point-set-topology

[honest narwhal]

If you're more advanced, which may or may not be the case depending on whether my read of "Homotopy stuff" is accurate, then I'm not sure. The Concise books are supposed to be pretty but lacking in geometric stuff for those who are into it

[shadow ermine]

xD ty<3

I took a module called Geometry (metrics) & Topology

and now I'm taking algebraic topology (final year) if that's a good indication on how little I know xD

[steel needle]

concise is scary for first course

rotman is good

hatcher is standard

[honest narwhal]

Okay yeah in that case go for Rotman. I was just thinking that you might've meant now advanced stuff if you talked about homotopy theory, in which case maybe More Concise or Davis/Kirk or something would've been good

[keen moth]

I used concise in my class for a week, didn't help anyone tbh

[pliant dragon]

I guess more of a soft question, but what's the cluster point as n goes to infinity?

IE the cluster point of {1,2,3, .... n-1, n}

[steel needle]

what's a cluster point

@pliant dragon

do you mean an accumulation point in some bigger space?

[gritty widget]

Hey guys, if I have (0, 15) U (20,infty), would my interior points be (0,15)U(20,infty) or (0,infty)?

[midnight jewel]

the former

16 is not even in that set

so it could not be an interior point

as a general rule, the union of (finitely many!) open sets is open

and for an open set, the interior points are the set itself

[pliant dragon]

@steel needle I guess limit point would be a similar definition

[gritty widget]

thanks @midnight jewel

[midnight jewel]

you know @steel needle this is actually more obnoxious than whatever is even going on

now all of my channels have unread messages because o fyou

there is for whole server which is good enough for me since it’s currently all freakout anyway

what even happened? a guy I know told me they were kicked here

[loud heart]

What changes and what remains the same if we equip every group with the discrete topology?

[small obsidian]

Changes and stays the same between the discrete and what?

What's the other option?

[loud heart]

compared to leaving the groups as is (without any additional structure)

[small obsidian]

You are legit talking about algebraic groups, oh gotchya

If you can relate the topology to the group structure, then you've gained a lot.

If the topology doesn't do anything with the group, then you have at least gained the theorems behind topology.

Discrete topologies capture absolutely nothing though

[loud heart]

ah cool

[torn spade]

Im trying to find a proof for a set S being closed iff the set equals its closure

[keen cliff]

what's your definition of closure

[torn spade]

but the definition of closure i have is: $ \overline{\rm S} = S \cup \partial S $

[gentle osprey]

NTK:

[keen cliff]

and boundary?

[torn spade]

is the collection of boundary points, which in turn are are points so that every neighborhood $ U(x,\epsilon ) $ contains at least one point of set S and one point of the complement of set S

[gentle osprey]

NTK:

[torn spade]

so ive been looking it up for a while and most proofs just use "the closure of a set S is the smallest closed set so that S is its subset"

[keen cliff]

well the goal will be to show that a closed set contains its limit points then

cuz then the boundary of S will be a subset of S

[torn spade]

agh im not sure have i seen what limit points mean, we only talked about accumulation points

[keen cliff]

x is a limit point of S if there exists a sequence in S that has limit x

or

could be the same thing

[torn spade]

no i think im just not supposed to know this yet

i mean, what a limit point is

[keen cliff]

lame

[torn spade]

lol

like i dont understand what "exists a sequence in S that has limit x" means

"Given a set S, we say a point P is a limit point of S if every neighborhood of P (not including the point P itself) always contains points in S, no matter how small the neighborhood is."

this is what i saw online. but this is the definition i have for accumulation points

[keen cliff]

ok then they're the same thing

[torn spade]

alright

lets see then

[keen cliff]

perhaps try contrapositive?

[torn spade]

whats that?

oops ok, language barrier hit me there lol

[steel needle]

the definitions have subtle differences sometimes

make sure to use the ones you are using in class

[torn spade]

Ok i fell asleep and now im back at doing this

So to restate: set S is closed iff it equals its closure

Im trying to work this from left to right first

So assuming that S is closed, i'll try to prove that it equals its closure.

[frigid patrol]

Closure is the smallest closed set containing the set

[torn spade]

we haven't had that definition in our lectures/slides so im not going to use it, i'd need to prove it first anyways @frigid patrol

but i got a proof together, someone can review it

[gentle osprey]

NTK:

[torn spade]

really took me 1.5 hours to get to this point jesus

[frigid patrol]

Why is it so long

What's the definition you're using

Ok I see

[torn spade]

Yea thats why p much

I saw proofs that used your definition but as i said, i had to/wanted to use this

• i couldve left some words out of there but this is my literal 3rd week in real analysis so i want it to be clear as possible mainly for my own sake

[frigid patrol]

It's only long because you explicitly wrote it all out

That's good

[torn spade]

Yeah

The fact that it took me 1.5h to do this tells smth about my knowledge and experience in this topic lol

[frigid patrol]

But now you know this and it's trivial

That's how math works

[torn spade]

Yeh lol

[gritty widget]

hey guys can anyone verify this is a correct proof?

Statement: Let $f: S^1 \to S^1$ be an even map, that is, $f(z) = f(-z)$. Then, $\text{deg}(f)$ is even

proof: conside the lift $\tilde{f}: I\to S^1$ where $I = [0, 1]$. Then, $\tilde{f}|{[0, 1/2]}$ and $\tilde{f}|{[1/2,1]}$ are equivalent maps that can be reparmeterized to maps $I\to S^1$. So, their composition is the sum of the degrees and so is even

[gentle osprey]

neolithic:

neolithic:

[steel needle]

looks fine

more than the sum of degrees I would say twice the degree of any of them

although that's implied

[midnight jewel]

Well, so far topology class is rather boring (we're proving, at a ridiculosuly slow speed, that the ε-δ notion of continuity on R from Analysis is equivalent to the open sets one we want to look at in this class, we've already done that in analysis)

[frigid patrol]

Yea

[midnight jewel]

Intuitively, a topology being finer than another defines a partial order. Can we use Zorn to find a "maximally fine nontrivial (ie not discrete)" topology on any set?

I assume so, it seems very similar to the proof of existence of maximal ideals

[gritty widget]

yes

apparently, they are connected to ultrafilters

https://www.jstor.org/stable/44236425

[frigid patrol]

Ultafilters!

[smoky meteor]

Can I have a bit of help understanding the K-Topology $\mathbb{R}_K$? I understand its basis $\mathcal{B}$ containing all open sets $(a,b)$ for $a,b\in\mathbb{R}$, but what do the sets of the form $(a,b)-K$ look like? I'm not even sure what that notation is describing tbh

[gentle osprey]

em:

[steel needle]

post context

[smoky meteor]

Er, where $K={\frac{1}{n} : n\in \mathbb{Z}_+$

[gentle osprey]

em:

[steel needle]

post definition of R_K

[smoky meteor]

Munkres, top of page 82

[steel needle]

ok so its the topology with basis (a,b) and (a,b) - K

[smoky meteor]

"Let K denote the set of all numbers of form (above), let B be the collection of all open intervals (a,b) along with all sets of form (a,b)-K, This topology generated is the K-Topology"

yea

sorry

[steel needle]

ask yourself this

what happens if (a,b) doesn't contain 0?

what does (a,b) - K look like?

is it anything new?

what happens if it does contain 0?

[smoky meteor]

I'm sorry, I don't understand specifically what (a,b)-K is. Is it all of the open intervals (a-k,b-k) for all elements k\in K?

[steel needle]

it's just set difference

(a,b) minus the points in K

[smoky meteor]

🤔 so we just remove all reciprocals of naturals from every interval

[steel needle]

yeah

[smoky meteor]

ohhh

gimme a minute to think about it

so, if your interval contains zero, you get a ton of lil holes in the neighborhood of zero

on the positive side

[steel needle]

yeah if your interval contains 0 you get something that isn't open in the usual topology

because it has 0 and no open interval around it

[smoky meteor]

does a open interval have to have every point in it

so like, (-1,3) but without the element 2 isn't an open interval?

[steel needle]

no, but it's an open set

[smoky meteor]

ye

[steel needle]

cuz around every point in the interval you can find a smaller interval

so it's an open of the standard topology

[smoky meteor]

riight. so you can't find one that contains zero in the (a,b)-K because you literally can't end on the pos side of zero

neato

one: open interval, not open set

(a,b)-K are all open sets

sorry im new at this

[steel needle]

I mean

I meant open set in the standard topology

not in the new one

but I think you get the idea

[smoky meteor]

oki

[steel needle]

we have some new open sets whijch are the ones that contain 0

and no intervals around it

[lucid turret]

Could anyone help me with adjunction spaces?

[keen moth]

@lucid turret what's an adjunction space?

[gritty widget]

@keen moth https://en.m.wikipedia.org/wiki/Adjunction_space

@lucid turret your question is vague. Please make clear what you'd like help with concerning adjunction spaces

[lucid turret]

Is it true that spaces (X coprod Y) x I / (a,t) ~ (f(a),t) and (X cup_f Y) x I are homeomorphic? f:A to Y is continuous map, A subspace of X. cup_f is adjoint space

Coprod meaning disjoint union

[fierce valley]

no

[lucid turret]

Wot about X/~ x I and (X x I)/~ ? We identify (x,t)'s with same t's only based on identification in X

[fierce valley]

no

[lucid turret]

Rip. Why tho

[fierce valley]

because im a three time fields medalist thats why

listen to what im saying

you speak too much

[lucid turret]

Ok

Anybody else? 😂

[gritty widget]

@lucid turret couldn't find "adjoint space" concerning general topology on internet or some books

how do you define X \cup_f Y?

btw Munkres seems to use both (X coprod Y) x I / (a,t) ~ (f(a),t) and (X cup_f Y) x I as being by definition the same thing

Pic taken from Munkres' Elements of Algebraic Topology, Chapter 4 page 210

Also don't care about what Divinité Hantée said. He's a troll and was banned

[frigid patrol]

What's the question

[gritty widget]

@frigid patrol Maxwell's question (scroll up):
"Is it true that spaces (X coprod Y) x I / (a,t) ~ (f(a),t) and (X cup_f Y) x I are homeomorphic? f:A to Y is continuous map, A subspace of X. cup_f is adjoint space"

[mighty needle]

Just to make sure I'm not losing my mind, the quotient space (S^1xS^1)/(S^1x{pt}) is homeomorphic to just S^1, correct?

pt is a point on S^1

[steel needle]

no

look at it like this

S1 x S1 is a torus

then you take a single circle S1 x pt

and squish it

it's gonna be a torus with a single circle squished into a point

[mighty needle]

hmm, okay I'll think about that more then.

But the fact that what I said was incorrect actually makes my life easier, so that's good. Thanks!

[midnight jewel]

This is probably well-known (it’s e.g. found in Proofs from the Book), but a really nice proof of the infinitude of primes that was on our homework sheet today. Needs only very basic notions of point set topology (essentially definitions of open & closed sets and bases of a topology)

[keen moth]

This was assigned for my topology homework when I took it a few years ago. Now that same professor is teaching point-set again this semester

@midnight jewel do you go to a university in chicago?

If so, there's a good chance we go to the same school

[midnight jewel]

never stepped foot in North America

[midnight jewel]

also @keen moth as I mentioned, this proof is well-known

and cool

so it’s not surprising to see it elsewhere

[frigid patrol]

@midnight jewel what's the proof for c

[midnight jewel]

If the union was finite, then it would be a closed set, since finite unions of closed sets are closed.
However, then its complement would be open.
Its complement is {-1,1}
But all open sets in this topology are infinite

thus, it must be an infinite union

(the complement is {-1, 1} because every other number is either prime or has a prime divisor by definition of what a prime is)

[frigid patrol]

Ah

That's a very cool argument

[frigid patrol]

Euclid's argument is better

It's actually the best kind of argument:
Not technical,
Short,
Intuitive

[midnight jewel]

Is it though? In Euclid we build a secuence of primes which must be infinite.
In this proof we show that the sieve of erastothenes cannot be generated by finitely many numbers using a topological contradiction

I don't really see the connection

Note, the version of Euclid's proof I'm referring to is the (imo simpler) contradictionless one, that is:

Given a finite list P of primes, take their product and add one. Then this number must have a prime factor not in P. Add that factor to the list.

(fun fact: it is unknown whether this algorithm generates all primes)

[midnight jewel]

you misunderstood the algorithm, then

you don’t take the next number as a prime

you state that it has a prime factor not in the list

(in fact it only has prime factors not in the list, but one’s enough)

the actual sequence I was referring to that we don’t know if it generates all primes specifically always takes the smallest one

https://oeis.org/A000945 this sequence

[austere meadow]

Hello, I am interested in learning more about this topology thingy, I've had recently a large assignment about it, but it's really a new thing for me, is it ok if i ask what's a, uh, ball, in a metric topology? I've been asked to draw a ball of radius 1, and uh, i really don't know what it is really (i'm not really asking for help, just to know more about what's a ball in this particular case, since its something new for me, if i need to go to a #help-1 like channel, then i will :) )

[midnight jewel]

a ball of radius 1 is the set of all points which have distance <1 from a given point. Distance being determined by the metric function of your metric space

e.g. in R with the standard metric it's just open intervals of length 2 with the chosen point in the middle, in R^2 with the standard metric it's circles of radius 1, but with other metrics it's other shapes

e.g. a diamond (1-norm) or a set of rays emanating from 0 (SNCF metric)

[warm hedge]

What's the difference between a metric and a norm?

[west ocean]

Definition.

If you have a norm then you can build/have a metric.

[west spindle]

a norm is something you can put on a real or complex vector space

metric is a more general concept

however, if you have a normed vector space then the natural metric to put on it is d(x,y) = ||x - y||

[midnight jewel]

let $X$ be your space

a norm is a function $X \to [0,\infty)$ which gives every element in $X$ a magnitude $||x||$

a metric is a function $X^2 \to [0, \infty)$ which gives every pair of elements in $X$ a distance $d(x,y)$

you can always make a metric out of a norm by defining $d(x,y) = ||x-y||$

[gentle osprey]

Sascha Baer:

[midnight jewel]

both have to fulfill a certain number of properties of course

for one, norms pretty much only make sense on vector spaces

[west spindle]

\| or \Vert for norm bars 😛

[midnight jewel]

thanks

[west spindle]

compare: $||x|| ||y||$ vs $\Vert x \Vert \Vert y \Vert$

[gentle osprey]

Ann:

[midnight jewel]

they have to fulfill
\begin{enumerate}
\item $|x| = 0$ if and only if $x = 0$ (positive definite)
\item $|\alpha x| = |\alpha| |x|$ (homogenous)
\item $|x + y| \leq |x| + |y|$ (triangle inequality)
\end{enumerate}

fuck wrong dollar signs

[west spindle]

single dollars 😛

[gentle osprey]

Sascha Baer:

[west spindle]

and #2 is more properly called absolutely homogeneous

or absolute homogeneity

bc homogeneous w/o qualifiers would be ||ax|| = a||x||

though of course we all know the triangle inequality is the most important 😛

[midnight jewel]

metrix meanwhile have to fulfill
\begin{itemize}
\item $d(x,y) = d(y,x)$ (symmetric)
\item $d(x,y) = 0$ if and only if $x=y$ (positive definite)
\item $d(x,z) \leq(x,y) + d(y,z)$ (triangle inequality)
\end{itemize}

…

brain

[west spindle]

wrong way around in the triangle ineq

[gentle osprey]

Sascha Baer:

[midnight jewel]

I noticed

[west spindle]

space between \leq and d

[midnight jewel]

ann please

you don’t have to tell me everything

[west spindle]

the bot sees \leqd and doesn't know what to do with it 😛

[midnight jewel]

I do have a brain

[west spindle]

ok

no worries

[west ocean]

👀.

[warm hedge]

Thank you for the answers

[west ocean]

You are welcome.

[loud heart]

Can you deformation retract a square to two perpendicular line segments intersecting at a point?

[west spindle]

yes

take the diagonals of the square

[loud heart]

what if the line segments intersect off center?

[west spindle]

i don't see how that changes anything? the space you're retracting onto is homeomorphic to the union of the square's diagonals

[random slate]

Can I potentially get a critique on whether this proof I think I'm done with seems correct?

https://snag.gy/YhDpXr.jpg

[fervent citrus]

What do $\bigcap A$ and $\bigcup A$ mean?

[gentle osprey]

Tuong:

[random slate]

The intersection and union, respectively, of all the sets in $\mathcal{A}$

[gentle osprey]

DMAshura:

[fervent citrus]

How would you justify with more details that the intersection of all of the sets in A is indeed the element of A indexed by min(I)?

It feels like you wanted to talk about min(A), the minimum of A for the inclusion relation

[random slate]

I'm not really sure how I could show that in more detail. Like it seems obvious from how the sets were constructed. :/ If A_n contains "everything up to n" and you intersect two of them, you should have the one with fewer elements.

Like at what point do you assume that something like that is apparent rather than having to prove something that should seem basic

And I've never heard of using "min" for an inclusion relationship, though I suppose that makes sense

...and I suppose I could note that $A_0 \subseteq A_1 \subseteq A_2 \subseteq \cdots \subseteq A_\infty$.

[gentle osprey]

DMAshura:

[fervent citrus]

Also, would max(I) still be infinity even if infinity isn't in I?

*in the case I is unbounded

when you write A_max(I), it feels like it's an element of A (because max(I) should be an element of I)

if A was everything but N, the union would be N but wouldn't be an element of A

Oops nevermind

'misread something

[timber pasture]

these proofs are perfectly adequate

[random slate]

@timber pasture -- do I not need to be more detailed in how I know that the intersections and unions are mins/maxes?

[timber pasture]

depends on who you're writing it for

if you wanted to be safe you could note that WLOG Ai1 subseteq Ai2 subseteq ...

or even just scrA is totally ordered by subseteq

[fervent citrus]

Actually I'm still puzzled, is max(I) an element of I?

[random slate]

This is for graduate level, but technically the class is Intro to Proofs and we're just looking at topology as a way of practicing proof writing with an unfamiliar subject

[timber pasture]

no, it's the max in (N union {inf}, <=)

grad level mathematicians or grad level not-mathematicians?

[random slate]

As in I'm taking a graduate level class

And yeah, shit, I see that the max may be an issue now

[timber pasture]

just replace it with sup

[random slate]

. . . hah. I could.

[timber pasture]

ez clap

[random slate]

I didn't even think of that.

That makes a ton of sense though.

[timber pasture]

i did it in my head mentally when i read the proof

because i breathe lattices

[random slate]

So I could use min and sup

Min is fine because we know there's a least element

[timber pasture]

i guess if you're being specifically marked on your carefulness then you should pay a little more lip service to the nature of the min and the sup here

[random slate]

The only thing that makes me wary is how I've written $\mathcal{A}=\left{A_{i_1},A_{i_2},A_{i_3},\ldots\right}$

[gentle osprey]

DMAshura:

[random slate]

Because it makes it seem like $\mathcal{A}$ has to be infinite

[gentle osprey]

DMAshura:

[random slate]

And it might be finite

But yeah I can definitely be more explicit

And the subset inclusion really helps me organize how I'd say it, so thanks.

[gentle osprey]

tubular:

tubular:

[timber pasture]

in the (2) version the subset-chain thing would be written like "A is totally ordered by the subseteq relation, that is, for any a,b in A either a subseteq b or b subseteq a"

you would probably find the (1) version more convenient though when you do that you should explicitly write $\bigcap_{i\in I}A_i$ instead of $\bigcap\mathcal A$ so as to be less confusing

[gentle osprey]

tubular:

[random slate]

Interesting

I see why that helps with the index set, that's much cleaner

[timber pasture]

in your case it makes sense because it's a good way to state that everything takes some particular form

and while it is applicalbe in general, you can sometimes be overcomplicating proofs if you use method (1) when method (2) could have sufficed

[random slate]

https://snag.gy/of0n4G.jpg Here's my revision for (a)

I feel a lot better about this

[honest narwhal]

Damn you actually write to be understood

:0

[random slate]

Thank you. That's absolutely something I care about, so it means a lot.

[honest narwhal]

This is in... sharp contrast to how I write

I'm lazy af

I remember in my measure theory class someone turned in a 6 page pset while mine was a page, we ended up with the same grade and the guy was like wuuuuut

[random slate]

Hah wow

[honest narwhal]

Recently it has started to change since I now email all psets so I'm less allergic to having equations on their own line

(When I had to print them out I would have each of my solutions be a single massive paragraph)

[random slate]

I wish I could understand any of that!

😛

[honest narwhal]

A Baire-1 function is a pointwise limit of continuous functions

[random slate]

Is there a special name for a topology that permits arbitrary intersections of open sets?

[honest narwhal]

Alexandroff space

[random slate]

. . . of course it's named after some guy

[honest narwhal]

They're actually kinda cool, the data of an Alexandroff space is the same as that of a... Pre-order?

If you assume the space is T_0, then it's the same as the data of a partial order

[random slate]

Oh hey that actually makes sense given my examples!

Since my open sets are totally ordered!

[honest narwhal]

Something like that. Let's restrict to the T_0 case since I'm happier with partial orderings

So you have an Alexandroff space and define U_x to be the intersection of open sets containing x. That's open by assumption

And then you say x≤y if U_x\subset U_y

And this is our pre-ordering. The condition that the space is T_0 is precisely the statement that this is a partial order: x≤y and y≤x => x=y

[timber pasture]

i was the short proof guy in my classes

and i had a friend a year above who was the 12 page guy

[random slate]

I would rather err on the side if verbosity if it means clarity

[steel needle]

I live on conciseness on the edge of unreadability

I routinely get taken 10% off because my stuff is too concise

AMA

[celest sparrow]

how much do you get taken off because you're being too concise

[random slate]

You might have a future in writing real analysis books then

[timber pasture]

i try to be clear but i work a lot at paring down my arguments once they're on the page

[frigid patrol]

I write long homeworks, hand written and stapled 6 pages

I'm old fashioned that way

None of that latex business

[steel needle]

6 pages isnt long

my psets nowadays are regularly around 6 pages

and I'm super concise

:(

also handwritten gross

[honest narwhal]

You're less concise than me lmao. I think my TAs just end up being thankful that my psets are quick to grade, also I choose the stuff I don't really prove so that if you read my statement it sounds right in your head

It's a narrative of true statements where you're like oh this implies this implies that yeah okay

Even though "implies that" requires a ton of justification

[honest narwhal]

Except my algebra TA

He knocked me so fucking hard

[keen moth]

I require my students to explain why 3n(n+1) is divisible by six if they haven't done so on a previous assignment

[midnight jewel]

if I write a longer thing I tend to factor out certain lemmas n stuff and do them before or after the actual main proof

like just mark an implication with a star and then at the end be like “proof of star”

so I can focus on the big picture first

[steel needle]

the solution I've found to my nagging TA who demands proof of obvious shit is to use footnotes

which is simultaneously clear and slightly passive aggressive

[midnight jewel]

until you want to put a footnote after an equation and notice that it looks like this: https://xkcd.com/1184/

xkcd: Circumference Formula

Assume r' refers to the radius of Earth Prime, and r'' means radius in inches.

[steel needle]

lmfao

never on an equation, usually
[formula]
as desired^1.

[midnight jewel]

you know what annoys me in latex though?

you have to do
$$equation,$$
as desired.

rather than
$$equation$$,
as desired.

…actually I could’ve left that for illustrative purposes I guess

but it was a bit big

[steel needle]

oh I wouldn't know, I don't use $$ $$

[midnight jewel]

it makes sense why you have to do it, but it feels so wrong to put punctuation into an equation

[steel needle]

I use \begin{align}

[midnight jewel]

yea but there too you have to put it into the environment

don’t you

[steel needle]

yeah but then it makes sense

you wouldn't try to put it after

[midnight jewel]

well, I don’t want to put punctuation into mathmode

you don’t put text into mathmode

[steel needle]

i usually put it in mathmode

but you could do \comma = \text{,}

[midnight jewel]

you have to for correct typesetting, but yea if you have font fuckery it might look wrong and you have to do that

otherwise the comma lands on the next line and looks bad :P

[steel needle]

i usually don't put comma between formula and "as desired"

[midnight jewel]

but grammar!

[steel needle]

is it bad?

"We get x as desired" seems fine

[midnight jewel]

also I tend to write my stuff in German, where that would be required

German requires a comma before subordinate clauses, always

(unless they begin the sentence, ofc)

[steel needle]

oh

[midnight jewel]

but even in English, I would want to put one

now, while we’re on the topic of topology

I’m currently stuck on a problem :P

I have to show that $\overline{A} \times \overline{B} = \overline{A \times B}$. I’ve already shown $\supseteq$

[gentle osprey]

Sascha Baer:

[midnight jewel]

but I don’t really know where to start on the other direction

I’ve also shwon that the product of two closed sets is closed, but apparently I was acutally supposed to conclude that from this exercise… I needed it in the other part though

there I considered the closure as the intersection of all closed supersets containing A×B

and could show it with only one hand-waved equation

which I’m actually not quite sure about

is this statement true?
$$\bigcap_{x} A_x \times \bigcap_{y} B_y = \bigcap_{x, y} A_x \times B_y$$
or if not equality then at least $\supseteq$?

(x and y range over some index set)

[gentle osprey]

Sascha Baer:

[midnight jewel]

should be true I think, yea

but anyway, that’s not really the point here; I think I’ve unhandwaved that now

I just don’t really know where to start to show $\overline{A} \times \overline{B} \subseteq \overline{A \times B}$

[gentle osprey]

Sascha Baer:

[timber pasture]

don't use $$-$$, use \[-\]

[midnight jewel]

honestly, I hate those

I use either $$ or \begin{align/align*/equation} depending on what I need

there’s really no reason to prefer one over the other apart from aesthetics

and \(, \[ are just … weird

[timber pasture]

the double dollar signs are archaic tex and they will not play nicely with the formatting extensions latex adds to the typesetting system

[midnight jewel]

never ran into issues

I know they’re tex

literally 100% of my math formatting issues are the align environment adding too much whitespace

anyway I had a topology question so latex’s not really what I wanna discuss here now

[west spindle]

cl(A) x cl(B) subset of cl(A x B)? is that what you wanted to show?

[midnight jewel]

yes

[west spindle]

hmmm

[midnight jewel]

I need to show equality, I’ve already shown subset in the other direction

[west spindle]

no assumptions on the spaces A and B are subsets of?

[midnight jewel]

I know only a few basic facts about the product topology

just topological spaces

[west spindle]

ok

i remember doing this

i think

[midnight jewel]

pretty much all I know about the product topology is that all the pairs O₁×O₂ where O₁ open in X and O₂ open in Y form a basis to the product topology

[west spindle]

yeah ok so mm

let's see

let (a,b) in cl(A) x cl(B)

thus a in cl(A) and b in cl(B)

[midnight jewel]

the only idea I had was I could try and do a case distinction depending on whether a, b are in the boundary or the interior

but that turns ugly

cause it’s three cases to consider

[west spindle]

for every open set U in X containing a, U intersect A is nonempty

[midnight jewel]

one is trivial though

[west spindle]

for every open set V in Y containing b, V intersect B is nonempty

[midnight jewel]

that uses nothing about closure does it, that’s just basic set theory, right?

or am I missing sth

[west spindle]

so far it's unfolding the defn of closure

for every open set $W \subseteq X \times Y$ that contains $(a,b)$, there exists an open rectangle $U_0 \times V_0$ s.t. $(a,b) \in U_0 \times V_0 \subseteq W$

[gentle osprey]

Ann:

[midnight jewel]

(I’m gonna go catch my tram but I’ll read what you’re writing dw if I’m not responding for a bit)

[west spindle]

but $(U_0 \times V_0) \cap (A \times B) = (U_0 \cap A) \times (V_0 \cap B) \neq \varnothing$

[gentle osprey]

Ann:

[west spindle]

therefore we have that $W \cap (A \times B) \neq \varnothing$

[gentle osprey]

Ann:

[west spindle]

for any open set W

therefore (a,b) in cl(A x B)

et voilà

[midnight jewel]

uh, that "any" should need some qualifier for sure

since this is eg definitely wrong foe W the empty set

[west spindle]

for every W that contains (a,b)

can't contain an element and be empty

eh

[midnight jewel]

oh I didn't see you introduce W earlier

ill ponder over it

thanks

[west spindle]

yw

[bold path]

Can I induce all Riemannian metrics by Euclidean metrics?

[steel needle]

Yes

Nash embedding thm

[bold path]

damn i really should have checked wikipedia given thats what i was looking for, it literally says it in the first line

thanks

[steel needle]

👌

[gilded shell]

For any set S does there exist a topological space Y they satisfies this?
We write X, for a topological space with base set S. We write C(X,Y) for the of continuous functions from X to Y.

For any set S does there exist a function F:X-> C( X,Y) is injective? So basically where the set of continuous functions determines the topology?

Maybe this trivially true by choosing some discrete topology or by some characteristic function but I'm not seeing it.

[steel needle]

just choose Y = X and F(x) = f: X -> {x}

i don't see how it determines the topology in any way

oh are you asking for F continuous?

then I guess it kinda does

no it kinda does determine the topology

yes it does

[gilded shell]

X is a variable, Y is fixed

[steel needle]

it's continuous too

[gilded shell]

But you mean the set

[steel needle]

yeah I mean the set

[gilded shell]

The set has no topology tough

[steel needle]

well if you want to vary X then you need the coarsest topology

uhh or the finest one?

it's either the trivial or the discrete topology

Y = S with one of those

[gilded shell]

That computes

[steel needle]

the trivial one I think

[gilded shell]

Yes

[steel needle]

so if you want a fixed Y for all X then you need the trivial topology

and this doesn't say anything

it's more interesting to fix X

Y discrete doesn't work i think

unless I have it backwards

Y discrete is bad because there's few functions in C(X,Y)

in particular if Y is discrete and X is trivial then only the identity is continuous

and you do not have an injection X -> C(X,Y)

no

cuz that's the point

it does

it's a continuous injection

you want it to be continuous as well

otherwise it doesn't make sense

yes

I assume so

but yeah you don't get anything

ifyou allow X to vary

you only get information on the set S

maybe you get that X has to be finer than Y

if the injection is reasonable

but dunno

seems like an odd question anyway

[gilded shell]

Yeah, i messed up the phrasing, sorry. Meant to say X|->C(X,Y) which is very different but oh well nvm. Thanks tough

[wind sierra]

Just a question about the name of a mathematical... object? More like a concept.

Anyway, the metric that tells you the least number of squares a rook must go through on a chessboard to go from one square to another one is called the taxicab metric. Is there a name for the metric that does the same, but on a hexagonal grid? I mean, if such a metric even exists.

[small obsidian]

Yeah, you could definitely make such a metric. Why do you want it?

[steel needle]

you couldn't

it's not linear

you can't tile a hexagon with hexagons

in R^2 i mean

if you mean as a distance function for a discrete set of points then sure

[wind sierra]

@small obsidian Well, I was wondering if such a metric had a name, just like the taxicab metric is called... the taxicab metric.

[steel needle]

the taxicab metric has a name cuz it's a metric on R^2

[small obsidian]

Wat? You can tile hexagons

[steel needle]

and I guess it's convenient

you can't tile a hexagon with hexagons

so linearity won't work

you can check it yourself as an exercise

try to define a metric in R^2 based on the hexagonal tiling

[wind sierra]

So, umm... does it have a special name or not?

[eager sage]

you can tile a triangle with triangles, which is very similar and very interesting as well

[small obsidian]

I'm confused. Which axiom doesn't hold? What's linearity?

[eager sage]

triangles and hexagons are similar in that the centers of hexagons in a hexagonal grid form a triangular grid

the way i'd start trying to figure out a metric for the hexagonal grid is to travel along the borders of triangles in a triangular grid

[steel needle]

The point is not every function is a metric

The triangle thing might work

I dont see obvious problems

[eager sage]

the question I have as a chess player is, on a hexagonal grid what precisely differentiates a rook from a bishop? the fact that each intersection of tiles has only 3 tiles around it means that you can't have "diagonals" in the same way as on a chessboard

motion on a hexagonal board is really interesting though, regardless

actually, this is an interesting way to demonstrate a weird type of 60-degree-grid orthogonality

bishops would be significantly less powerful on a board like this, than on a square chessboard, but it's a thought at least

[wind sierra]

Just to make things clear, I'm talking about an infinite hexagonal grid on a plan, in which you stand at the center of a hexagon and can move to another one only if they share a common edge. Like I said, like chess but on a different kind of board.

I know that rooks on a chessboard are only one of the many ways to visualize the taxicab metric, but I need to know the name for the metric that you can get on this type of grid instead. If such a metric exists at all, I mean, it seems to me that you're still debating that.

[eager sage]

i believe it should be possible for a metric to exist but i don't know if there's a name for it

[wind sierra]

Oh. I assumed this problem was already a well-studied one, so I thought someone had to have given it a name.

[warm junco]

oof.... if i played chess on that kind of board, if you let rooks move in any straight line along cells which have adjacent edges, they become queen like.... obviously that isn't good.... and you cant go diagonal

but there's 2 solutions

one is that rooks can move along "straight" lines of precisely two colors... don't know what this would mean for bishops tbh...

but the simpler and much more reasonable solution is this

in a square

you have two bidirectional motions

that are perpindicular to each cell edge

and diagonal, which traverse along the lines which connect corners....

How about a rook may move in one of 3 directions, defined as perhaps the white on black or grey on black as i believe any rook can cross any color but they would have different motions on a black tile

as for the bishop

what would one do with a bishop? Well just as squares have those diagonal bisecting directions, a hexagon has 3 bisecting directions

and with 3 tile colors

you are left with a grey bishop

a white bishop

and a black bishop

@wind sierra @eager sage

I think this would make for a VERY interesting 3 person chess game

or a 2 person, forming a diamond shape instead

heck, chinese checkers board shape and you can have 6

I'd love to make this actually, don't know where this convo came from but thanks for the idea :)))

oh wait... it's been done.... but they made it so that rooks have the 6 directions... i think that's way too powerful

LOL

[wind sierra]

Pinging me is useless, I don't even know what you're talking about. 😆 Honestly, I'm not even asking this for myself.

A friend of mine sent me a message today, saying: "I'm creating my own RPG system and wanted to use the words 'taxicab metric' to describe the way characters move on the grid, can I do that?", and I was like "Sure, I guess, if the grid is like that of a chessboard", and then he said "Nevermind, I'm making my grid hexagonal so I guess it has another metric. Hey, does it have its own name or can I still call it 'taxicab metric' in the manual?", and I was like "Dude, I don't even like math, but I'll be asking in a certain server I know".

And that's why I wrote here.

[warm junco]

haha

gotcha

[wind sierra]

Then I read that people here weren't even sure it was possible to build a metric on a grid like that, and the only relevant article on the topic I could find online dates back to 1976 and I can't even download it.

I didn't think a concept as simple as this would actually be largely unexplored... at this point, I guess my friend is free to come up with his own name for this metric. I hope he comes up with something stupid and silly, so when this topic suddenly becomes a hot topic in research in 20 years, everyone will be forced to use the name he chose.

[warm junco]

tell him to call it the honeycomb-metric

obviously

[midnight jewel]

I remember a homework exercise about metrics on different grids

[warm junco]

actually, you can do a metric circle in honeycomb metrics

not in 2dimensional math

but perhaps.... in 3

except the dimensions are not orthognal

it'd be complicated

actually hell.... if a taxicab metric circle is like 1/sintheta plus cos theta

what aboutchanging the period

hmmm

or..... ah yes, use a tetrahedron

no

scratch that

i dunno but i nede to do homework, i think ill ponder this later

[wind sierra]

Anyway, even though I still hate math with a passion, I'm a bit curious about the stuff you were talking about earlier: can you actually create a metric on a hexagonal grid (with the points of the space being the vertices of the grid), by defining the distance between two points to be the least number of edges you must travel on in order to go from the first point to the second one?

Because if you can't, all this talk about giving it a name would be pointless. Though the honeycomb metric is a cute name. I like it. 💜

[warm junco]

in taxicab they do it by projecting the vector onto the axis

there's square syymetry there anways

and happens to be that absolute values also create squares

easy and cool

however, i think there might be a way

it might involve something a little more complex

instead of 2d

some sort of 3d

where if you go up or you go down, you can then turn right (or left) respectively and end up back in the middle

in other words, if you go "up" and turn right, you end up going right and down

so equilateral

in math, usually up has no correlation to right, and right no correlation to down

however here they are both

AND YET you can still go right without going up or down

so it's like instead of having a y axis on an x axis, you have a y1 and y2 pointing out at 60 degrees

i think then that if you project any vector onto the surrounding directions of movement you can find it

i thiiiink

[wind sierra]

...Couldn't I just check whether the triangle inequality holds to know if the stuff I defined above is a metric? Distance being 0 iff the two points are the same and symmetry are obviously true already.

[warm junco]

the thing is deciding which y axis to use to define your distance

or if you should use a y at all

you have to define which sector you are in

theres 6 sectors

sextors? hextors? idk

[steel needle]

it's definitely a metric on the vertices

any graph induces a metric on its vertices in that way

it's not unexplored

it's just not very interesting

[warm junco]

yeh

tbh

its kinda

like

oh

hexagons

but i think its cool in the application

[wind sierra]

Well, that's why math is boring. Everything you can actually do and understand on your own always turns out to be uninteresting to everyone else.

[steel needle]

no

but if you wanna do something interesting you have to actually get a result

[wind sierra]

Yeah, and with the current status of math research, that would mean spending 80 years of your life to advance mathematics one inch forward with some new lemma nobody's ever going to use. Math is now so big that everything my little mind is capable of thinking of has already been thought by somebody else.

[steel needle]

doesn't take 80 years

more like 2-5 or so

depending on the area

[wind sierra]

I admit that if you're a super-genius only five years would be enough. I wish I were like you. Anyway, thanks for your answer. I kinda suspected that, but it's cool to have my idea confirmed by someone else who knows more on the topic than I do.

[steel needle]

you don't need to be a super genius

all you need to do is learn math

[wind sierra]

If you say so. Anyway, we can't talk about this here because this is the topology channel, and now that I know that stuff is actually a metric I have no more questions on that matter.

[eager sage]

if anyone wants to talk more about hexagonal grid board games dm me

i’ve got experience designing board games on hexagonal boards and i love talking about it

[frigid patrol]

Is it the case that every single vertex that is neither the top or bottom is identified?

[frigid patrol]

Can someone help me see why the second Homology group of this lens space is 0?

[brisk shadow]

yo i dont get topology

i have no intuition for anythign

i will give an example

so I have the definition of a limit point:

[ \begin{array}{l}{\text { 3.1.1 Definition. Let } A \text { be a subset of a topological space }(X, \mathcal{T}) . \text { A point }} \ {x \in X \text { is said to be a limit point (or accumulation point or cluster point) of }} \ {A \text { if every open set, } U \text { , containing } x \text { contains a point of } A \text { different from } x}\end{array} ]

[gentle osprey]

89_87:

[brisk shadow]

nice

and this example:

[ \mathcal{T}={X, \emptyset,{a},{c, d},{a, c, d},{b, c, d, e}}, \text { and } A = {a,b,c} ]

[gentle osprey]

89_87:

[brisk shadow]

X = {a,b,c,d,e}

anyways

clearly b,d,e are our limit points

a,c are not

this makes sense

also the next example states that discrete topologies have no limit points --- this makes sense also, because they have a singleton for each x \in X.

so I get that too

then they claim without any reasoning that the limit points of [a,b) are all x \in [a,b]

which makes sense based on how i think of limit points

but I thought the standard topology on an interval S in this case would be the set of open intervals (a,b), a<b, for all x \in (a,b)

and that doesn't seem to agree with this definition of limit points at all

because the e-ball or whatever you want to call it here around an element has elements not in the e-ball for other elements or whatever

idk i dont get this subject friends

life is rough

[brisk shadow]

anyways if someone here can help me debug my broken brain i'd greatly appreciate it!

[frigid patrol]

X is R right?

[brisk shadow]

yeah

[frigid patrol]

The open set U is open in R

Any open set containing b intersects [a,b)
Any open set containing a intersects (a,b)
There for they are both limit points of [a,b)

[brisk shadow]

ah yeah i think i get it

thx

[brisk shadow]

[ \begin{array}{l}{\text { 3.1.4 Example. Consider the subset } A=[a, b) \text { of } \mathbb{R} . \text { Then it is easily verified }} \ {\text { that every element in }[a, b) \text { is a limit point of } A . \text { The point } b \text { is also a limit point of }} \ {A .}\end{array} ]

[gentle osprey]

89_87:

[brisk shadow]

[ \mathcal{T}={X, \emptyset,{a},{c, d},{a, c, d},{b, c, d, e}}| ]

[gentle osprey]

89_87:

[brisk shadow]

I want find closure({b})

so closure({b}) is limit points \cup {b}

so we want points which always share a set with b, right?

so in this case I look at {b,c,d,e}

c \in {c,d}so that's out

d \in {c,d} so also out

so closure b = {b,e}

something like closure({a,c}) would be all elements which are always in the same open sets as either one of "a" or "c"

so {a}, {d},{c},{b},{e} all work

and closure {a,c} would be X

ok i got it i think cool

[brisk shadow]

So to prove \mathbb{Q} is dense in \mathbb{R}

[you suppose there is x \in \mathbb{R} - \bar \mathbb{Q} ]

so then there is some open interval (a,b) \in R such that x \in (a,b)

wait not in R

in \mathbb{R} - \bar \mathbb{Q}

and every interval of the reals contains a rational according to this book, which seems like cheating here?

hmm

but anyways that's a contradiction

[honest narwhal]

Shit I thought that was you and wondered why you forgot this stuff when you're supposed to be the person who loves point-set

See, I've heard at some point that you didn't know TeX but I also heard that you were past first year of undergrad

And the latter dominates my conception of you, so I keep forgetting you don't know TeX

[brisk shadow]

lmaooo

i've had this pfp for like half a year now

hmm

i have absolutely no attachment to this pfp

yeah i just don't want to confuse others

lmao

tfw u suddenly become a brainlet

painful

2nd

[steel needle]

wtf

[brisk shadow]

tbh tho

as soon as I type this stuff out in a public forum I realize quickly that it's really easy before anyone even responds

I guess it's like rubber ducking code or whatever

[honest narwhal]

Jan's talking to himself

[brisk shadow]

So the finite closed topology:

is this another word for the finite complement topology

or is this a topology where the open sets are the complements of the elements in the finite complement topology

ok the former I think

but then I get questions asking about like

[ A={1,2,3, \ldots, 10} ]

[gentle osprey]

89_87:

[brisk shadow]

which is not an open set in the finite complement topology I don't think?

hmm ok I think I figured it out

[wind sierra]

@honest narwhal I'm here.

[gentle osprey]

Darisal:

[wind sierra]

Yeah, but it's hard to understand what's the point of knowing this. Covering maps look completely useless to me, starting from the fact that the definition of a covering space looks totally random and unnecessarily complicated.

[honest narwhal]

So, the initial motivation is that covering spaces have a lot to do with the fundamental group

[wind sierra]

Yeah, I know. If I'm not wrong, my book uses covering spaces to prove that the fundamental group of S^1 is Z. But that's pretty much it, it's the only useful application of covering spaces I've seen in the whole book. And even that seems unnecessary, because the fact the fundamental group of S^1 should be Z looks so obvious to me, that I can't comprehend why one can't prove it without using a concept as difficult as covering spaces.

[honest narwhal]

There's a lot more to it, you actually have that covering spaces of a space correspond to subgroups of the fundamental group

And it's not so much that you can't prove it without reference to covering spaces so much as that I don't think there's a way to do it which is strictly easier, like you still have to show that any map S^1 -> S^1 is homotopic to one of the form z^n and the idea of saying alright, I can think of such a path instead as one of the form e^{i\alpha(t)} and then show that homotopy downstairs depends precisely on basepoints upstairs is exactly covering spaces

And also covering spaces just happen to be important for much more general reasons so it sorta becomes the best candidate for how to present it.

[wind sierra]

Hmm, I see. Studying this stuff is just very hard for me because I feel like I have no intuition of why covering spaces are needed in the first place... and then there's the fact the theorems are so difficult, yeah. That covering-spaces-correspond-to-subgroups thing has a huge and super-detailed proof, if I'm not wrong. And I don't know how to keep all of that in my mind.

Especially if you consider that proof is 1% of all there is to say about covering spaces, and that itself is only 1/4 of the whole subject (other topics are fundamental groups, curves and differential geometry). And I can't even remember my own cellphone number...

[honest narwhal]

Yeah AT is gonna involve a bunch more stuff lmao. Tbh I never really learned the subject properly, I know it in bits and pieces. I never had a class that really went through all the details about pi_1/covering spaces/Van Kampen so while I think I know most of the theorem statements, I don't know the proofs

And I think about the stuff more in reference to fancier concepts

e.g. covering spaces are fiber bundles with discrete fibers, you have the LES of homotopy groups, free actions, etc. is how I keep it in my head

[wind sierra]

Oh, you're so lucky. I have to study all theorems and am supposed to remember all the proofs. I've got to study almost 400 pages, it's crazy.

[honest narwhal]

The stuff my AT class did cover was like, Whitehead's theorem, homology/cohomology

Hurewicz

Poincare duality

And my professor wasn't always that great at explaining + it was a 9:30AM class so I wasn't always perfect about attending.

Which book are you using for this class by the way?

[wind sierra]

I call it a 'book' but I'm actually studying on my professor's notes. I wish I had an actual book about covering spaces or other related topics.

[honest narwhal]

Hatcher seems to be the standard book but I don't like it that much somehow

Rotman is sorta easy and good for spelling out all the point-set topology details

Bredon's also less handwavy I think, and is more advanced (also includes stuff on differential topology/smooth manifolds)

Concise is rough

[wind sierra]

Aw, thank you for your help. Now I've got to go, but I'll try looking for those books.

[steel needle]

The fundamental group is a very important invariant

And characterizing it by coverings is very convenient

Example: the uniformization theorem in complex analysis says the all riemann surfaces have either the disk, sphere or complex plane as universal cover

@wind sierra

Which means they are all realized as quotients of these by discrete groups

covering spaces may look daunting, but one thinks of them in this way: a covering space is a quotient by a discrete group

you can be very explicit about this of course, the correspondence between coverings and the fundamental group tells you that the quotient M/G of the universal cover by the discrete group G will have fundamental group G

another result that comes to mind: a non orientable connected manifold has a double cover which is an orientable connected manifold

so you may work on its orientable cover

as for the technical details: the lifting properties are the central result

you should focus on looking at the lifting lemmas first

they are the main tool to build the rest of the theory

[frigid patrol]

Viro's Elementary Topology

I like

[west spindle]

sutherland's Introduction to Metric and Topological Spaces is good

[lucid turret]

Yeah pls hint on how to easily prove this?

[steel needle]

it's what you obtain if you quotient only the inner points of the sphere

and leave a circle intact

(the right part)

[lucid turret]

WELL visually it makes sense

[chrome locust]

What do you think about Gamelin? @frigid patrol

[frigid patrol]

Is that a book?

[loud heart]

Given a covering map p:E->B and E simply connected with no restriction on B other than it's covered by E, can we say p is a homeomorphism?

trying to think of a counter example...

maybe B is some nasty disconnected space

[fleet trench]

consider the projection of a helix going around the z axis of R^3 onto the unit circle

the helix is simply connected and the projection is a covering map but it isn’t bijective

[loud heart]

can I think of that as R^1 wrapping around the circle a bunch?

[steel needle]

why would p be a homeomorphism

yes you can

that's the best example

every good top space has a universal cover p: E -> B

with E simply connected

and a local homeo

and they are only homeos when B is simply connected

[loud heart]

oh there should be a ton of counter examples then

lol my bad

[steel needle]

yeah

any non simply connected manifold and its universal cover is an example

e.g. RP2 and the sphere

torus and R^2

etc

[lucid turret]

Hey, I'm in trouble again. Consider ctous map f:X -> Y, it's easy to get homo of hlgy groups H_n(f). How do I get homo of reduced homology groups tho??? 😦

[lucid turret]

Ah I have possibly figured this bullshit out. The map f:X->Y induces morphism of augmented chain complexes C(X), C(Y), where for n = -1 I take identity Z -> Z

If someone knows what I'm talking about pls comment

[steel needle]

it does

[lucid turret]

Ty. Lazy prof at it again

[gentle osprey]

89_87:

[brisk shadow]

does this just mean that if a set is closed, the complement of that set has an open neighbourhood for every point?

ok yes i think so nice

[west spindle]

the complement of a closed set is open

[gentle osprey]

89_87:

[brisk shadow]

is this not just the definition of connectedness? like, straight up?

hmm i think maybe they want me to prove it from the "X, \emptyset" are only clopen subsets definition

idk how

I guess A would be open

and A^c = B would be open

so A is closed?

is that all?

[gloomy plover]

That's one direction, but it makes sense

Now you have to prove the other direction

[brisk shadow]

ok so i proved disjoint $A,B$, with $A \cup B = X$ implies $X$ is disjoint, so now I need to show X is disjoint implies there exists disjoint $A,B$, with $A \cup B = X$.

[gentle osprey]

89_87:

[brisk shadow]

X is disconnected *

sry

but wait, so A is disconnected $\implies$ X, \emptyset are the only clopen sets $\implies$ there exists A, B as above

which is what I just showed

so I need to go the other way

and show A \cup B = X for disjoint A,B implies that X, \empty set are not the only clopen sets

which I already did too?

idgi dude 😦

[gloomy plover]

What you wrote in that piece of latex is precisely flipped

You assumed A, B disjoint open sets with union X and proved X is disconnected

Wait, am I being stupid here?

Yes I am

I'm sorry for the confusion

What you wrote is correct

[brisk shadow]

no it's all good, what i wrote was not a clear proof initially at all, thanks for your help tho!

[gloomy plover]

No problem 🍮

I need to brush up on topology again at some point...

[brisk shadow]

hello ok

[gentle osprey]

89_87:
Compile Error! Click the reaction for details. (You may edit your message)

[brisk shadow]

i am new to this topic

I know continuous maps are continuous iff the inverse map brings open sets to open sets

and you get elements in a topology though (arbitrary?) union of the basis elements

[gentle osprey]

89_87:

[brisk shadow]

which is one direction i think

to go the other way, that is , to show that the inverse map brining basis elements to open sets in T, is more confusing for me

because I'm new to these types of proofs

can you split up a function of unions into a union of functions?

like, f(A \cup B) = f(A) \cup f(B)

?

i think it would follow from that immediately but idk if that's legal

[gilded shell]

That's not true

Imagine a function that is always 1

Where the domain is {0,1}

And choose B={0}

[brisk shadow]

ah

Can someone help me understand compactness?

My book says "The formal definition is that a space is a compact whenever it is the subset of a union of an infinite number of open sets then it is also a subset of a union of a finite number of these open sets"

or like elsewhere i've seen "every infinite cover permits a finite subcover" i think

and I know compact sets in R are sets which are closed and bounded

[midnight jewel]

what do you mean with understand? do you want intuition for which spaces are compact and which aren’t?

[brisk shadow]

yeah i guess

[midnight jewel]

or do you not understand the definition itself?

the definition is technical

there ar emany equivalent definitions if you e.g. demand X to be a metric space

this definition is just one that worked for all more specific spaces and generalizes well to topology

the “every open cover admits a finite subcover” one I mean

another one is via the finite intersection property, but that seems even more technical

if X is a metric space, then all the following are equivalent (and some more). Not all of these need X to be metric but I know they’re all true in that case

1. Every open cover of X admits a finite subcover
2. For every family F of closed sets in X, if all finite subsets of F have nonempty intersection, the intersection over all of F is also nonempty (finite intersection property, this one’s general too)
3. Every function from X to ℂ is bounded
4. Every function from X to ℝ takes on min and max
5. Every sequence in X has a convergent subsequence
6. X is complete and for every ε>0, there exists an N such taht N balls of radius ε cover X

the last one’s the only really intuitive one imo

7. Every infinite subset of X has a limit/accumulation point

[brisk shadow]

(im not ignoring you im just working though some problems at the same time)

this is a great answer though, thanks

i really appreciate it

[midnight jewel]

the general intuition is that compact spaces are “small”

compact is something like the analogue of “finite” in topology

also “closed and bounded” extends to ℝⁿ

not just ℝ

[brisk shadow]

ah

that is very useful to know

so why do we need the infinite cover part — can we not just say that a compact space has a finite cover and leave that out?

[midnight jewel]

ℝ has a finite cover: {ℝ}

the point there is that no matter how stupid you try to make your cover, you can still achieve it with finitely many of them

finite covers themselves aren’t particularly interesting, every space has them

it’s much more interesting that even in, say, an uncountable cover of [0,1] (say, arbitrarily small intervals centered around every point), there is a finite subset of them that covers it all

[brisk shadow]

ahhhh

[gilded shell]

Can i ask a question or am i interrupting?

(inb4 )

[midnight jewel]

we are kinda discussing sth still

I was gonna make another example

[gilded shell]

Ok, good luck : )

[midnight jewel]

take [0,1] again, and here’s an attempt at covering it:

for n∈ℕ, take the intervals (1/n, 1]

now, there’s an issue

the point 0 isn’t covered yet

so we have to put sth around it, say, [0,ε)

but now for some k∈ℕ, 1/k < ε, and so we can just take [0,ε) ∪ (1/k, 1] and that’ll cover it all

and no matter what else you try, it’ll always play out like that

so in a sense, the point is “even with very small open sets, you can still cover it with finitely many”

but because a general top. space has no notion of “size” except for inclusion, this is the best we got

[brisk shadow]

ah, alright — I think I get it now.

[midnight jewel]

now you can ask a question :P

[gilded shell]

Also check this if you feel like it https://www.drmaciver.com/2015/03/topological-compactness-is-an-induction-principle/

@brisk shadow

Thanks, i think i figured it out though

Induction is also connected to topological connectedness i think

[steel needle]

compactness in practice means that local properties become global easily

as in, if you can prove things in small neighborhoods, cover your space in finitely many of these, and you can probably extend your property to all the space

[midnight jewel]

that’s essentially what that blog post’s about

[steel needle]

nice

[midnight jewel]

does anyone have a “favourite” proof of Tychonoff’s theorem in the general case (not restricted to countable products)? I wanna get a second look at it

we did a proof in class but it was a bit weird, imo

I mean, I guess that’s to be expected since any proof of it has to invoke AoC in some way

it shouldn’t require too heavy topological machinery (e.g. no algebraic topology or what not, I’m still in intro to pointset)

it looks like we did the one that was OP’s favourite in this thread: https://mathoverflow.net/questions/26416/what-is-your-favorite-proof-of-tychonoffs-theorem

the other ones mentioned there (e.g. the ultrafilters one) all use stuff I’m not familiar with (e.g. ultrafilters)

(though actually we did briefly cover filters in analysis)

[loud heart]

Can I "quotient" a cube by identifying all vertices connected by a line?

[west spindle]

what do you mean exactly

you want to quotient by an equivalence relation, and the relation "x ~ y iff x=y or x and y are both vertices of the cube and are connected by an edge" isn't one

[loud heart]

hmm

I mean if they're endpoints of the same line segment, let's say

[west spindle]

then the whole thing collapses to a single point

bc you can always connect two points within the cube by a line segment

[loud heart]

ah good, thanks

what would be a good way to visualize this?

I was trying to think of actually pinching a cube together

[west spindle]

are you thinking of identifying the 8 vertices of the cube & nothing else

bc then i think you get something like... a 3-sphere with six blobs removed? just going off the analogy one dimension lower

it's a 3 manifold with boundary a wedge sum of six 2-spheres

[loud heart]

yes

for some reason I was thinking it would be like a higher dimensional torus since you can think of gluing endpoints as getting a circle then basically taking the product of 3 circles

[sleek canyon]

Gluing the whole edge would be that

Like

You have to identify opposite faces

Not just the 8 pts

Not sure what it is

In dim2 its a sphere @west spindle ?

[west spindle]

you don't have to identify anything.

it depends on what you want

[sleek canyon]

To get a torus I mean

Okay I see what you mean by sphere with blobs removed

Does look like that

[iron socket]

Oke

[west spindle]

oooffff

sorry

i was cooking dinner

@iron socket you still there?

[dim meadow]

Is a function $f:X \times Y \to Z$ continuous iff for every continuous $g:X \to X \times Y$ and $h:Y \to X\times Y$ we have $f \circ g$ and $f \circ h$ is continuous?

[gentle osprey]

Liquid:

[iron socket]

@west spindle lol

Am back

If you write out me an example ill read it

[dim meadow]

Oof

[gentle osprey]

Liquid:

[dim meadow]

Yeah

Hmm

I don't really see how I would do that

I was able to prove this is true in R^n but only through the fact that you can kind of emulate limits with curves

That's good

[keen cliff]

[dim meadow]

What's a counterexample?

[keen cliff]

which one is which again?

is only if → this direction

[dim meadow]

Composition of continuous functions is continuous

Is only if

[keen cliff]

ok

[dim meadow]

@gritty widget why is it false?

[dim meadow]

I don't think such a function exists

Assuming the discrete topology on {1, 2}

@gritty widget is that what you're assuming?

Cause if you take the preimage of an open set in R it splits into open sets in R\times 1 and R \times 2.

Assuming each restriction of the function is continuous

Maybe if you impose the indiscrete topology on {1, 2} it would work

I guess I'm content if it works for sufficiently nice spaces

Assuming it does

Not really

I get your intuition but I would like to think that if you prod your function with enough functions then you should be able to say whether or not it's continuous.

That would be nice

Thanks

[sleek canyon]

it's an interesting question

post it on math stack

[dim meadow]

just did

I've actually never posted on math stack before

and had to make an account just for this

[small obsidian]

Good luck, that place can be a bit of a gauntlet

But there's some very smart people there if you have a good question

[dim meadow]

@gritty widget I think if the space {1/n} \cup 0 imbeds in X and Y you can make a limit argument

does that make sense?

[sleek canyon]

yeah if you have that, then your function will be continuous on sequences

[dim meadow]

In metric spaces we have that sequential continuity = continuity

[sleek canyon]

yeah

[dim meadow]

and that space imbedding really just means that theres at least one convergent sequence which achieves it's limit in your space

which, if not true makes every function continuous

I think

at least in sufficiently nice spaces

[sleek canyon]

yeah it works immediately

for metric spaces which contain a converging sequence

okay no

hang on

it's more subtle

cuz you need to connect the points of the sequence

by whatever connects them in the space

[dim meadow]

yeah

so theres a bit more

[sleek canyon]

but continuity is local so i'd imagine it still works

[gritty widget]

I feel you man

But I'm very new to topology so I'm no help

[dim meadow]

If your spaces X and Y are path connected (and normal) then you can define a map from your sequence in X to the sequence {1/n} union 0 in R, and then use the tietze extension theorem to extend that to a map from X to \mathbb{R}

And then you can use the path connectedness of X \times Y to define a map from \mathbb{R} to X \times Y which when restricted to [0, 1] looks like you're connecting the sequence you want to converge with line segments

@gritty widget that's my idea right now

And all metric spaces are normal so this works for all metric spaces that are path connected

@sleek canyon what do you think?

[sleek canyon]

agreed

my issue was exactly with metric spaces that are not path connected, like the long line

asking for locally path connected is enough however

since continuity is a local notion