first topology, you know the rules of combining are still valid (schnitts and unions). Now, the cross term of old open + new open, and new opens are what you've to check

the properties of the new opens you can do with the usage of properties of compact sets. Eg closed subset of compact is compact (I think you need hausdorf for this property) and intersect of two compacts is compact

[tender halo]

a topology on N given by {x} with x < a, a fixed, then add {a, a + 1} as an open set, then add add {a} and {a+1} as open sets

[rugged escarp]

the real problem is proving the required properties of the old opens x new opens

[tender halo]

you don't need hausdorfness to prove that a closed subset of a compact space is compact

also to prove that the one-point compactification is hausdorff you just do it by hand really

[rugged escarp]

oh oops

I completly misread

I thought you meant show it's a topology first

[unreal stratus]

One point quasicompactification

[tender halo]

x has a nbhd whose closure is compact, its complement is a nbhd of y

(if y is the infinity point)

done

[uneven bronze]

yes 👍

[uneven bronze]

Theorem. If X, X* and T are LCH, the one-point compactification and the topology on X* respectively, then (X*,T) is compact Hausdorff and the inclusion map i: X -> X* is an embedding.

What exactly do I have to show in order to prove i: X -> X* is an embedding?

[rancid umbra]

show that it’s continuous and is a homeomorphism onto its image

[uneven bronze]

but isn't this always the case?

[rancid umbra]

wdym

[uneven bronze]

I mean that the inclusion is a homeomorphism onto its image. Always.

[rancid umbra]

yes

well

the inclusion of X with the subspace topology from X* is an embedding

but i guess the point of the exercise is to say that X with its original topology and the subspace topology from X* are the same

[uneven bronze]

yeah

[uneven bronze]

Ok, here's my attempt. If U is open in X, and i(X) is given the subspace topology, then i(U)=U =U\cap i(X), and U is open in T. If V is open in i(X) with the subspace topology from X*, then i^{-1}(V)=...?

[rancid umbra]

i believe it’s just V again

[uneven bronze]

yeah me too kind of, though I'm hesitant

[rancid umbra]

V doesn’t contain the point at oo so it’s a subset of X

[uneven bronze]

how do you know it doesn't contain oo?

[rancid umbra]

cus V is open in i(X) and i(X) doesn’t contain oo. so V = i(X) \cap V’ for some V’ open in X*

[uneven bronze]

i(X) = X, right?

[rancid umbra]

mhm

[uneven bronze]

ok 😅

[rancid umbra]

all good

[uneven bronze]

Consider again \

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

In particular, the last sentence. I feel like this extension of $f$ should follow from some previous result I've brought up in chat, though I'm not sure which one.

[gentle osprey]

psie

[tender halo]

whats your definition of a one point compactification?

because like

[uneven bronze]

$$\mathcal{T} = {U\subseteq X^\ast : U\in\mathcal{T}_X \lor (\infty\in U \land X\setminus U\text{ compact and closed})}$$

[gentle osprey]

psie

[uneven bronze]

If X is LCH (which it is!), we can drop the closed condition.

[tender halo]

well uhh if you restrict that to X then it will be exactly the topology on X

its not so much a theorem as it is just looking at the definition really hard

[uneven bronze]

hmm, but I don't understand why there's a continuous g in C_0(X) (which by the way are the functions vanishing at infinity)

I understand that the subspace topology on X from X* is just the original one on X.

[tender halo]

ah ok sorry

well i mean its the same thing really

vanishing at infinity means its infinitely small outside of compact subsets

which is converging to zero at \inf

its just rewriting the definition of the point at infinity a little

and what it means to take a limit at that point

[uneven bronze]

\begin{framed}
... if $f\in C(X)$, then $f$ extends continuously to $X^\ast$ iff $f=g+c$ where $g\in C_0(X)$ and $c$ is a constant, in which case the continuous extension is given by $f(\infty)=c$.
\end{framed}
Given that $f\in C(X)$ extends continuously to $X^\ast$, let $F:X^\ast\to\mathbb{R}$ be such that $g:=F|X=f$ and $c:=F(\infty)$, where $c$ is some constant. Why would such an extension exist? I'm not really sure what it is I need to show in this direction.\

Conversely, if $f=g+c$ for $g\in C_0(X)$ and $c$ a constant such that $f(\infty)=c$, then I think I need to show $f$ is continuous. We have $f^{-1}(U)=g^{-1}(U)$ if $c\notin U$, which is open, and $f^{-1}(U)=\ldots$ something something (but I'm not sure what exactly?).

[gentle osprey]

psie

[brazen sierra]

does this work? the exercise is finding a counterexample to the theorem (d) below when U isnt saturated

[ruby delta]

I don't think this works

[0,0.5) u (0.5,1) is actually homeomorphic to (0.5, 1.5)

because the topology of R/Z isn't the same as the unit interval, the two ends are connected

and quotient maps do not have to preserve open and closedness anwyays

[brazen sierra]

ah got it thanks

[rancid umbra]

try restricting to (0,oo)

[brazen sierra]

thanks! ill try that

[ruby delta]

does that work though?

[rancid umbra]

im pretty sure?

i could be misremembering

[ruby delta]

oh, I think it does (?)

[rancid umbra]

let me check again

[ruby delta]

proving that it's not a quotient map seems kind of like a pain though

[rancid umbra]

im not sure if it works or not, but i misremembered that the restriction of this map to (0,oo) is not a covering map

not that its not a quotient map

[tender halo]

I -> S1 restricted to [0; 1) is not quotient

[rancid umbra]

i was just working through this one

[uneven bronze]

I think I understand the $\implies$ implication. Here's a renewed attempt at $\impliedby$. \

If $f=g+c$ for $g\in C_0(X)$ and $c$ a constant such that $f(\infty)=c$, then we need to show $f$ is continuous. It suffices to check the preimages of open intervals of the form $(a,b)\subset\mathbb{R}$, since these form a base for the topology on $\mathbb{R}$. We have $f^{-1}((a,b))=g^{-1}((a,b))$ if $c\notin (a,b)$, which is open. If $c\in(a,b)$...this is where I'm stuck. Somehow I need to use $g\in C_0(X)$, though I'm not sure how.

[gentle osprey]

psie

[tardy carbon]

given $f \in C(X)$ with $f = g + c$ with $g \in C_0(X)$, your are supposed to show that $f$ extends continuously to $X^*$

[gentle osprey]

L

[uneven bronze]

Right. And above, I was treating cases to show that f actually is continuous. Makes sense, right?

[uneven bronze]

When is $C(X)$ a closed subspace of $\mathbb{C}^X$, where $X$ is some topological space? Apparently this occurs when $\mathbb{C}^X$ has the topology of uniform convergence, though shouldn't we require $X$ to be compact to speak of this topology since the uniform metric is only defined for bounded functions?

[gentle osprey]

psie

[tender halo]

uniform metric is defined for all functions

the norm isnt

[uneven bronze]

Are you sure? The Wikipedia article only defines it for bounded functions.

[tender halo]

hmm maybe im wrong

[soft wharf]

whats all this guys

im still in grade 12

[tender halo]

i dont think i am though

you just set the distance between functions to 1 if the difference is unbounded

[lucid ocean]

that should work

[uneven bronze]

but is that the uniform metric then? You have modified the metric.

[tender halo]

it is the uniform metric in the sense that it induces the topology of uniform convergence

[uneven bronze]

oh ok 👍

[tender halo]

a different approach is to say that all metrics involved are bounded

which leads to the same

[cerulean oriole]

The uniform metric requires the bound but the uniform topology makes sense regardless (you can think of it as a metric that can take the value ∞ if you want).

[cerulean oriole]

Different kinds of convergence

[cerulean oriole]

Bump.

[uneven bronze]

Lemma. Let $X$ be LCH and $E\subset X$. Then $E$ is closed iff $E\cap K$ is closed for every compact $K\subset X$.\

Consider $\impliedby$. If $E$ is not closed, pick $x\in \overline{E}\setminus E$ and let $K$ be a compact neighborhood of $x$. Then $x$ is an accumulation point of $E\cap K$ but is not in $E\cap K$. So $E\cap K$ is not closed.\

I don't understand why $x$ is an accumulation point of $E\cap K$. If $x\in \overline{E}$, then every open set $U$ containing $x$ intersects $E$ nontrivially. But why are we guaranteed that $(E\cap K)\cap U \setminus{x}\neq\varnothing$?

[gentle osprey]

psie

[uneven bronze]

You see, all we know is that K and U contain x, but at the same time, this is the very point we are removing.

[rancid umbra]

an open set U containing x will intersect E at some other point x' in E because x isn't in E.

[uneven bronze]

ah yes, good point 🙂

[uneven bronze]

what about K and U though? could they not intersect E at different points such that their intersection is empty?

sounds impossible, though I'm not sure how to actually show they can't

[rancid umbra]

K is a compact neighborhood of x. is your definition that there is an open neighborhood V of x contained in K?

[uneven bronze]

Yes, indeed.

[rancid umbra]

okay, so take your open set U around x. put U' = U ∩ V. then U' is an open neighborhood of x contained in K, so U' - {x} meets E ∩ K at some other point x', as x is not in E

then U - {x} meets E ∩ K at x' because U' - {x} is a subset of U - {x}

[uneven bronze]

ah yes, makes sense, thank you

[uneven bronze]

Consider the topology of uniform convergence on compact sets, i.e. the topology generated by sets of the form$$\left{g\in \mathbb{C}^X:\sup_{x\in K}|g(x)-f(x)|<n^{-1}\right}\quad(n\in\mathbb{N},f\in \mathbb{C}^X,K\subset X\text{ compact}).$$Now I'm reading a proof of $C(X)$ being closed in $\mathbb{C}^X$ under this topology. The first sentence reads\

If $f$ is in the closure of $C(X)$, then $f$ is a uniform limit of continuous functions on each compact set $K\subset X$.\

Is this topology induced by some metric? Why would $f\in\overline{C(X)}$ be the uniform limit of continuous functions on compact sets?

[gentle osprey]

psie

[tardy carbon]

Is it true that a net $f_{\alpha}$ converges to $f$ in the topology of uniform convergence on compact sets if and only if $f_{\alpha} \to f$ uniformly on every compact set $K \subset X$?

[gentle osprey]

L

[uneven bronze]

Well, it's certainly true that $f\in \overline{E}$ iff there is a net ${f_\alpha}\subset E$ that converges to $f$ (we can't say if it's a sequence though).

[gentle osprey]

psie

[tardy carbon]

that's fine. All we need to show is that if $f_{\alpha} \in C(X)$ and $f_{\alpha} \to f$, then $f \in C(X)$.

[gentle osprey]

L

[uneven bronze]

Still struggling with this claim. I'm trying to work from definitions. If f is in the closure of C(X), every open U containing f intersects C(X). I'm not really sure how to proceed here to show f is the uniform limit of continuous functions on each compact set K subset X. The subbasic open sets of C^X with the compact-open topology are specified to the message I'm replying to.

[uneven bronze]

I would prove it like this. Let me know if you think otherwise or have any suggestions for improvements. \

Let $B_K(n^{-1},f)$ denote the subbasic open sets I specified above. If we fix some compact set $K$, and $f\in\overline{C(X)}$, then $B_K(n^{-1},f)$ is open and contains $f$, so it intersects with $C(X)$. So for each $n$, there are $f_n\in C(X)\cap B_K(n^{-1},f)$. Then $f_n\to f$ uniformly on $K$.

[gentle osprey]

psie

[tardy carbon]

I think this works. My proof was that since $f \in \overline{C(X)}$, there is a net $f_{\alpha}$ in $C(X)$ with $f_{\alpha} \to f$. Since the topology is generated by seminorms $\rho_K$, we get $f_{\alpha} \to f$ uniformly on every compact $K$.

[gentle osprey]

L

[tardy carbon]

This ties into your question about the metric. The topology isn't defined by a metric, rather it's defined by the seminorms $\rho_{K}(f) = \sup_{x \in K}|f(x)|$. If countably many $K$ suffice (this depends on $X$), then it is metrizable.

[gentle osprey]

L

[tardy carbon]

e.g. when $X$ is an open subset of $\mathbb{R}^n$

[gentle osprey]

L

[uneven bronze]

Ok, thanks. 👍

[rain gyro]

Can I make this proof tighter?

[tender halo]

tighter?

[rain gyro]

Like if there’s a gap

[tender halo]

its just juggling the definitions, there is no technique here to be made tighter

[ruby delta]

looks good

[rain gyro]

Thanks then 🥰

[unreal stratus]

Only thing is I would not say "said to be open"

[rain gyro]

I will just write U is open if and only if U^c is closed is that better?

Thanks 🥰

[uneven bronze]

I thought I had this worked out, but I realized my solution failed. Let $X$ be LCH. Consider $f\in C(X)$. Why is every extension of $f$ to the one-point compactification continuous? \

Attempt; I think there are two cases to consider. Points in $X$ and $\infty$. Someone told me; every extension of $f\in C(X)$ to $X^\ast$ is continuous at all points of $X$ since these have a neighborhood (e.g. $X$) not containing $\infty$. Why does this show continuity for $y\in X$? We need to show that $f^{-1}(V)$ is a neighborhood of $y$ for every neighborhood $V$ of $f(y)$. Moreover, I'm totally stuck on showing $f$ is continuous at $\infty$. Any help is very appreciated.

[gentle osprey]

psie

[ruby delta]

you can equivalently show that the preimage of closed sets is closed

and since every point other than infinity has a compact (ie closed) neighbourhood, it necessarily cannot contain infinity

so its preimage under f* is just that under f

[uneven bronze]

hmm, I remain doubtful. Is every closed set a compact neighborhood? This is what I gather from your arguments. Yes, it is true that every point has a compact neighborhood, and as a subset of the X*, which is compact, it is closed.

[uneven bronze]

you see, we need to show that the preimage of every closed set is closed.

[ruby delta]

hmm, good point, I think it can be fixed by instead looking at the interior of the compact neighbourhood

[steady gorge]

i'm a little unsure of where to start with exercise 4. what should i try to show first?

[ruby delta]

an embedding is a continuous injective map

so, is there any 1 of the 2 that you're having trouble with?

[steady gorge]

i'd say the 1st one. am i needing to show that both are continuous injective maps?

i think i was confused on if i had to do something with both

[ruby delta]

well by symmetry you only need to show one of them is

[steady gorge]

ah ok

so then i'll try it with f(x). and then symmetry will imply that g(y) is also an embedding?

[hidden abyss]

Must also be a homeomorphism onto its image

[rancid umbra]

an embedding is a homeomorphism onto its image.

there are continuous injections that are not homeomorphisms onto their image. for example, consider a dense curve in the torus. it is the image of a continuous injection R —> S^1 x S^1 which is not an open map.

if your continuous injection is proper or closed or open, or the domain is compact, then they are the same *(see below, this is slightly incorrect)*

or even simpler, [0,1) —> S^1 is a continuous bijection whose inverse is not continuous

[hidden abyss]

Just compact domain isnt sufficient either, since you can send a finite discrete set into an indiscrete one

You'll need to assume hausdorff codomain too

[rancid umbra]

ah, yea.

but all spaces are hausdorff

[steady gorge]

so i gotta make sure to add that to the things to prove about f(x)

[rancid umbra]

i think proper also needs both spaces to be hausdorff lol

one more reason to forget about non-hausdorff spaces

[hidden abyss]

Yeah

[steady gorge]

so for injectivity, i know that we gotta show that if $f(x_1)=f(x_2)$, then $x_1=x_2$ but how would i go about it for this problem? am i relying on some properties of a product topology here?

[gentle osprey]

ushygushytoes

[hidden abyss]

Injectivity is purely set-theoretic

You'll just need to apply the definition of f and of the cartesian product as sets

[steady gorge]

ah ok so no topology here yet

[uneven bronze]

The whole statement reads

Statement: If f in C(X), then f extends continuously to X* iff f = g+c where g vanishes at infinity on X and c is a constant, in which case the continuous extension is given by f(oo)=c.

I can prove the forward direction, i.e. showing g vanishes at infinity. However, showing that f = g + c is continuous eludes me.

[hidden abyss]

How you should interpret this is that the image of f (which turns out to be X×{y}) is the topological space X

when you consider it with the subspace topology of X×Y

[steady gorge]

so $(x\times x_1)=(x\times x_2)$ would only be true if $x=x$ and $x_1=x_2$ by the equality of ordered pairs? did i understand this step correctly?

[gentle osprey]

ushygushytoes

[uneven bronze]

I have a basic question. If I have a function that is continuous and I extend it by a single point, does it suffice to prove continuity of the extension by simply proving continuity at the extended point?

[warped shore]

Yes

Functions are continuous if continuous at all points

[unreal stratus]

One subtlety is that the extra point you add may be in neighbourhoods of one of the original points

So if the space is T1 this works but ye

[warped shore]

Wdym

[unreal stratus]

Consider the space {0,1} where the open sets are empty set, {0}, and {0,1}. Consider the map {1} -> R. sending 1 to 1. Obviously this is continuous. But then extend to send 0 to 0 The extension is obviously continuous at 0, since {0} is a neighbourhood of 0, but not continuous at 1 anymore

[uneven bronze]

So in an LCH space this can not happen? 🙂

[unreal stratus]

Well, you only need something like T1

Because then every other point has a neighbourhood that doesn't contain the point you added

[tender halo]

all spaces are either hausdorff or the zariski topology

[ruby delta]

"or the zariski topology"

[unreal stratus]

Lol

All spaces are either Hausdorff or spectral

😼

[warped shore]

Aha

Okay neat

[rain gyro]

Hiii i wonder is the argument actually redundant or loose?

Is it actually better if I use disk(x, [a,b]) to make it a bit tighter if it’s necessary

[unreal stratus]

The notation you have is better lol

[tender halo]

i am unsure about what you are proving tbh

the proposition reads like a definition

[rain gyro]

It’s on the book probably just an equivalent definition

I just started studying this

It’s fun

[ruby delta]

looks good

[unreal stratus]

Well as with the last one presumably you don't want the "said to be a"

But also like

[rain gyro]

I will change that part 😭😭 it was done yesterday

[unreal stratus]

The statement you are proving sounds like "Any interval which contains all limit points of itself is a closed interval", but what you prove is the converse of that

[rain gyro]

Is it better? I still find the logic a bit unnatural (just started studying) so for proposition 5 I was proving converse? Is it because of proof by contradiction or it’s just my strategy entirely wrong

[tender halo]

the wording implies that you are proving that I is a closed interval, which is not what the proof actually about, you are proving that I is a closed set

closed interval means interval of the form [a; b] as opposed to the open interval (a; b)

[ruby delta]

you're proving that closed intervals are closed

[rain gyro]

I noted that.. how should I fix it? Like to prove that it contains all limits points, if I want to prove by contradiction then should I still assume that limit point x is not I?

I still got a bit confused here

[brazen sierra]

what is I here? the closed unit interval?

so basically i need to find some V in S1 whose preimage is open in either the subspace or quotient topology and not open in the other?

[brazen sierra]

the problem in question for reference

[ruby delta]

well, it is still a continuous map

just no longer a quotient map, which is more subtle

[brazen sierra]

oh sorry it should be the other way around

q^{-1}(V) open but V isnt open

[tender halo]

[0; a) is open, its image is not

since its an injection, being a quotient map is equivalent to being a homeomorphism

[brazen sierra]

what is the topology on I here

[unreal stratus]

The usual one

[brazen sierra]

shouldnt the image be open in S1 by definition of the quotient topology? like V subset S1 is open iff its preimage is open

[ruby delta]

yes

and that's a contradiction

[brazen sierra]

so which topology is it not open under

sorry im just having a weird mental block around this idk why

[ruby delta]

its image is not open in S1

[unreal stratus]

Actually lol it was a question in an exam I took to determine for which a the map [0,a) -> S^1 is a quotient map

But I guess it is obviously not for a < 1 and then by periodicity you reduce to studying a = 1

[brazen sierra]

as a subspace of R^2?

got it, thanks

[grim knot]

Does someone know how many sheets the universal covers have?

I think it's infinity, but is it contabile?

[red yoke]

As many as there are elements of π1 of the base space

[unreal stratus]

For the same reason as what Arki said, it isn't always infinite either

Indeed the universal cover of a point is just itself

[grim knot]

What if we consider the one if S1 v S1

[rancid umbra]

𝛑_1(S^1 v S^1) = Z * Z is countable since Z * Z is finitely generated

[unreal stratus]

ye and for the same reason pi_1 of any finite cw complex will satisfy this (so you'll need a more "hairy" space)

[rain gyro]

Guys are these identities proven correctly? I tried to let AI to verify it but I got too many macros and it doesn’t see picture very nicely. And I kinda need an organized note for myself 😭

It’s by equation (2) for the reference of last claim

[rancid umbra]

it looks good to me. maybe justifying/citing why int(A) his equal to the complement of the closure of the complement, same with ext(A), in (ii) could improve it.
referring to the fact that cl(A) = bd(A) U int(A) may speed things up too, since ext(A) = X - cl(A), so cl(A) U ext(A) = X, and you can replace cl(A) with bd(A) U int(A)

[uneven bronze]

Just want to check a doubt of mine.\

Proposition. If $X$ is LCH, $U\subset X$ is open, and $x\in U$, there is a compact neighborhood $N$ of $x$ such that $N\subset U$.\

This proposition is saying that an open subset of a locally compact Hausdorff space is locally compact (perhaps even LCH), or?

[gentle osprey]

psie

[quick crane]

Yes it says $U$ is LCH

[gentle osprey]

L

[uneven bronze]

👍

[tender halo]

a subset of a locally compact space is locally compact iff its an intersection of an open subspace and a closed subspace

as every LCH is Tychonoff and therefore embeddable in a compact hausdorff space, a space is locally compact iff it can be represented as an open subspace of a compact space

[uneven bronze]

Prop. 1. If $X$ is a $\sigma$-compact LCH space, there is a sequence ${U_n}$ of precompact open sets such that $\overline{U}n\subset U{n+1}$ for all $n$ and $X=\bigcup_1^\infty U_n$.\

Prop. 2. If $X$ is a $\sigma$-compact LCH space and ${U_n}$ is as in Proposition 1, then for each $f\in \mathbb{C}^X$ the sets $${g\in \mathbb{C}^X:\sup_{x\in\overline{U}_n}|g(x)-f(x)|<m^{-1}}\quad(m,n\in\mathbb{N})$$form a neighborhood base for $f$ in the topology of uniform convergence on compact sets. Moreover, $f_j\to f$ uniformly on compact sets iff $f_j\to f$ uniformly on each $\overline{U}_n$. \

I want to prove Prop. 2. We need to show that (i) $f$ is in all the sets (obvious) and (ii) if $U$ is open, then there exist a neighborhood basic set $V$ such that $V\subset U$ (how do I do this?). Moreover, I'm not sure how to approach the statement about the uniform convergence. Help is appreciated.\

Hint: If $K\subset X$ is compact, then ${U_n}_1^\infty$ is an open cover of $K$ and hence $K\subset\overline{U}_n$ for some $n$. (I get the statement in this hint, but not how to use it.)

[gentle osprey]

psie

[gentle osprey]

psie

[uneven bronze]

Does anyone understand how to use the hint?

[uneven bronze]

Unrelated but somehow related to what I asked above. If $x\in U$ for some open $U$ and I know the subbasic elements of the topology, does it imply that $x\in V\subset U$ for some subbasic element $V$? I know this holds for base elements, but I'm tempted to say it holds for subbase elements too. Am I correct?

[gentle osprey]

psie

[uneven bronze]

Well, perhaps the result about bases does not copy over to subbases. But we should be able to say that if $x\in U$, then $x\in\bigcap_{\alpha\in A}V_\alpha\subset U$ where $A$ is finite and $V_\alpha$ is a subbase element. I guess we can't say that $V_\alpha$ is a subset of $U$, or?

[gentle osprey]

psie

[unreal stratus]

It's not correct because of the condition that V is contained in U

Like for example: consider the topology on R generated by the subbasis (0,2), (1,4). Then the set (2, 4) is open and contains 3, but (2,4) doesn't contain any subbasis sets

[uneven bronze]

Ah yes, makes sense.

[unreal stratus]

Technically it's also false because subbases are often/usually defined in a way such that that the subbasis doesn't have to cover the whole space. Then you can pick a point that's not in any subbasis element

Like you could consider the topology generated by the empty subbase or something silly like that

Lol

[uneven bronze]

ah yes 😄

[unreal stratus]

Good question though, not something I've thought about really

[gentle osprey]

esca

[unreal stratus]

The topologise bit is not really part of the question

It's just like part of how they are defining CP^n as a topological space

If you want you could briefly mention why that quotient can be identified with lines in C^(n+1), but there's nothing else to do - the main thing is the next part

[brazen sierra]

okay got it, thank you!

[uneven bronze]

Consider the topology of uniform convergence on compact sets on $\mathbb{C}^X$, generated by sets of the form $$\left{g\in \mathbb{C}^X:\sup_{x\in K}|g(x)-f(x)|<n^{-1}\right}\quad(n\in\mathbb{N},f\in \mathbb{C}^X,K\subset X\text{ compact}).$$Let's denote these sets by $B_K(f,n^{-1})$. Are these sets, for fixed $f$, a neighborhood base for $f$?\

What we need to check is that each $B_K(f,n^{-1})$ contains $f$ (obvious) and that for open $U\ni f$ in the topology, there exists a compact $K$ and an $m$ such that $B_K(f,n^{-1})\subset U$ (how do I do that?).

[gentle osprey]

psie

[uneven bronze]

I am sort of considering locally compact Hausdorff spaces X, don’t know if that helps.

[uneven bronze]

Any help would be very appreciated. 😔

By the way, U is open in C^X, whereas K is compact in X.

[uneven bronze]

Still working on this one. 😔

[rancid umbra]

so, this collection def forms a neighborhood subbasis of x

if you can show they form a basis, you would be done

but i don’t have much more insight

[uneven bronze]

Yes, that's the claim that I want to prove.

[rancid umbra]

you are trying to prove something stronger

[uneven bronze]

Well, the sets B_K(f,m^{-1}) form a subbase, where f in C^X, K is compact and m is some natural number, for the compact-open topology of C^X. What I'm trying to prove is that, for fixed f, B_K(f,m^{-1}) forms a local base (or a neighborhood base) for f. Are you familiar with this concept? A neighborhood base at f for a topology is a subset of the topology such that each set in this subset contains f (this is obvious in regards to the sets B_K(f,m^{-1}) ) and that if U is open in the compact-open topology containing f, there exists K and m such that B_K(f,m^{-1}) is a subset of U.

[rancid umbra]

yes, i am familiar. but i am saying that it forms a neighborhood subbasis for f, that is, the collection of finite intersections of subbasic sets forms a neighborhood base of f

this is weaker than a neighborhood basis for f

[uneven bronze]

Ok. Someone gave me a hint that I should consider "taking the union of all K and the minimum of all m", but I didn't get this hint. Taking the union of what exactly? I fear some arbitrary union of compact sets won't be compact, but perhaps that's not what they meant.

[rancid umbra]

well, every open neighborhood U of f is a union of finite intersections of the B(K,f,n)’s

[rancid umbra]

further restated by this

so f is in the intersection of finitely many B(K,f,n)’s

maybe they meant take the union of those finitely many K’s

[uneven bronze]

ah yes, perhaps that's what they meant 👍

[rancid umbra]

hope that works lol

[uneven bronze]

So if I understood you correctly, if we consider B_K(f,m^{-1}) for fixed f, then you are saying here that the collection of finite intersections of these sets forms a neighborhood base of f. How do you...like know this? 🙂

[rancid umbra]

this

[uneven bronze]

hmm, ok, but you were saying yourself that was just a restatement of your claim

[rancid umbra]

hmm. i hate when i do this, but i may have overlooked something…

[uneven bronze]

no worries 🙂 I'm not sure my original question is even true

[rancid umbra]

sorry about that. will keep thinking on it

[uneven bronze]

We know that $B_K(g,m^{-1})$ are subbasic sets for the topology, for $K\subset X$ compact, $g\in \mathbb{C}^X$ and $m\in\mathbb{N}$. This means if $f\in U$, then we know from subbase definition that $$f\in \bigcap_{(K,m,g)\in \mathcal{A}}B_K(g,m^{-1})\subset U\quad(\mathcal{A}\text{ a finite index set}).$$If we only could switch $g$ for $f$ in this intersection, then taking the union of those finite $K$ (call it $C$) and the minimum of all the $m$'s (call it $n$), then indeed I think $$f\in B_C(f,n^{-1})\subset\bigcap_{(K,m,g)\in \mathcal{A}}B_K(g,m^{-1})\subset U,$$right?

[gentle osprey]

psie

[rancid umbra]

yea, this is the issue i overlooked

[uneven bronze]

Ok. Do you see a way to switch all the g's in A for fixed f? Currently I don't see a way. I see perhaps the possibility in doing this by also adjusting the K's and m's too. But that complicates things I think.

Perhaps triangle inequality has a say in this. 😄

[rain gyro]

Is the proofs for these basic identities sufficiently rigorous? I took advice from the server yesterday and proved them again with identities I don’t know if the proof is a bit more refined now

[ruby delta]

A° is the largest open set contained in A

Just a semantic thing

[rancid umbra]

wdym?

[plush folio]

Writing "pairwise disjoint" when there are only two sets doesn't make much sense, you can just write disjoint. We usually write pairwise disjoint when there is a collection of sets, and each pair of sets are disjoint

[brazen sierra]

does this idea work?

[heady skiff]

Sorry what do you mean by this? Like we find some map from \R^2 \ {0} to S^1 and show that its kernel is equal to the equivalence relation which defines RP^1?

[unreal stratus]

How are you defining RP^1?

\R^2 \ {0} modulo scaling?

[heady skiff]

Yeah

[unreal stratus]

Ah okay well this is an important point for me to make then i guess lol like

so more generally like you can define $\mathbf R P^n = (\mathbf R^{n+1} \setminus {0})/\mathbf R^\times$, and there's a canonical quotient map $\mathbf R^{n+1} \setminus {0} \to \mathbf R P^n$ which you can think of as sending a point to the line through it. But we can restrict this to the unit sphere to get a map $f: S^n \to \mathbf R P^n$. You can check that $f(x) = f(y)$ if and only if $x = \pm y$ and that $f$ induces a homeomorphism $S^n/(x \sim - x) \to \mathbf RP^n$

[gentle osprey]

Prismatic Potato

[unreal stratus]

(Basically just cause every line has 2 points on the unit sphere)

Then for S^1, note that squaring as a map S^1 -> S^1 induces a map S^1/(x ~ -x) -> S^1 which is a homeomorphism

[heady skiff]

Ah I see, that makes more sense then

Thanks I'll try to work out the details

Wait so would the canonical quotient map just be the projection

[unreal stratus]

ye

[heady skiff]

By squaring do you mean $(x_1, x_2) \mapsto \frac{(x_1^2, x_2^2)}{\norm{(x_1^2, x_2^2)}}$ lol

[gentle osprey]

okeyokay

[plush folio]

Hmm, I think it would be complex squaring, because that function always outputs non-negative x_1, x_2

[unreal stratus]

Yes

[brazen sierra]

does this work?

[pseudo marsh]

In the definition of a covering map if you allow the preimages to be not necessarily disjoint unions of open sets that map as homeomorphisms, does that give you more maps than just covering maps?

[unreal stratus]

What precisely do you mean?

[opaque scroll]

You'd get fiber bundles as well at least

No wait, you still have open

[pseudo marsh]

Like I’m trying to convince myself that including disjoint is necessary in the definition of covering map

[quartz horizon]

well ig a covering map is just a locally trivial bundle of discrete spaces

[opaque scroll]

Take the line with two origins for example

[pseudo marsh]

True

What if we want both spaces to be Hausdorff

[opaque scroll]

How about this:

Consider RxN with an extra point x. Thinking of the open sets of RxN as indexed open sets {Ui}, and open set is either an open set of RxN that doesn't contain x or an open set such that
lim_i -> infinity Ui is a neighborhood of 0 together with x.

This should be like Hausdorff version of the line with two origins.

Nvm, it doesn't quite have the desired property. But maybe you can modify it...

[pseudo marsh]

So you’re thinking of that being a covering space for RxN without x right

Or “covering”

So is there some nice conditions you could put on the spaces so that you could prove the definitions are equivalent?

Like maybe manifolds for example

[hidden abyss]

Cant the preimage have any topology then

Or what do you want it to be isomorphic to

[pseudo marsh]

but it still needs to be a homeomorphism on each of those open sets

[hidden abyss]

I dont think I understand

What should p^{-1}(U) be homeomorphic to

[opaque scroll]

It should be a union of open sets each of which are homeomorphic to U (by p)

[unreal stratus]

Tbh idk what you are left with otherwise lol

Ah okay some non-disjoint union

[hidden abyss]

Something like two copies of R glued along [0,1]?

[unreal stratus]

Good point

[hidden abyss]

Would also be interesting to force the union to be set-theoretically disjoint but not necessarily have the topology of the disjoint union

[pseudo marsh]

Nice, that's a good example

and that's a manifold too right?

[hidden abyss]

No its not Euclidean at the points 0 and 1

Since theres three directions to go

-<

[pseudo marsh]

but at 0 there's a neighborhood only on one copy of R right?

[unreal stratus]

If you draw it like

[hidden abyss]

You can go from 0 either to the right along the glued part or left onto either copy of R

[pseudo marsh]

nvm that doesn't work

[unreal stratus]

It looks like

I guess

And kinda evidently not a 1-manifold cause of those two problem points as said

I guess a fun way to make this rigorous is that R minus a point has two connected components, but every (small enough) punctured neighbourhood of those two problem points has 3 connected components

[uneven bronze]

What does "locally compact in the relative topology" mean for a subset E of a locally compact Hausdorff space?

I thought that compact sets are compact not relative to any subset.

[rancid umbra]

im assuming it means that if a subset A of E is LC in the relative topology on E, then for every x in A there is an open subset U of E and a compact subset K contained in E such that x in U subset K

[uneven bronze]

Ok, this leads to my second question kind of 🙂 I know that if E is open in X, then x in E has a compact neighborhood N_x contained in E. What more do I have to show in order for E to be locally compact in the relative topology? Locally compact according to my definition means every point in the space has a compact neighborhood.

[rancid umbra]

i think you need K subset E

[uneven bronze]

Perhaps I should show that N_x is actually a (compact) neighborhood in E, i.e. contains an open set (relative to E) of x.

[rancid umbra]

i thought N_x is in E already. are you saying you haven't shown that?

also, this feels silly

locally locally compact

[uneven bronze]

sorry, I meant only that N_x is a subset of E. I don't know if its a neighborhood in E, but perhaps that's trivial? According to the proof of the statement "E open in X, where X is locally compact Hausdorff space, there exists a compact neighborhood N_x subset E of x in E", one chooses N_x to be a closed (relative to X) subset of X. I'm a bit confused.

[unreal stratus]

You're done

[uneven bronze]

I'm currently working a problem where I need to use that $$K=\overline{K^\circ}.$$How do I prove this equality? One inclusion always holds I believe, namely since $$K^\circ\subset K\implies K\supset\overline{K^\circ}.$$What about the other inclusion? More background: $X$ is LCH and $U\subset X$ is open. Then for any $x\in U$, there exists a compact neighborhood $K$ of $x$. So $K$ is closed.

[gentle osprey]

psie

[harsh moat]

isnt [0,1) = K. a counterexample for even the first implication

[uneven bronze]

Yes, I should have added that K is always closed.

[tender halo]

what do you mean prove the equality? it says that K is a regular closed set

[harsh moat]

i think you are sayng that K = the closed set of int(K)

[tender halo]

not all set are regular closed

[harsh moat]

busty beaver

☺️

[uneven bronze]

Right. Is that equality true?

[tender halo]

in general no

[uneven bronze]

when is it true?

[tender halo]

when K is a closure of some open set

[harsh moat]

not in general i think, because as bussy beaver said: not all sets are regular closed

[tender halo]

all closed sets are regular closed (in a T1 space) iff space is discrete

[uneven bronze]

So if K=cl(U), where U is open. This leads me to the question. Is U then the interior of K?

[uneven bronze]

Since you said "some open set", this made me a bit uncertain whether or not U is actually the interior of K.

[uneven bronze]

I don't think this is true. We have that K = [0,2] is the closure of the open set U = (0,1) U (1,2). But U is not the interior of [0,2] since (0,2) != (0,1) U (1,2).

I get it now I think.

[tender halo]

it need not be

U will be a subset of the interior

[rain gyro]

Is this argument considered correct though it’s not standard proof using ivt

I kinda felt it could be doable but not sure.

ChatGPT says it’s wrong though..

No but summer breaks I can’t really rely on anyone 😭

[tiny obsidian]

What is z

[rain gyro]

Boundary

Between the image of f over interior of A and interior of complement of A

[tiny obsidian]

Is your A of a specific form

[rain gyro]

A set of the real line

[tiny obsidian]

The boundary of the interval (0,1) is {0,1}? Not a single point?

[radiant stone]

dont rely on chatgpt, it's wrong in a lot of sides

it just says nonsense thingy

[rain gyro]

No I should have made it into a collection

[hidden abyss]

Also the fact that for any y1, y2 you have either y1>y2 or y2>y1 doesnt say anything about monotonicity

[rain gyro]

Would it fix the argument if I say since f is continuous for every f^(y_1) and f^(-1)(y_2)

We have either y_1>y_2 or y_2>y_1

[tender halo]

i dont understand where the decomposition is coming from at all

[rain gyro]

I think there must be slips I didn’t realize 😭

[tender halo]

int(f(A^c)) wont even intersect f(A) in certain cases, much less be contained inside

[rain gyro]

You’re right I should just stick to ivt for it😭😭

I thought I could’ve phrased anything in pointset topology

[alpine nest]

That would be very difficult because monotonicity isn't really a topological notion Retracted, see below

[tender halo]

you can like do stuff with it

[rain gyro]

Yes I realized when jussari mentioned that having inequality doesn’t give monotonicity…

[ruby delta]

no, there's a way to do this without IVT, I see a proof idea using epsilon delta

[hidden abyss]

The fundamental problem is that the statement "y1>y2 or y2>y1" is just equivalent to y1≠y2, which is just a rephrasing of injectivity

[rain gyro]

That’s exactly I felt it can’t be right anymore… though it’s a fun attempt compared to other silly mistakes of me 😭

[tiny obsidian]

It has to use completeness (i.e. more than metric / order properties): there are continuous injective functions Q to R that aren't monotonic

[ruby delta]

sure, I just say that there's a way to avoid IVT

of course we will be using the properties of C(R,R)

[alpine nest]

It could probably use connectedness

[tender halo]

a map is said to be monotone if a preimage of a connected set is connected

[hidden abyss]

Is there an easy proof of invariance of domain in just R

[tender halo]

which results in the same definition for the functions from R to R

[alpine nest]

Oh, neat

[tender halo]

for a closed mapping it is enough that all fibers are connected

[alpine nest]

I retract my earlier remark, then

Retract: another topological notion

[tiny obsidian]

😩 this doesn't agree with the order on Q

[alpine nest]

Every injective map from Q to Q is monotone!

And no map from Q to R is!

[tender halo]

makes sense yea

[alpine nest]

For a given value of "sense"

[rain gyro]

This is great theorem…

[tender halo]

its easier to just use ivt for you

[alpine nest]

It's not a theorem, it's a definition (but the fact that it's equivalent to the standard definition of monotonicity for functions from R to R, is a theorem)

[tender halo]

you find a peak or a valley (values that go ⬆️⬇️ or ⬇️⬆️) and that gives you a contradiction with injectivity

the end

[tender halo]

the order on R is "more topological" than the one on Q

Q you can shake around and it stays the same shape, topologically

[tiny obsidian]

I am aware, but I would still have liked "monotone" to agree with the 'monotone' of a (insert relevant type) ordered set

(though I'm sure it is useful to not enforce that)

[tender halo]

meh topology and order dont play well with each other, they dont even agree on something as basic as topology on a subspace

i like this definition because its more natural than the one that says "either monotonically rising or falling"

[granite crane]

what is the intuition behind (ii)?

i cannot see any reasonable intuition

[tender halo]

it is there to guarantee that intersection of basic sets is open

namely, that there is another basic set covering a given point that lies inside the intersection

[granite crane]

oh

this

[tender halo]

kinda yeah

[granite crane]

oh ok, i got it i guess.

thank you beaver

[uneven bronze]

When you have to show two metrics define the same topology on some space, what is it you have to show?

[untold lily]

one way is: given any open ball around a pt x in one metric, try to find an open ball around x in the other metric that is contained in it and vice versa

I mean, in a general sense, you want to show open set in one metric is open set in another metric

[uneven bronze]

Ah ok, makes sense.

[uneven bronze]

Define $\Phi:[0,\infty]\to[0,1]$ by $\Phi(t)=t/(t+1)$ for $t\in[0,\infty)$ and $\Phi(\infty)=1$. Then one can show $$\rho(f,g)=\Phi(\sup_{x\in X}|f(x)-g(x)|)$$is a metric on $\mathbb{C}^X$ whose associated topology is the topology of uniform convergence. That topology is generated by sets of the form $${g\in\mathbb{C}^X:\sup_{x\in X}|g(x)-f(x)|<n^{-1}}\quad(n\in\mathbb{N},f\in\mathbb{C}^X).$$I struggle with showing the metric $\rho$ induces the topology of uniform convergence. Any help is appreciated.

[gentle osprey]

psie

[uneven bronze]

Ok, here's a big realization: \begin{align*}&{g\in \mathbb{C}^X:\sup_{x\in X}|f(x)-g(x)|<(n-1)^{-1}}\ &\phantom{==}=\left{g\in \mathbb{C}^X:\rho(f,g)<n^{-1}\right}=B(1/(n-1),f).\end{align*}But what about $${g\in \mathbb{C}^X:\sup_{x\in X}|f(x)-g(x)|<\infty}={g\in\mathbb{C}^X:\rho(f,g)<1}=B(1,f).$$Is ${g\in \mathbb{C}^X:\sup_{x\in X}|f(x)-g(x)|<\infty}$ open in the topology of uniform convergence?

[gentle osprey]

psie

[uneven bronze]

I have a typo. The first display here should end with B(n^{-1},f) and not B(1/(n-1),f).

[quick crane]

it seems easy to show that $\rho(f_a, f) \to 0$ if and only if $f_a \to f$ uniformly. Since the convergence of nets determines the topology, you are done.

[gentle osprey]

L

[quick crane]

you could also do this using the open set stuff, but it may be a little more writing

[uneven bronze]

Ok, thanks. On a related note, how do you show the topology associated with the metric $\rho(f,g)=\sup_{s\in S}\min{1,|f(s)-g(s)|}$ on $\mathbb{C}^S$ is the topology of uniform convergence? \

I once worked an exercise where, if $S$ is nonempty and $(X,d)$ is a metric space, then $\rho'(f,g)=\sup_{s\in S}\min{1,d(f(s),g(s))}$ is a metric on $X^S$ and that a sequence $(f_n)$ converges to $f$ in the metric space $(X^S,\rho')$ iff $(f_n)$ converges uniformly to $f$ on $X$. So...this exercise I worked is saying $\rho$ induces the topology of uniform convergence on $\mathbb{C}^S$, right?

[gentle osprey]

psie

[tender halo]

yea

you proved that the two topologies have the same closure operator

which means they are equal

[uneven bronze]

is closure operator a special function? 😅 (never heard this word before)

ah wait, I think you just mean the closure with respect to one topology equals the closure with respect to the other topology

[tender halo]

yeah

closure operator is a function from P(X) -> P(X) with certain properties

it sends sets to their closure

the sets whose closure is themselves are called closed and form a topology

and in the other direction, the function that sends a set to its closure in a topology is a closure operator

[unreal stratus]

Yee you can characrwrisĺe topological spaces this way and it's extremely useful to all algebraic topology etc

[uneven bronze]

There is a generalization of this concept I believe; Kuratowski closure operator. I don't know what it is, just saw it once.

[tender halo]

i mean kuratowski closure operator is what im talking about

[uneven bronze]

ah ok

[tender halo]

there is like a generalization for general posets

that is useful in lattice theory

[uneven bronze]

I have a stupid doubt. The metric $\rho(f,g)=\Phi(\sup_{x\in X}|f(x)-g(x)|)$, where $\Phi:[0,\infty]\to[0,1]$ by $\Phi(t)=t/(t+1)$ for $t\in[0,\infty)$ and $\Phi(\infty)=1$ is supposed to induce the topology of uniform convergence on $\mathbb{C}^X$.\

Now I know, if $(Y,d)$ is a metric space, then $\Phi\circ d$ is a bounded metric on $Y$ that induces the same topology on $Y$. I want to use this to show $\rho$ induces the topology of uniform convergence, but $(f,g)\mapsto \sup_{x\in X}|f(x)-g(x)|$ is not a metric. How can I reason differently?

[gentle osprey]

psie

[quick crane]

just reason directly

[uneven bronze]

Hmm, how would you reason directly?

[quick crane]

show that it is a metric, then show that a net f_a converges to f in the metric topology if and only if it converges to f in the uniform convergence topology

[uneven bronze]

Hmm, I don't know how to do that. I thought I could use what I had previously worked on, i.e. that Phi o rho is a metric, though the supremum norm is not a metric in this case on C^X.