point-set-topology

[red yoke]

And quotient it out

Imma go eepy now

[ebon galleon]

[rancid umbra]

ah, so like this:

[tiny obsidian]

\overline{V} is also metrisable, and in metric spaces lindelof <=> second countable (as opposed to just <= in general)

[gritty widget]

hello i was trying to understand these terms

exterior, interior, closure and boundary

so from my understanding a subset S of topological space X has its interior defined as the union of all open sets in X that are subsets of S

so

$$(X, \tau)\text{ is a topological space.}$$
$$\text{int}X S := \bigcup{K\in \tau, K\subseteq S} K$$

Now, it is to my understanding that the boundary is defined by

$$\partial_X K := \text{cl}_X K \backslash \text{int}_X S$$

[gentle osprey]

imaginebeingtired

imaginebeingtired

[gritty widget]

what ik from closure is that

$$\text{cl}_X S := \left{s\in X\big| \forall M \left[s \in M \exists P \in \tau [P\subseteq M]\right] \exists n \in S [n \in M] \right}$$

i think

i think this completes it yea

[crude apex]

Yes, these are also the definitions I had

[gritty widget]

not sure if the closure def is correct

[gritty widget]

can u check the closure def

wait shit

no

[gentle osprey]

imaginebeingtired

[crude apex]

The last one seems correct to me

[gritty widget]

i think this might be wrong if K is not closed maybe

$$\partial_X S := \text{cl}_X S \backslash \text{int}_X S$$

this might be better

[crude apex]

Why is there a K and a S ?

[gritty widget]

this is better yea

[gentle osprey]

imaginebeingtired

[gritty widget]

sorry lol

[crude apex]

Ok this one is good

[tender halo]

but whats your question lol

[gritty widget]

now

wth is an exterior

[gritty widget]

im learning defs

cuz idk them

i cant wrap my head around an adherent point btw @crude apex

can u help

[crude apex]

An point P is adherent to a subset S if each neighbourhood of P intersects S

Then you can define the closure of a subset as the set of all its adherent points, that's what you wrote formally earlier

(in french we even say "adhérence" for closure)

Does this definition seems clear to you ?

[gritty widget]

ah

so

just needs to be defined like the point s

gotcha

[gritty widget]

.

whats an exterior btw

cuz like

im genuinely rlly confused

hm

ig that is fair

it should be that true that $$\text{ext}_X S \cup \partial_X S \cup \text{int}_X S = X$$ but i wonder how youd prove that

[gentle osprey]

imaginebeingtired

[gritty widget]

yh but im working w/o one

[merry geode]

How did you define the exterior

These proofs depend wildly on the definition.

[lusty trench]

Isn't the the exterior of A (in X) the complement of A's closure (in X)?

[merry geode]

It's the same thing, but it could be defined as interior of A^c

Or similar

[gritty widget]

Nvm

It's early

I read your msg wrong

Yeah interior of complement of A in X is fine

[gritty widget]

Can you check if this is correct

The definition of the closure of A.

Well S not A mb

[merry geode]

I mean, identifying this and seeing relations with different definitions is the difficult part.

[gritty widget]

Seems like it's the official def

[lusty trench]

Notice that $int(S)$ is the largest open contained in $S$. In other words, it consists of those points that are sufficiently separated from $X \setminus S$.

Similarly, $ext(S)$ is the largest open contained in $X \setminus S$. In other words, it consists of those points that are sufficiently separated from $S$.

Then $\partial S$, i.e., the complement of $int(S) \cup ext(S)$, consists of those points that aren't sufficiently separated from either $S$ or $X \setminus S$. In other words, it's the boundary between $S$ and $X \setminus S$.

[gentle osprey]

Eduardo León

[pearl trail]

Can someone check my proof for this problem?

[viral atlas]

Rough

[full merlin]

I have a quick sanity-check question: Assume someone tells me they have a regular (the boundary maps are homeomorphisms) CW-complex (finite). They tell me how many cells there are in each dimension, and they tell me which cells are contained in the boundary of each cell.

Questions:

1. This uniquely determines the CW-complex up to cellular isomorphism, right?
2. Is there some easy computation which tells me if the information they tell me actually corresponds to a CW-complex?
3. Is it easy to calculate the homology from this information?

[quick bough]

why do homotopy groups commute with finite cw complexes? what i mean precisely is: if X = colim_J X^(n) for J some finite subset of N, then pi_k(X) = colim_J pi_k(X^(n)), i know that there should be some argument where finite cw complexes are compact, but i dont see why that implies my question

[red yoke]

Is X^(n) the n-skeleton

The colimit of a finite sequence is just the last element right

[unreal stratus]

Ydah

Anyway you can always do this, don't need X to be finite

One way is just to use the fact that S^k and S^k x I are compact

Then the argument can be generalised beyond CW complexes (under suitable hypotheses)

Basically note any f: S^k -> X has finite image and the image must be contained in X^(n) for some n

But then also if f,g : S^k -> X are two homotopic maps, this is witnessed by a homotopy h: S^k x I -> X, which has image lying in X^(m) for some m by the same argument

This is enough to show that pi_k X = colim pi_k X^(n)

You can immediarely generalise this to, say, filtrations X_0 -> X_1 -> ... of X where every compact subspace of X is contained in some X_k

Which is helpful

[full merlin]

You probably already forgot about it but the statement is wrong (I think). For example Just take S^1 again with 2 1-cells and multiplication by 5. Thanks for your input though!

[umbral panther]

What is your definition of a cellular map? That sounds like it isn’t cellular

[full merlin]

sends the k skeleton into the k skeleton https://ncatlab.org/nlab/show/cellular+map

[full merlin]

Question: Is there such a thing as an abstract (regular) CW-complex? I feel like what I am describing in my question should be its definition (it is purely combinatorial information), and my question should be phrased:

1. Does an abstract CW-complex define the isomorphism type of the realizations?
2. How can you tell if some combinatorial information defines an abstract CW-complex?
3. Can you calculate the homology groups using only the combinatorial information about the abstract CW-complex?

But a quick google search for "abstract cw-complex" doesnt lead me anywhere...

Any hints/ideas?

[red yoke]

Are you looking for simplicial complex

Without the homeomorphism criteria perhaps it might be something like simplicial set?

[full merlin]

I am not sure... but I don't think so (correct me if I am wrong). For example: I have a regular CW-complex X (spoiler: its S^1) with

• 2 0-cells {e_1^0, e_2^0} and 2 1-cells {e_1^1, e_2^1}
• boundary of e_1^1 = e_1^0 \cup e_2^0
• boundary of e_2^1 = e_1^0 \cup e_2^0

this combinatorial informaion determines X. But this is not a simplicial complex, right (or is it)?

[red yoke]

Yea I think you're looking for simplicial set

[umbral panther]

There is a functor from simplicial sets to simplicial sets known as the subdivision. That is a geometric description, but there is another description: form the partially ordered set of inclusions and take the nerve of this category

Do you have enough combinatorial information to form this partially ordered set? Does this give you a topological space homeomorphic to your original space?

I googled those words and found this
https://mathoverflow.net/questions/61948/are-there-two-non-homeomorphic-finite-regular-cw-complexes-with-isomorphic-face

[full merlin]

Thanks a lot for the input guys (or girls)! I will look into it

[full merlin]

Sadly I don't feel like I have the background to understand this yet ^^

[umbral panther]

The MO answer might be lower tech

[full merlin]

Hmmm... I am a bit confused why the answer is so complicated. Doesn't it follow almost trivially (which is why I initially started my question with "sanity check") as follows:

-The 0 skeleton is obviously unique. There is no choice for the attaching map.

-The 1 skeleton is also obviously unique. The only choice is which end of S^0 I attach to which of the two points in the boundary, which doesnt change the resulting space.

• Similarly, when attaching a k-cell e^k and I have uniquely constructed the k-1 skeleton, From the well definedness I know that the boundary of e^k is isomorphic to S^k-1. So the only choice I have is choosing an automorphism of S^k-1 for the attaching map. But this choice should not affect the resulting space (maybe using that any automorphism of S^k-1 can be extended to an automorphism of e^k.
  Thus, the attaching of the cell is basically unique. And thus X is unique.

Does this argument check out?

[quick bough]

oh yikes, yeah, in that case, that's trivial, i misread my case, the cw complex wasn't finite, sorry

[quick bough]

i see, thank you, this argument makes sense to me

[unreal stratus]

There's also a more refined argument for X a cw complex: S^k is k dimensional and S^k x I is (k+1)-dimensional, so by applying the cellular approximation theorem you can show that pi_k(X^(n)) -> pi_k(X) is surjective for k <= n and a bijection for k < n

[quick bough]

isn't this somehow also a corollary from whiteheads theorem?

(we haven't actually proved cellular approximation, that's like next lecture lol)

actually no, hm

but i see your argument definitely

[rancid umbra]

given a discrete closed subset X of R^n with n>=3, is there an open cover of X which is locally finite?

oh, R^n is paracompact

nice

[unreal stratus]

Idk how this has whitehead

[unreal stratus]

Also like

X

Or R^n depending on what you mean by open coer

[rancid umbra]

i think i left out some details:
i have an open cover of X (since X is discrete) and i wanted to make sure there was a locally finite refinement

[rancid umbra]

i think this is using the fact that X is closed, right?

[unreal stratus]

Yes that makes sense

[rancid umbra]

and this works for n = 0, 1, 2 as well

[unreal stratus]

But you merely said an open cover so X then works lol

But yeah

[rancid umbra]

yea lol

[unreal stratus]

Anyway since it is discrete like

For each point you can find some open ball about it which doesn't contain any other points

[rancid umbra]

right, but the open balls might intersect poorly

[unreal stratus]

Those open balls give you a refinement

Oh yeah sorry

[rancid umbra]

all g

i was trying to think of something along those lines at first too, but it doesn't work because of that

unless there is a different argument

[unreal stratus]

I think the set must necessarily be countable though

So then you can just do the balls one by one

No scratch that idk

[rancid umbra]

but this is an intermediate step, because since the cover is locally finite, i can get a mutually disjoint open cover of X

sorry, talking is hard.

this exercise:

[quick bough]

nah, i dont see it either, it just looked like somethingm, where i can apply whitehead at first sight, but doesn't work

[rancid umbra]

uh, just to make sure im understanding, correctly, R^n is obtained from R^n - U(disjoint open balls) by attaching 3-cells, so the inclusion of R^n - X induces an isomorphism with the trivial group

[quick bough]

theres a solution on math stacksexchange, i rmb seeing it at some point

[rancid umbra]

i think im pretty close already

yea, i think thats right

since for each ball, you can glue it back to R^n - U(disjoint open balls) in the cavity that it creates

[tender halo]

all metric spaces are

so ur safe

[opaque zodiac]

that's probably my favorite one-page paper
https://www.ams.org/journals/proc/1969-020-02/S0002-9939-1969-0236876-3/S0002-9939-1969-0236876-3.pdf

this wasn't the original proof but it was a massive simplification of the existing proof

[tender halo]

oh thats the proof i learned

from Engelking

also it doesnt only prove that they are paracompact, but also that any cover has a \sigma-discrete refinement

[opaque zodiac]

what is sigma-discrete?

[tender halo]

a countable union of discrete

[opaque zodiac]

and discrete means no itersections?

[tender halo]

means every point has a nbhd that touches at most one member of the family

you see it in the bing metrization theorem

bing says that metrizable iff regular and has a sigma-discrete basis

nagata smirnov says metrizable iff regular and has a sigma-locally-finite basis

bing has a funnier proof though despite being strictly weaker

you embed the space into the the hedgehog which is metrizable

so its stronger in entertainment value

[tender halo]

in fact all metric spaces are realized as subspaces of a product of hedgehogs

[opaque zodiac]

https://www.youtube.com/watch?v=nfHuZ5qrYX4

(thanks)

[gritty widget]

hi, what's the difference between point-set topology , and just Topology?

are the same?

[opaque scroll]

Point-set topology is a somewhat loose term referring to the basic constructions and properties of topological spaces.

It's somewhat similar to how the word 'calculus' is often used in relation to 'analysis'

So typically one might take a class on point-set topology before diving into a more specialized field of topology.

[gritty widget]

so munkres topology book is actually Point-set topology?

[opaque scroll]

I'm not super familiar with the book, but looking at the table of contents at least the first 7 chapters is something one might call point-set topology yeah

[gritty widget]

thanks

🫡

[limpid fern]

he has alg top too but at the end

ch9-13

[gritty widget]

better ,I will learn more then

[fringe thorn]

higher! Learns Point Set Topology :lisayay:

[rancid umbra]

wait what uh. what space is this?

it’s not RP^2

[unreal gull]

klein bottle

https://en.wikipedia.org/wiki/Klein_bottle

oh wait

nvm

[rancid umbra]

nbm?

[unreal gull]

misread the arrows

[merry geode]

It is, iirc

You can cut in between to change the arrows

[merry geode]

Like, cut in diagonal and attach.

[rancid umbra]

i don’t think diagonal works? lemme try again

[merry geode]

Ah wait, it is projective plane. Nvm

My memory is not serving me well..

So diagonal cutting may give you RP^2.

[rancid umbra]

i think its a sphere

oh wait

dam

uh

i previously was cutting across the wrong diagonal i guess

it is RP^2

bruj

[merry geode]

Yes

[rancid umbra]

wait. so what am i doing wrong with van kampen’s here?

this doesn’t look like Z/2Z to me

the orange set deformation retracts to a figure 8 i believe

and the green set is contractile

[red yoke]

It's a circle I believe

And the grey loop loops twice in the circle

[rancid umbra]

like this?

[red yoke]

Uh yea

Or you can look at the equivalence classes on the square directly

Opposite points are glued together

[rancid umbra]

oh right

[red yoke]

So it's like R/(2Z) → R/Z

[rancid umbra]

so its just RP^1

the 1-skeleton

[red yoke]

Yea

[rancid umbra]

i kept wondering where i was getting a figure eight from

i was like, alr, ig there’s a figure eight in RP^n that i wasn’t aware of

[red yoke]

I think you assumed the 4 vertices are glued

[rancid umbra]

yea

[rancid umbra]

initially i was gluing the two pink points as well

[red yoke]

This has happened several times in this channel

[rancid umbra]

alr, so then its just <x | x^2>

by me? 😿

[red yoke]

No by others

[rancid umbra]

cutting and pasting is a difficult skill 😭

[red yoke]

You'll get used to it

[pearl trail]

Any hints on this question? Part 1?

[lapis monolith]

We know that the composition of two continuous functions is continuous. But is it necessarily true that if we have a continuous function that is the composition of two functions both of those functions are continuous?

[unreal stratus]

No

Given any f: X -> Y, consider the (unique) map g:Y -> * where * is a 1 point space

gf is always continuous

[quartz horizon]

Ooh I’ve never seen it called corestriction

Sounds like a good name

[plucky veldt]

what happens when you remove a set from the topology on A and then look at the preimage of that set in the inclusion function?

[pearl trail]

It’s not open in A?

[plucky veldt]

right, so is the inclusion continuous? and do you know why the preimage is not open?

[pearl trail]

Are you making a contrapositive argument?

[pearl trail]

So consider a set W not in the topology of A. Then W is not open in A because sets open in A can be written as U = A cap V for some V in T_x.

[echo folio]

whats the differnece between low-dimensional topology and point-set topology

[plain raven]

point-set topology is a foundational field that establishes the technical devices and definitions which are needed support and formalize the definitions and rigorous arguments carried out in geometry, low-dimensional topology, algebraic topology and functional analysis. for a student, a basic understanding of point-set topology is a prerequisite for any of those other fields. as a research topic, questions in point-set topology are primarily questions of set theory, and are no longer closely connected to research questions in geometry and analysis.

low-dimensional topology is the study of manifolds of dimension n <= 4. This includes curves, surfaces, and three-dimensional Euclidean space. You can read the wikipedia page. The Poincare conjecture is a question about low-dimensional topology.

[sonic crane]

Chat gpt?

[umbral panther]

Nah, ChatGPT says that 4d is only sometimes low dimensional topology

[gaunt linden]

Hmm, I'd gotten the vague impression that low-dimensional meant the dimension was too high to intuit directly about (so larger than 3) yet still low enough to have idiosyncracies that don't generalize to generic large dimensions...

[opaque zodiac]

generic large dimensions like 5

[umbral panther]

Claude agrees that low dimensional uniformly includes 4

[rain mason]

I don't think you can really intuit much for 3 manifolds beyond R^3 (and maybe S^3) which you can't for higher dimensional manifolds tbh

low-dimensional topology usually refers to 3 and 4 manifolds from my experience

I'm not sure anyone does research on just surfaces, people do study them in areas like symplectic topology

I guess 4 manifolds people also care about embedded surfaces

[opaque zodiac]

Chat gpt uses capital letters

[umbral panther]

A lot of work on the mapping class group is considered 2d topology

[plain raven]

This is a really unkind thing to say to someone who gives their valuable time to answer a question. I get these comments a lot because I write long and detailed answers like ChatGPT does.

[sonic crane]

Im sorry i did not mean offense

[plain raven]

Okay. No harm done but please don't say that kind of thing.

Assessing writing on overall volume and shape of the paragraphs, disregarding the content, is also how they grade the SAT/GRE writing exam, funnily enough. https://www.nytimes.com/2005/05/04/education/sat-essay-test-rewards-length-and-ignores-errors.html

[rain mason]

okay yeah I guess I'm showing my bias here lol, cuz I was mainly thinking about Symp(M) and Ham(M), but yeah, people do think a lot about Diff(M) and Homeo(M), and there are really interesting questions there. Although I think a lot of recent progress is coming from symplectic topology?

[umbral panther]

People try to reduce symplectic topology questions, especially in 4d, to questions about the mapping class group. But do they contribute to understanding the mapping class group?

[rain mason]

maybe? I'm not sure tbh, this would've been related to what I'd do of I accepted my symplectic PhD offer lol, there's a bunch of people here interested in (symplectic) mapping class groups

But I think the symplectic geometry side is mostly just interesting subgroups of the mapping class group?

At least from the perspective of people who think mostly about MCGs

[tame arrow]

My guess is that there is a slow upward creep in the dimensions of the definition, since 3-manifolds are fairly well understood now. 4-manifolds require different techniques that other dimensions because of the failure of the Whitney trick and embedded surfaces (usually knotted surfaces) in 4D is a very active area of research. There are still basic open questions in knot theory like the Slice Ribbon Conjecture and there's a lot of adapting techniques from 3D to 4D. Low dimensional topologists do spend some of their time in 5D.

[umbral panther]

No, there’s a downward creep. The first couple proofs of the h-cobordism theorem only worked in dimension at least 7

[umbral panther]

Or maybe it wasn’t the h-cobordism theorem, but the generalized Poincaré conjecture. I think Smale announced that a smooth homotopy sphere of dimension 7+ is PL homeo to the standard sphere. But he didn’t publish a proof. Stallings published a proof using the technique of engulfing that perfectly matched the restrictions. Then Smale published his proof, which over the years got modified to be sharp

Smale’s argument was to take the manifold, remove two disks, apply the h-cobordism theorem to conclude that it is a cylinder and glue back the disks. Thus exotic spheres are the same as automorphisms of the lower dimensional sphere. In the PL category there are no exotic automorphisms. But the h-cobordism theorem was special to the smooth category, so the hypothesis and conclusion are in different categories. So you just have to redo the proof in the PL category. I don’t remember why Smale proved the h-cobordism theorem in dimension 7+ rather than 6+. To get the sphere down to dimension 5, you need a new argument, surgery theory, so that your h-cobordism is a dimension larger than your starting sphere. But this argument is very similar to the proof of the h-cobordism theorem

[hollow thorn]

have you guys heard of the gdp per capita (i know this has zero connection with math i am just acting like i am albert einstein idk why)

[merry geode]

It's quite interesting how higher dimensional manifolds can be easier to understand

[tight oracle]

I don't know how to create the homeomorphism Tao talks about in his hint.

From his measure theory book (but this is a topology question so it goes here)

[solemn temple]

Lay out your closed disjoint intervals on [0,1). There are infinitely many of them. Pick an interval somewhere in the middle such that there are infinitely many more on either side, remove it. Now you have two half open intervals covered by countably many disjoint intervals…

[tender halo]

wait how does that help

[warm hedge]

If we have a strong choquet space does it mean that it has no points which are isolated ?

[red yoke]

A point is strong Choquet

Actually if X ⊂ Y is any set of isolated points then Y is strong Choquet iff Y\X is strong Choquet right

[warm hedge]

so i think i cant understand what strong choquet actually gives to a space

[red yoke]

What are the applications of this

[opaque zodiac]

this shows up in set-theoretic topology stuff often

strong Choquet spaces are like a 'good' generalization of Baire spaces iirc

or maybe not generalization

but like this property here that it passes to G_delta subsets, that's a characteristic property of Polish spaces

any G_delta subset of a Polish space is a Polish space

I think maybe it's like Baire and Choquet are too weak to be useful in full generality, but the strong Choquet property gives you more of a handle on stuff

[opaque scroll]

I'm not sure how to make the homeomorphism, but a different solution could be:

||Consider the initial endpoints of the closed intervals. Ignoring 0, it's a countable dense order without endpoints, so is order isomorphic to the rational numbers.||

||Consider some irrational Dedekind cut and the two sets you get as the union of the closed intervals above and beyond the cut.||

||It's not hard to see both sets are open, so [0, 1) would be disconnected.||

[unreal stratus]

Interesting as I thought this would be easier than it actually is

[plush folio]

Isn't this just a result of the fact that you can't cover any part of the real line by two or more closed intervals without them overlapping at some point?

For example, consider the covering of closed intervals {C_0, C_1, C_2, ....}, ordered by their lowest endpoints. Then C_0 and C_1 cannot overlap, but there also cannot be a gap between them

[opaque scroll]

You can't assume that the ordering is well-ordered

[plush folio]

Hmm, meaning there might be an infinite amount of intervals between C_0 and C_1?

[opaque scroll]

Yeah sure

[plush folio]

but since the covering is countable, surely there must be some pair of intervals that are "next to" each other?

[opaque scroll]

Consider the rational numbers

They're countable, yet no pair of rational numbers are next to one another

[plush folio]

ah, I see 👍

[night heron]

But since we’re talking about intervals specially, wouldn’t that entail well ordering?

[tender halo]

no

[opaque scroll]

Why would it?

[tender halo]

you can think about the intervals as a whole really

define order on the intervals if one is more left then the other

it will be a countable dense order with a least element

so its isomorphic to 1 + Q

[night heron]

I see

Thx

[tender halo]

so the whole order is a (1 + Q)-indexed sum of either points or copies of [0; 1]

which is impossible because that is not a complete order

because Q is not complete

[night heron]

I was assuming intervals on the real line where the well ordering principle applies. Didn’t read the message properly and so I didn’t realize you were talking about the ordering of the intervals and not the ordering in the intervals

[plush folio]

How about this proof: ||consider a disjoint covering S of [0, 1) consisting of closed intervals. S must contain atleast two intervals, [a, b] and [c, d]. Since b != c, there must be atleast one interval in (b, c), call it [x, y]. By the same argument there must be an interval in (b, x) and (y, c). Continuing recursively, we see that S has cardinality atleast 2^N, so S is not countable.||

WLOG b<c I guess I should add

[opaque scroll]

Why would this imply S has cardinality 2^N?

Like take the rational numbers covered by the closed intervals [x, x]. What would change in your proof?

[plush folio]

I thought you could show that S would be in bijection with the set of infinite bitstrings, like at each step you have two branches, each of which has an infinite amount of elements

but I'm not 100% sure what the bijection would look like

[tender halo]

same argument would work on Q

[plush folio]

and I see your point about the rationals, you can cover the rationals in [0, 1) with countably many singletons

[tender halo]

I mean you can make it kinda work

i will be working with (0; 1) cuz its easier

on each step you have some interval (a; b) that is realized as a countable union of closed intervals A_i.
fix a closed subinterval [c; d] that properly contains at least two A_i (and therefore contains countable amount of them). inside [c; d] find two open intervals (a'_1; b'_1), (a'_2; b'_2) that are also a countable union of A_i. do the same again recusrively on them

uhh actually no, you do need two

sorry

if you go down the tree, you will find a point in the intersection of all [c; d]s

the interval it lies in must be a point

but there are continuum branches in the tree, so there are continuum different intervals that must be points

uhh maybe you dont even need that

well doesnt matter its a contradiction either way

[coral pawn]

Is there any relationship between an sing(X) and pi_infinity X?

They are both infinity groupoids

Are they equivalent?

[unreal stratus]

To me, at least, the former is the definition of the latter, provided you take oo-groupoids to mean N(Kan)

[quartz horizon]

The latter is the definition of the latter…?

[umbral panther]

Yeah, I would say the former is a concrete model of the latter, not vice versa

[unreal stratus]

Yeah

[unreal stratus]

Thanks for pointing out my typo lol

[unreal gull]

I know that normal means that for each pair of disjoint closed subsets A, B of X, we can find disjoint open sets U, V of X containing A, B, respectively. However, does this mean that given disjoint closed subsets A, B of X and an open set U containing A (which is also disjoint from B), we can find an open set V contaning B such that U, V are disjoint

[tender halo]

no

consider the complement of B

[unreal gull]

oh if u let U be the complement of B?

i see

[rancid umbra]

---->----
| |
\/ o /\
| |
---->----

klein bottle with an open disk missing.
how do you compute its fundamental group?
i can't quite figure out how to use van kampen's.
i can give it a cell structure, but i dont see how thats helpful.

any hints?

[plush folio]

So basically you construct a tree where the branches are closed intervals, and the leaves are points in [0, 1)? And since there are uncountably many leaves, there must be uncountably many branches. But why must there be uncountably many leaves? Not all the points of [0, 1) are leaves in the tree

[tender halo]

there are no leaves, its an infinite tree

points on the tree are closed intervals, for each infinite branch there is a point on the interval, all those points are distinct

the tree is dyadic (each point of the tree has two successors), so there are continuum infinite branches (on each step you choose 1 or 2, countable amount of times, its just the cantor set)

[plush folio]

I think I get it, but I don't quite see how it differs from the proof I gave. Also, I'm slightly confused by the terminology. A dyadic tree is what computer scientists would call a binary tree, right?

[tender halo]

well there is a distinction in that a binary generally has children with a fixed order

also i expect binary trees to be finite

but really its minutiae

[tender halo]

it is different because you need to make sure to actually find a point in the interval that corresponds to an infinite branch

in your proof you just find the tree, the tree itself is countable

[plush folio]

aha, I was proposing to count the points in the tree, which would be 1 + 2 + 4 + ..., which is of course countable, but you're proposing to count the number of paths in the tree (what you call infinite branches), which is uncountable (with cardinality 2^N?)

[unreal gull]

given some metric d(y, x), if we fix y (to be 0, for example), then is f(x) = d(0, x) a continuous function?

[tender halo]

yes, its the amount of countable sequences of 0 and 1

[tender halo]

yes

in fact d(x, A) where A is any set is continuous wrt x

[plush folio]

this is where my familiarity with finite binary trees is unfortunate, because for a finite tree, the number of paths in the tree is equal to the number of leaves, which is less than the number of points in the tree

kind of unintuitive actually

no wait

[tender halo]

well this tree has no leaves

[plush folio]

how many paths are in a finite tree...?

[tender halo]

if you are talking about maximum paths from root, then its the amount of leaves

[plush folio]

lol, yes, I got confused for a second

but just a final question, could you explain how your proof fails for the rational numbers in [0, 1)?

[tender halo]

because the rational numbers arent complete, for them a sequence of closed intervals doesnt have to have a nonempty intersection

[plush folio]

Right, I see. Thanks! 🙏

[lethal oxide]

if f:A--->B a cofibration then (B',A) a topological pair and HEP and F:B'---->B a homotopy equivalence and F(x)=f(x) because F a contractible B' x {0}=B' and B' x {1}={x} and {x} point of B' and A and B for every element of A Is it enough to proof?

[unreal gull]

is any continuous function from X -> R some metric d(x, -)

[quartz horizon]

No cause all such d(x, -) are zero somewhere I think

If I’ve understood you correctly

[unreal gull]

yeah at x

cuz d(x, x) = 0

oh sorry my question was deifntely not very clear

but its a pretty stupid question

I meant to ask if every continuous function from X -> R is really a function of the form d(x, -) for some metric d which induces the topology of X

but that is not true

[quartz horizon]

Mhm

[opaque scroll]

I guess interesting follow up is, does X embed into a metric space Y such that f = d(x, -)

Feel like the answer should still be no, but it's trickier

[quartz horizon]

You do need f to be nonnegative at least

[opaque scroll]

Yes yes

[quartz horizon]

And if it has multiple zeroes..

[opaque scroll]

Maybe make f strictly positive

Or even bigger than 1 or something

[quartz horizon]

This feels a lot more possible

Is X a metric space?

[opaque scroll]

Well, we'd want it to be metrizable at least

[quartz horizon]

Right

I’m thinking something like

Take the graph of f

So X x R

Disjoint union with a single point

And make it so that the distance from this point to the graph of f gives f

[opaque scroll]

I guess some work is needed for X to be homeomorphic to its graph maybe

[quartz horizon]

Hmm like

[opaque scroll]

Also some care to make sure the triangle inequality holds

[quartz horizon]

The map from X to its graph is x to (x, f(x))

This is continuous

And has a continuous inverse, the projection to the first coordinate

So that should be a homeomorphism

So you can embed X into X x R

Then f(x) is the distance from the “X” axis (hehe)

You just need to somehow make this equivalent to the distance from a single point

Hence my suggestion of taking the disjoint union with a point

Which is say the same distance from every point on the X axis

And for general points in X x R, the distance slowly increases

As you move away from the X axis

Ooh how about this

You can embed X into X x [0, infty) this way

And then shrink the X axis to a single point, making a kind of cone shape

Kind of feels like a mapping cylinder type thing..

Ok hmm can one make the “cone over X” into a topological space?

Then you just make f(x) the distance from the apex

[tight oracle]

I think the hint might be a typo

It might be that the set of end points of the cover (maybe excluding 0) are homeomorphic to the cantor set

Cause the set of end points are closed and for every point x in the set, every neighborhood of x contains another element in the set, hence the set of end points for the cover is homeomorphic to the cantor set.

I don't know if you can extend the set of end points to [0,1) (intuitively you can't) or find a contradiction somewhere else

Thank you for your solution as well

I come back after a year and you are still helping everyone

[tight oracle]

Ah, you have a contradiction on the cardinality of sets. Cantor set is uncountable, whilst your set of end points is countable

[inland thistle]

I need help for an (lexicographically) ordered square [0,1] x [0,1]. How can I find the least upper bound for this?

[gentle girder]

do you mean you want a least upper bound for any subset of the lexicographically ordered square?

[inland thistle]

oh yes, I want to show taht this ordered square has a least upper bound property

so taking any subset of it, I want to show how we find the supremum

[gentle girder]

okay, let S be the set, and consider the set S' = {x \in [0,1]: there is some y such that (x,y) \in S}. Show (or conclude, if you just believe me) that S' has a least upper bound (as a subset of [0,1]), and we'll use that to show the entire result

[inland thistle]

I see, can I also talk about what I have in mind as well?

[gentle girder]

yeah of course

[inland thistle]

for any subset of an ordered square, I think there are two cases. The first case is that for the x-coordinate, let's take the supremum for it and let it be "b". Then if this b coordinate is contained in this subset of an ordered square, then we can just take the least upper bound of the intersection of {b} x [0, 1] and this subset, because R has a least upper bound property, so this case is easier.

But the second case is when this supremum of the x-coordinate is not contained in this subset. Then, some may think {b} x {0} is the least upper bound, but I dont know if it is that simple for the case lexicographically ordered square.

So here, what I want to ask is, is my first case even correct? and how do we do the second case

[gentle girder]

this is good, and you're gonna do it like you'd show any other least upper bound thing, you already have a guess for the least upper bound, and you're gonna show two things (I'll say it for the first case, and you can generalize to the second case) Let (b,c) be the least upper bound for {b} \times [0,1] \cap S: 1. show that (b,c) is an upper bound for S, by picking an element x of S and showing that (b,c) \geq x, and also that any other upper bound y of S has (b,c) \leq y (or showing that if y < (b,c), that there's an element of S such that x > y)

if the set {b} \times [0,1] \cap S is empty, you will try to show this for (b,0), as you predicted

it's clear that in the second case, (b,0) is an upper bound of S, but to show it's the least upper bound, you're going to use the motto "there's always a bigger x-coordinate"

[inland thistle]

alright thank you so much for your help

what you said aligns with what I had in mind I think

[gentle girder]

good to hear!

[hollow geyser]

This proof is too long.

[prime elbow]

In order topology it is not necessary that complement of dense set is dense, right?

Let X = R×R and A = Q×Q since it is diagonal set, hence it is closed so its Closure will be Q×Q. But X/A has closure R×R so it is dense.

Is it correct? Maybe not

[tender halo]

uhh

the whole set is certainly dense and its complement is certainly not dense

[prime elbow]

Proper set example

[tender halo]

ok, delete one point

[prime elbow]

Got it, thank you

[gaunt linden]

But Q×Q is definitely not closed in R×R. (Not in the usual topology; neither in say, the order topology derived from a lexicographic order).

[alpine nest]

It is closed in the discrete topology

[gaunt linden]

(Exercise: explicitly define a total order on R² such that the order topology is discrete).

[prime elbow]

Yes

[unreal gull]

what is an example of an open subset of a Lindelof space that is not Lindelof?

[opaque zodiac]

Take an uncountable set X and an extra point y. Look at the topology where countable subsets of X are clopen

[unreal gull]

Because given any open covering ${U_i}{i \in I}$ of some Lindelof space $X$, then there exists a countable subcovering ${U{i_n}}$. If we simply take the union of elements $$ \cup_n (U_{i_n} \cap A) = A \cap (\cup_n U_{i_n} ) = A \cap X = A$$

[opaque zodiac]

X is an open set that is not Lindelof but the whole space is Lindelof

[gentle osprey]

Henry

[sonic crane]

how do we form the topology when we have a simplicial complex?

Is it just like, a subset is open if that subset is a simplex in the simplicial complex?

[tiny obsidian]

If its intersection with every simplex is open

your proposed definition would be unreasonable: a single 2-simplex (a triangle/disc) then only has 7 open sets, 3 of which are single points, and it also misses the empty set

[sonic crane]

Its also unreasonable because the union isnt a simplex maybe? Line union a line is 4 points which is not a simplex

So thats not closed under unions

[tiny obsidian]

that too

[sonic crane]

Thanks, for context i am trying to learn basics of topological data analysis

And it seems like with the data set they have some distance function and they create a simplicial complex

Then form the topology based on that somehow

[tiny obsidian]

if you can understand what simplicial complexes look like, the definition of the topology is the one that agrees with that geometric intuition

aka if you put the complex into some (sufficiently large dimension to avoid self intersection where you don't want it) R^n, the same as the induced topology

[sonic crane]

I see, yes I asked chat gpt (lol) and it did say something like that. i will read what gpt said

[tiny obsidian]

something this general it's more likely to not be wrong

that said

!nogpt

[lucid geyser]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[light herald]

That doesn't sound right
This would mean that there are no open sets within simplices of dimension < n

[tiny obsidian]

why not?

[light herald]

With the induced topology from R^n from metric

[tiny obsidian]

oh

[sonic crane]

Chat gpt talks about subspace topology

[tiny obsidian]

n is something larger than the dimension of the simplicial complex e.g. klein bottle sits in R^4

(and the relevant simplicial complex has the same topology via a homeomorphism with it)

[light herald]

I mean that for example for a 1-dimensional simplex in R^2 doesn't have any open subsets
Since no 2-dimensional ball can be contained in it

[tiny obsidian]

it doesn't have to be?

we're talking about subspace topology

[light herald]

Ahhh okay I see
My bad

[tiny obsidian]

do you have questions about it?

[sonic crane]

So if this was in R^3

this is our simplicial complex embedded in R^3 lets say

[ebon galleon]

Bruh

[sonic crane]

what are the open sets?

Oh I was confused at first cause i was like, is this just the discrete topology?

[quartz horizon]

why would it be discrete…?

[sonic crane]

it wouldnt bro im just a noob

bear with me big dawg

[quartz horizon]

please don’t refer to me with those terms

[sonic crane]

lol

ok certified zoomer

[quartz horizon]

now that’s perfectly fine

[sonic crane]

🥰

[sonic crane]

is this space connected ?

under subspace topology

[gentle girder]

no, the point on the left is open and closed

for example

[sonic crane]

Its open because you can form a ball around it

why is it closed?

[quartz horizon]

cause the complement is open

[night heron]

It’s clopen

[light herald]

Every connected component is closed and open right

[night heron]

I believe so

[ebon galleon]

No. Consider Q.

Connected components are singletons.

[light herald]

Oh I mean the complex here

[ebon galleon]

Oh sure

Okay yeah I didn't read the context that'll be true then

My b

[limber ravine]

w = v|_S^1 is homotopic to the inclusion j : S^1 -> R^2 - 0 because we assumed at first that v does not point directly inward? Could we have any map on S^1 that is not homotopic to the inclusion?

i don't understand the picture

Ok, I understood the picture

but still not understanding if we need the assumption that v does not point directly inward to get the homotopy to the inclusion map

because if we have an inward pointing vector then we would get something like <- -> and it is not possible to find a homotopy between these two?

[sonic crane]

I'm getting confused on the open sets here again (I am a beginner on topology). This is a subset of R^3, so the topology on this we consider is the subspace topology. So, a subset S of this complex is open if you can take an open set of R^3, intersect it with the whole complex and you get S. Is this true so far?

[gentle girder]

yeah that's fine

[sonic crane]

Im not seeing something, it seems like every subset is open to me

[gentle girder]

pick a single point on the interior of one of the triangles, that is not open!

[sonic crane]

Ohh

[gentle girder]

in fact, if you take just the interior of one of the triangles, you can pick any "non-open" subset as though it was R^2, and generally that won't be open in the subspace topology

[sonic crane]

If you pick a line segment along a face on the triangle, that is not open, but a line segment thats just hanging out on its own, that would be open right

[gentle girder]

yeah that's right, as long as its endpoints aren't doing something funny

[sonic crane]

Ok thanks this makes more sense

[light herald]

Since it's a punctured R^2 space, any loop that has the origin at the exterior wouldn't be homotopic to inclusion
I think

[gentle girder]

like if i take the very top left line segment in your picture, that isn't open, for example

[sonic crane]

the one along the triangle?

[gentle girder]

the one that's hanging off at an endpoint

at the top of the triangle that you're thinking of

that's also not open

[sonic crane]

oh ok cause that endpoint is connected to the triangle

[gentle girder]

yeah, because you have to have a neighborhood around that point contained in any open set

[sonic crane]

so open set in R^3 cant capture that point without also taking some triangle

[gentle girder]

yep

[sonic crane]

Thx a lot for the help

[limber ravine]

I didn't understand, could you elaborate please??

[light herald]

So, what the proof actually says is that since w is a restriction of an embedding of disk, it has to have origin at the exterior
On the other hand, then the "homotopy path" of at least one point p in homotopy from w to j would cross the origin, and if you draw a ray between the points j(p) and w(p), it would mean that there is a vector pointing inwards

I would draw a picture but I'm on mobile

[limber ravine]

Humm ok, I will think on that

thank you!!

[scenic ridge]

Does anyone have an idea of what the topology of [0, infinity)^n / S_n looks like, where it is equipped with the quotient topology

[light herald]

I can send you an illustration in an hour if you still have a problem then

[limber ravine]

Sure, I can wait for it. thanks

[night heron]

[tiny obsidian]

I assume you mean the action that permutes the coordinates?

[scenic ridge]

yes

so like for n=2 the equivalence classes are {(a, b), (b, a)}

[tiny obsidian]

choose the point where a <= b

(higher dimensions this is extended the obvious way)

are any of those points identified?

[scenic ridge]

wdym

[tiny obsidian]

can you have two elements of an equivalence class where the coordinates are in increasing order

[scenic ridge]

I don't really get your question

each equivalence class is basically just all possible permutations of a n number set

[tiny obsidian]

take the function from an equivalence class to the unique point inside it where the coordinates are in increasing order

[scenic ridge]

increasing or non decreasing?

[tiny obsidian]

non decreasing

[scenic ridge]

since the point could be like 1 1 1

ok

[tiny obsidian]

I believe this is an homeomorphism onto its image

or at least I can't see any reason it wouldn't be

[scenic ridge]

could you explain why?

also the image is with the subspace topology correct?

[tiny obsidian]

by looking at it 🗿 /half-joking. I'm trying to figure out an actual proof now

[scenic ridge]

lol

im trying rn also

[robust drum]

I don’t think this should matter because of subspace topology so you can still have some intersection with an actual open subset of R^n giving you one of the lower dim simplexes

Oh someone alr said that

[light herald]

yeah mb

[fair tusk]

Can’t lie, I’m totally lost on this. Can someone explain? (Hatcher pg. 51)

[novel acorn]

which part exactly is confusing you?

[fair tusk]

Well I’m not sure exactly what’s meant by attaching cells by a word

[novel acorn]

ah

that means you attach the boundary of a cell by the word
Each letter is a circle

so a^2 means you wrap the boundary of a cell around the circle twice

[tiny obsidian]

(circle with orientation)

[fair tusk]

Ah ok

Wait, so is that a disk that’s overlapping itself? Sorry, still a bit confused lol

[gentle girder]

the interior of the disk is not overlapping, you're really identifying points on the boundary

[robust drum]

You can’t picture it in R^3

[novel acorn]

it's a bit more complicated than that
It's sorta impossible to visualize and that's the point of the algebra
You sorta abstract these impossible to visualize shapes into more managable algebraic expressions

[robust drum]

It’s best to just picture it as identification spaces

Meaning view the disc and the thing it’s being attached to as disjoint and imagine that getting near an edge of one teleports you to where it’s attached

That’s at least somewhat doable for 2 and 3 manifolds but once you go beyond that not even worth trying to picture them lol

[fair tusk]

Ah, I guess that diagram was a little misleading then lol

[gentle girder]

by a "word" he means "there is a map from S^1 = boundary of D^2 -> X^1 given by 'first go around a for a small amount of the circle, then go around b' etc.."

[fair tusk]

But I might have also skipped some steps - when we say we attach a 2-cell to a circle, does it become an interior and make it D^2 in general, or is the shape more complicated?

[novel acorn]

it depends on how you attach it

[robust drum]

Depends how we attach it. The interior is always still just the interior of D^2, by the definition of a CW complex. But the boundary can make the whole cell more complicated

Ex: the classic cw decomposition of the 2 sphere

The 2 cell is actually just the entire sphere

[fair tusk]

So does the 2-cell have to be attached along the entire boundary of S^1, or can it just be attached to an arc?

[gentle girder]

one could, for example, in the simplest case, attach D^2 via the empty word, and the resullting complex will be a sphere (or a wedge sum of a sphere with some number of circles if you start with a wedge sum of circles), since the entire boundary of D^2 will be identified with a point

[robust drum]

But the interior is still a disc

[robust drum]

It can be just attached to an arc

That’s legal

[tiny obsidian]

I think an actual proof is painful because you have to separately treat at least n+1 different classes of points to show continuity

[fair tusk]

Ohhh that makes so much more sense

[robust drum]

Just needs to get attached into the 1 skeleton

[fair tusk]

I had always thought it was the entire boundary lol

[robust drum]

Ye CW complexes have a lot of freedom

[fair tusk]

Does the cell also have to be totally bounded by that space, or can it just be attached to the point and have a boundary elsewhere?

Or no boundary at all, like an open disk containing the point

[tiny obsidian]

the entire boundary of the cell you're attaching needs to be attached to something

[gentle girder]

well, I'm not exactly sure what you mean. You have to have an attaching map that takes your boundary to the 1-skeleton

[fair tusk]

Oh ok

[robust drum]

The boundary of a cell is connected and the attaching maps are continuous so they have to attach to exactly one connected component

Unsure if that’s what you’re asking

[fair tusk]

Yeah I just wasn’t sure if you could have some trailing edge of a 2-cell when it was attached, that answers my question though

[gentle girder]

a motto might be "you'll always be able to reach the "boundary" of the 2-cell (even if it doesn't look like a boundary in the complex), and furthermore when you reach the boundary, you will be in the 1-skeleton of X"

[scenic ridge]

Which different classes? Could u give me an outline/idea of what u thought the proof would look like. I’ll try to do the nitty gritty details

[tiny obsidian]

For simplicity I'll give them for n=3 because I don't have a nice description of what they are

So the image $Y$ is the set ${x\in\R^n:0\leq x_1\leq x_2 \leq\dots\leq x_n}$ where we write $x=(x_1,\dots,x_n)$. Consider a point in open set $U$ in $Y$, take sufficiently small ball around $x$ so that it's in $U$, we will show that (potentially shrinking further) the preimage of the ball is open. For points were $0<x_1<\dots<x_n$ this is easy: shrink the ball even further so that every point in the ball also satisfies this, then the preimage is six disjoint open balls and so open.

I've just come up with a nice description when writing it down

[gentle osprey]

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

[tiny obsidian]

this is the easy case

[merry geode]

Maybe simple case like n = 2 helps.

[tiny obsidian]

then further classes are those where we replace one < with an =; two <s with =, etc. until we have $0=x_1=\dots=x_n$ that is the $n+1$st class of unique point $(0,\dots,0)$

[gentle osprey]

Edward II

[merry geode]

You just fold a plane by half.

[tiny obsidian]

essentially proving it for the further ones is proving that preimages of the parts of balls of [0,infty)^n that actually lie in Y glue together into complete open balls, or as complete as they get in [0,\infty)^n

in R^3 the map collapses six copies of Y rotated and flipped around x=y=z in a certain way down into just Y

etc.

[merry geode]

Isn't this just basically the subspace 0 <= x1 <= .. <= xn?

I might be missing something

[tiny obsidian]

I mean the original space decomposes into six of these infinite pyramid like things, all of which are basically Y reflected across a bunch of planes

and they all meet in the line x=y=z

[merry geode]

Yeah so quotienting would give 0 <= x <= y <= z

[tiny obsidian]

this has been established yes

prove that this quotienting is actually continuous

I would stop here tbh and this is going to be a constant in topology that things are continuous by just looking at them

[merry geode]

Constant in topology?

(Well I mean I am too lazy to write down why things are continuous)

[tiny obsidian]

as jujumumu keeps doing topology, proving continuity will often be a lot of hassle, too much to be really feasible

[merry geode]

Ah, it is, compared to the "it is obviously continuous".

I think in this case, you may consider a sequence converging to the "boundary" point and see that it is still convergent on the target space.

[tiny obsidian]

sequences 💀 I didn't even think of those

this is absolutely far easier

[scenic ridge]

wait what

this works to show that the two spaces are homeomorphic?

[merry geode]

Hm actually you also have to show the converse, IIRC.

And then invoke universal property of quotient space

[scenic ridge]

Is this easier than caseworking like Edward II said

[merry geode]

I do not know the method Edward suggested, sorry.

[tiny obsidian]

showing the preimage of basic open sets aka open balls is open

that requires justifying everything fits together nicely which to me seems annoying (extremely very much so)

[quartz horizon]

Yes you can use the universal property here

All you have to do is show the following

Continuous maps out of the increasing subset of [0, infty)^n

Naturally correspond to continuous maps out of [0, infty)^n which are S_n-equivariant

One direction is easy - given a continuous map out of [0, infty)^n which is S_n-equivariant, just restrict it to the increasing subset

The other direction is harder but also doable

Note that [0, infty)^n works as an S_n-space, i.e. S_n has a group action by homeomorphisms

So one can look at the orbit of the increasing subset of [0, infty)^n under this group action, you get a bunch of different (but homeomorphic) subsets of [0, infty)^n

A nice way to picture this is by considering the action groupoid

The objects are these homeomorphic subsets of the increasing subset of [0, infty)^n

And the arrows are the homeomorphisms induced by S_n

It being a group action means this forms a mini-category (indeed a mini-groupoid) inside Top

Given a map out of the increasing subset, you just use precomposition to get a family of maps out of the subsets

You can then check that these agree on the overlaps of the subsets (where at least two coordinates are equal)

Then just apply the gluing lemma to get a continuous map out of [0, infty)^n. By construction this is S_n-equivariant!

Also easy to check this process is natural

So that’s how I’d use universal properties to show that the quotient map is indeed a homeomorphism

The gluing lemma is crucial here

As well as considering this mini-groupoid inside Top

@scenic ridge happy to explain further if you want

[urban zinc]

Hello everyone! Topology-alg-topology has now been split into two channels. This channel will remain for point-set topology, and #alg-top-geo-top will be the new channel for discussing algebraic and geometric topology. Enjoy!

[gentle girder]

first

[fringe thorn]

second

[opaque zodiac]

countable

[urban zinc]

What's everyone's favorite finite topological space?

I feel like finite topologies probably have interesting combinatorial properties, but I've never looked into it

[opaque zodiac]

Well they're equivalent to finite pre-orders, right?

[quartz horizon]

Sierspinski space probably

Represents the topology functor

[opaque zodiac]

oh yeah I love (2,3)

[quartz horizon]

Huh..

[urban zinc]

I hadn't realized that, how so?

[gentle girder]

pseudocircle

[opaque zodiac]

Specialization pre-order determines the topology if arbitrary intersections of open sets are open

So you can recover the topology from the relation x \in cl({y})

[urban zinc]

Oh cool, I'd never heard of that

[craggy cedar]

This is a really good one

[quartz horizon]

i wonder where pointless topology would go now

[opaque zodiac]

I think classically every locale is a sublocale of a topological space

[red yoke]

You can visualize this by taking [0,∞)^min(n,m) and drawing all the hyperplanes {xi = xj}

Then the group action is generated by reflection across the hyperplanes

The subspace consisting of increasing coordinates is one of the regions cut out by all the hyperplanes, and intuitively should be homeomorphic to the quotient since walking through a mirror is the same thing as bouncing off the mirror

[quartz horizon]

yeah and I gave an argument using the universal property

[scenic ridge]

I was gone for a bit but yeah, i get the proof. Thanks!

[quartz horizon]

wait you do?

that’s surprising to me

[scenic ridge]

👍

[quartz horizon]

wow

well im happy

[scenic ridge]

it was more detailed than most discord messages so

[quartz horizon]

i feel like i would’ve given up quite quickly if i had to construct a direct homeomorphusm

[scenic ridge]

@quartz horizon do u know what darnymuckhi

was saying about hyperplanes

{xi = xj}

is this just all points where two coordinates are equal?

[quartz horizon]

yeah those are the overlaps!

[scenic ridge]

oh

ic

got it

[quartz horizon]

you need to consider those for the gluing lemma

[broken nacelle]

wrong choice

should've kept this for AT since most discussion here are for AT

[red yoke]

Are they

[broken nacelle]

are they not?

every time I check on this channel I see one of you AT wizards conjuring up some math beyond of my wildest imaginations

[alpine nest]

I think it varies, but I'm glad the splint went that way because this is the channel I've been following

[proud tulip]

Is (0, 1) open in C

[red yoke]

No

Any neighbourhood of a point should contain some non-real numbers

[scenic ridge]

no

[scenic ridge]

doesnt this also counter our hypothesis

on D being open in H

if D is open in [0, ...

[red yoke]

Yea D isn't open

But U(m) D U(n) should be

[scenic ridge]

So i need to find new way to go from D being open in [0, infinity) ^ min(m, n) / S to U(m) D U(n) being open in H?

[red yoke]

Yea

[scenic ridge]

ok i'll look into the schnatten norm thing u sent

to see if that helps

[red yoke]

Wait this is even closer https://qchu.wordpress.com/2017/03/13/singular-value-decomposition/

[scenic ridge]

what do these mean?

[red yoke]

Orthogonal group

This is for the real matrices case

Though complex is probably similar

[scenic ridge]

Hmm, this article doesnt really seem to help with our problem tho

[red yoke]

This shows H → {increasing components} is continuous

[scenic ridge]

ah

[scenic ridge]

Ok, so if we can continously get the kth highest singular value we get a continous function from H -> [0, infinity)^min(m, n)/S right?

[red yoke]

Yea

[scenic ridge]

and then

inverse of D, if open,

which is uDv

is open in H

so proof finished right

😃

[scenic ridge]

@red yoke I went over the whole thing and I'm just a little confused on the proof for the other direction

You said: Since there is a continuous map [0,∞)^min(n,m) → H → H/~, we get a map [0,∞)^min(n,m)/S → H/~

Why is the first arrow in the map [0,∞)^min(n,m) → H which is embedding as diagonal matrices continous?

And how does [0,∞)^min(n,m) → H → H/~ being continous imply [0,∞)^min(n,m)/S → H/~ this being cont?

[red yoke]

This comes from universal property of quotient

A continuous map out of [0,∞)^min(n,m) that respects the equivalence relation induces a continuous map from the quotient

[scenic ridge]

nvm, i understand this part now. I thought we were looking at the universal property of H/~

But still, how do we know the first arrow in the map [0,∞)^min(n,m) → H which is embedding as diagonal matrices continous?

[unreal gull]

how do we know that each U_i may be imbedded in R^m?

[novel acorn]

an m-dimensional manifold can be covered by open sets homeomorphic to some subset of R^m and because it's compact we can choose finitely many such open sets

[unreal gull]

this is the definition of an m-manifold in the book

\textbf{Definition - $m$-Manifold}: An \textit{$m$-manifold} is a Hausdorff space $X$ with a countable basis such that each point $x$ of $X$ has a neighbourhood that is a homeomorphic with an open subset of $\mathbb{R}^m$. A 1-manifold is often called a \textit{curve}, a 2-manifold is called a \textit{surface}.

[gentle osprey]

Henry

[unreal gull]

but this says that each point x has a neighborhood that is homeomorphic with a n open subset of R^m, not that every open set in X is homeomorphic to some subset of R^m

or am I misunderstanding something

[novel acorn]

that isn't what I said

I said it can be covered by open sets that are homeomorphic

not that every open set is homeomorphic

the idea is you take a point x
Then you know that there exists an open nbhd of x that is homeomorphic

so you take all such open sets as you vary x

and that's your cover

[unreal gull]

ahhhhh

i see

ty

[proud tulip]

If I have functions f_1, ..., f_n : X -> Y, all of which are continous

is (f_1, f_2, ..., f_n) : X -> Y^n continous? basically the new function is one where each component was one of the old ones

[limpid fern]

no this isnt actually true if i remember correctly

hold on let me come up with an example

wait

[quartz horizon]

Yes!!

[proud tulip]

?

[limpid fern]

yes its fine

[quartz horizon]

This is the universal property of the product

[limpid fern]

composition with projections

[rain ether]

recall how the product topology is defined

[limpid fern]

sorry

brainfarted

lol

[rain ether]

you're probably thinking about a similar thing where you consider component-wise continuity in the domain rather than the codomain

[proud tulip]

and this works for any x and y right

[quartz horizon]

Yep!

In fact, this gives my favourite proof of the following fact

Let $f, g : X \to \mathbb{R}$ be continuous, where $X$ is a topological space. Then $f \times g : X \to \mathbb{R}$ is continuous

[gentle osprey]

Pseudonium

[limpid fern]

codomain is R^2

[quartz horizon]

I mean pointwise multiplication

Essentially, showing that the product of continuous functions is continuous

But when the domain is a topological space and not a metric space

So $(f \times g)(x) = f(x) g(x)$

[gentle osprey]

Pseudonium

[limpid fern]

usually f x g refers to the map (f(x),g(x))

lol

but yes

[quartz horizon]

Right, I usually denote that by (f, g)

[proud tulip]

wait in my question what if the new function was X^n to Y^n, not just X -> Y^n

[limpid fern]

this doesn't save too much effort though

[proud tulip]

so each component has paramater from one X

is this still cont?

[quartz horizon]

How would you do it otherwise?

[limpid fern]

but i agree its shorter

[quartz horizon]

Yeah totally!

[limpid fern]

you still make use of the continuity of the multiplication function anyway

[quartz horizon]

Again, you can use the universal property of the product to show this

[rain ether]

If each of the components X^n -> Y is continuous, yes

[quartz horizon]

I remember when I tried to do it before, I was working directly with preimages

And it was pretty disgusting…

[proud tulip]

I only know each component function X -> Y is cont..

[limpid fern]

okay that is painful

[rain ether]

But not if you only have continuity along each direction in X^n

[rain ether]

You can try to find a counter example for this, e.g., for a function R^2 -> R

[quartz horizon]

So, if I understand you right.

Suppose you have f, g : X -> Y. The function you’re proposing is:

(x_1, x_2) -> (f(x_1), g(x_2))

Is that right?

[proud tulip]

yes

and general n

[quartz horizon]

That’s guaranteed to be continuous!

[proud tulip]

is this cont?

YAY

[quartz horizon]

And you can use the universal property of the product to show this

So in our example

[proud tulip]

also i dont know how to phrase this but

[quartz horizon]

We have two functions X^2 -> Y

The first sends (x_1, x_2) -> f(x_1)

This is continuous

We also have (x_1, x_2) -> g(x_2)

This is continuous

So you can package them together to get (x_1, x_2) -> (f(x_1), g(x_2)) as a continuous function

[proud tulip]

can u like just add an extra dimension to the codomain, and make it always constnat. does that keep it still continous?

Like f: R -> R. New function:

R->R^2: x -> (f(x), constnat)

is this cont?

[quartz horizon]

Yes, and again you can use universal property of the product

You have two functions R -> R

The first maps x -> f(x)

The second maps x -> constant

Both continuous

So you can package them together

[proud tulip]

ok ill read the universal property thing, seems it solves every problem in the world

[quartz horizon]

To get x -> (f(x), constant)

[quartz horizon]

The universal property is essentially exactly what you stated

What the product “does” is let you package together a family of continuous functions with the same domain

Into a single continuous function

Except it also works when the codomain varies

So you can have f : X -> Y and g : X -> Z continuous

And you can package them together into (f, g) : X -> Y x Z continuous

In fact, this property determines the product up to unique homeomorphism

[quartz horizon]

And also reverse to get back f and g

[proud tulip]

can i also just add an extra dimension do domain, and basically keep function same, just not care about that paramater? is this also cont?

[quartz horizon]

Could you give an example of what you mean?

[proud tulip]

As in f: R -> R the new function being

R^2 -> R
(x, y) -> f(x)

[quartz horizon]

Yep, that’s continuous!

[proud tulip]

by the universal thing?

[quartz horizon]

You can write it as a composition of continuous functions

Not actually universal this time lol

The first is the projection (x, y) -> x

Which is continuous

And the second is x -> f(x), which you already know is continuous

And then a composition of continuous functions is continuous

[proud tulip]

comprendo ty

[quartz horizon]

Universal properties in general are described by an area of maths called category theory

[scenic ridge]

All finite dimensional spaces of same dimension are equivalent right

[proud tulip]

is this topology related

[quartz horizon]

It’s related to a lot of things

But topology has a few universal properties, yeah

The universal property of the product is one

[rain ether]

What do you mean by equivalent?

[scenic ridge]

isomorphic

i guess

over same field

also

[rain ether]

yes, as vector spaces

[scenic ridge]

can this be extended to infinite dimensions

like if their bases have the same cardinality

then they are equal or something like that

[rain ether]

I do think so, yes

[quartz horizon]

Yep because

A basis is just a choice of isomorphism with a free vector space

And you can compose isomorphisms

To get isomorphisms

[red yoke]

A function R^a → C^b is continuous iff all b components are continuous

[scenic ridge]

ty

[robust drum]

Trolling the undergrads by telling them to read category theory to do point set topology is a classic

[robust drum]

Maybe you’re thinking of the box topology

In which case this fails, which explains why you have to be careful with infinite products and not just naively take products of open sets as your basis

[quartz horizon]

I mean… idk, it’s genuinely useful for a lotta things

[robust drum]

Depends why you’re doing point set topology imo… unless you’re an algebraic topologist or geometer or just general algebraist or smt category theory is probably a big waste of time

[quartz horizon]

I mean, it was genuinely useful here

In describing the universal property of the product

[robust drum]

You don’t need to describe the universal property to remember what the product topology is lol

[umbral panther]

Mentioning category theory is excessive. Emphasizing the universal property of products is the right thing to do

[robust drum]

It isn’t even particularly faster to reconstruct the topology from the universal property

Than to just learn the topology

[quartz horizon]

Right, there’s two things here

What the product topology is, and what it lets you do

Both are important

I’m not necessarily suggesting defining the product topology by its universal property

It’s pretty straightforward to just describe the topology directly in terms of opens

But i think it’s useful to explicitly say that it satisfies a universal property

[wispy veldt]

you should be careful about the type of isomorphism in infinite dimension, since in most cases, infinite dimensional spaces vector arent topologically isomorphic

[quick crane]

I think the best definition of the product topology is it's the smallest topology that makes the projection maps continuous.

which is given in terms of open sets

[quartz horizon]

Hmm I’ve never particularly liked the smallest/largest definitions for these things

[limpid fern]

it's useful for certain definitions

eg sigma algebra generated by some family of sets

[quartz horizon]

The main thing is that

When I have a topological space, I have a single topology on it

I don’t usually consider other topologies and compare whether they’re coarser or finer

[ebon galleon]

i mean

at that point you don't have a topology on it

and it's easy to show that there's a coarsest/finest topology satisfying whatever conditions you want (projections being continuous)

so then that condition gives you a reasonable choice for a topology.

Not an unreasonable process. In fact, generally it is actually how you form (co)limits in Top: You take the (co)limit of the underlying set and then give it the coarsest/finest topology that makes the (co)limit cone continuous. This makes U: Top --> Set into a topological category in the sense of Joy of Cats (although this is a fairly niche definition and competes with the other more commonly used (and not at all equivalent) meaning of topological category)

But this concept generalizes to a variety of categories. Pseudotop or pretop spaces, convergence spaces, uniform spaces, diffeological spaces, etc.

[quartz horizon]

I see…

I guess I just prefer the representability def

The coarsest/finest thing always felt a little tricky to work with iirc

Or maybe said differently

I think it’s fine as a construction

But for actually working with the topological spaces you get, I prefer the universal property

Cause obviously the universal property doesn’t tell you such a space exists

[quick crane]

no, because such spaces have a generalization of your universal property

it's very good to know which maps generate the topology

[quartz horizon]

Wdym no

Subjectively, when I was working with them, they felt tricky to work with

[quick crane]

this is very useful

it includes the universal property of the product topology

[quartz horizon]

I’m a little uncomfortable with you saying “your universal property”

It’s not like I came up with it or anything

And yes I’m aware of the weak topology

[quick crane]

true

[quartz horizon]

Hmm hang on

I think you can phrase this in terms of representability too!

I hadn’t realised this

Ok so

Let $X$ be a set

[gentle osprey]

Pseudonium

[quick crane]

Actually, this is exactly the universal property of the weak topology

[quartz horizon]

And let $f : X \to Y$ be a function

[gentle osprey]

Pseudonium

[quartz horizon]

And $Y$ a topological space

[gentle osprey]

Pseudonium

[quartz horizon]

Then, you can try to topologise $X$ as follows

[gentle osprey]

Pseudonium

[quartz horizon]

You want Top(-, X) to be naturally isomorphic to the following functor

It takes in a topological space Z, and spits out the set of functions $g : Z \to X$ such that $f \circ g$ is continuous

[gentle osprey]

Pseudonium

[quartz horizon]

And actually yes this should work with a family of functions $f_i : X \to Y_i$

[gentle osprey]

Pseudonium

[merry geode]

Sorry to intervene, but how does the commutative diagram looks like?

[quartz horizon]

So essentially - you want some topological space TX which represents a particular subfunctor of Set(-, X)

That’s very cool!

And - you can construct such a topological space by taking the coarsest topology on X making all the f_i continuous

And final topologies are about wanting a topological space TX which represents a particular subfunctor of Set(X, -)

And you can construct such a topological space by taking the finest topology on X making all the f_i continuous

That is very very cool

Now I know how initial/final topologies work categorically

[quartz horizon]

So did you understand what I just wrote out

[quick crane]

I found this on wikipedia:

[quartz horizon]

Yep yep exactly

[merry geode]

Ah, initial topology

[quartz horizon]

Little strange they don’t call it a universal property though

[merry geode]

Ah wait, so weak topology is a initial topology here, right?

[quartz horizon]

Yep

Quotient topology is a final one

[quick crane]

yea same thing

[quartz horizon]

Indiscrete is like a degenerate case of an initial topology

[merry geode]

Different terminologies confuse me time to time

[quartz horizon]

Discrete is a degenerate case of a final one

That’s cause of the adjunction

Ok so now I can say what’s properly happening categorically

Let $X$ be a set, $(Y_i)_{i \in I}$ a family of topological spaces, and $f_i : X \to Y_i$ a family of set-functions

[gentle osprey]

Pseudonium

[merry geode]

So this means you only need to look at behavior of sequences N -> X, N -> Yi.

[quartz horizon]

We can define a functor $\mathbf{Top}^{\text{op} } \to \mathbf{Set}$

[gentle osprey]

Pseudonium

[quartz horizon]

Sending a topological space Z to the set of functions U(Z) -> X such that every composite with f_i lifts to a continuous function Z -> Y_i

This is a subfunctor of $\mathbf{Set}(U(-), X)$

[gentle osprey]

Pseudonium

[quartz horizon]

The initial topology on X is what represents this functor

And can be constructed as the coarsest topology making all the f_i continuous

So in general, initial topologies should be about representing particular subfunctors of Set(U(-), X)

And final topologies, dually, are about representing particular subfunctors of Set(X, U(-))

And also have a construction - the finest topology making the f_I continuous

[scenic ridge]

do people use ≅ to say two topological spaces are homeomorphic?

[limpid fern]

sure

[fringe thorn]

sometimes, yes

[unreal gull]

is that not the most standard notation?

[fringe thorn]

I believe it is, but I often see no notation at all

[tender halo]

I just use =

[night heron]

That’s crazy

[opaque zodiac]

til R = (0,1)

[robust drum]

It’s extremely common to just use = in my experience

Probably not for point set stuff but the moment you start actually using topology for stuff

[night heron]

Might I ask where?

[robust drum]

I see it a lot in low dimensional topology

Eg: annuli are always equal to S^1 x I

Not using the homeo sign

Tori are equal to S^1 x S^1

Etc etc

[night heron]

Weird, we used ≅ whenever we talked about homeomorphisms. Even in complex analysis where topology was only loosely referenced

[quartz horizon]

Yeah I’ve seen $\cong$ used more

[gentle osprey]

Pseudonium

[robust drum]

I think that the more frequently topology is used the looser people get with identifying homeo things not the less it’s mentioned

There’s also nothing wrong with identifying homeo things honestly

[opaque zodiac]

Well I mean it depends on context. This is the same as the way people talk about groups. They'll say that some subgroup is Z/4Z for instance. But on the other hand nobody but the HoTTiest of HoTT-heads are writing (R,+) = (R_+, *)

[night heron]

Yes