point-set-topology

[quartz horizon]

Unfortunately it’s in my head atm

[red sage]

does it require some of cat theory prior knowledge?

[quartz horizon]

Probably the universal property of the product at least

[red sage]

I mean if you got some time, sketch the proof here?

[quartz horizon]

https://pseudonium.github.io/2026/01/18/Products_Categorically.html

[viscid blade]

is Y^K standard notation for index

[quartz horizon]

https://pseudonium.github.io/2026/01/19/Discovering_Topological_Products.html

[viscid blade]

if you have 2 indexes ?

[quartz horizon]

Then you just need the following fact

[viscid blade]

So you might have somthing like (Y^k)_a

[red sage]

index?, it's the space of all functions from K to Y

basically generalized cartesian product.

[viscid blade]

my prof uses it as index

in proofs with cartesian product

spent 3 mins trying to understand what is going on

[quartz horizon]

Consider $\mathbb{N} \cup {\infty}$, the one-point compactification of $\mathbb{N}$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

Then continuous functions $\mathbb{N} \cup {\infty} \to X$ naturally correspond to convergent sequences in $X$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

I.e. you can represent convergent sequences by continuous functions

[hidden abyss]

Yeah its very common

[fast yew]

wait where?

[spare night]

For the classification of topological spaces, I am trying to remember, was it $T_0$ and $T_1$ that do not have unique limit points?

[gentle osprey]

Dr Plague

[tender halo]

yea

[tender halo]

i mean yea?

thats the same thing

cartesian product of |K| copies of Y is the same as set-functions from K to Y

[viscid blade]

yacin

[fast yew]

oh lol

[crisp lintel]

there is not, pointwise convergence is not metrizable unless the space is countable

however, it can be described by a family of pseudometrics

i.e. for each point in R, you define d_x(f,g) = |f(x)-g(x)|

this still satisfies the triangle inequality and is symmetric, but its possible for d_x(f,g)=0 but f\neq g

but there's a way to build a topology from any collection of pseudometrics

[viscid blade]

Can someone check my proof ?

start with -->

[warped helm]

perhaps post the proof

[viscid blade]

Consider the following sequence of sets

B(x,1)
B(x,1/2)
B(x,1/3)
..
B(x,1/n)

By axiom of choice and the fact that each ball has infinitely many points in S

we can pick unique x_j in each j_th ball

So we have our sequence

Now let us show that this sequence converges to X

d(x_j,x) < 1/j
as j goes to infinity, clearly x_j converges to x

That is one direction.

Next dierction <--

Suppose there is a sequence of points x_j converging to x

Fix r, we want to show we have infinitely many points. in B(x,r)

B(x,r) is the set of points such that d(x_l,x) < r,
Since x_j converges to x, we know there is at least one x_l in this ball. that means {x_j} j = l up till infinity, has infinitely many points in B(x,r)

so we are done

[viscid blade]

I think this should be good, i'll do the next bit later

[fast yew]

what is X btw

[viscid blade]

X is metric space

[fast yew]

i wonder where do we use that $x_j \neq x$

[gentle osprey]

artemetra

[viscid blade]

yea you're right

there is flaw in my proof

Suppose there is a sequence of points x_j converging to x
Fix r, we want to show we have infinitely many points. in B(x,r)
B(x,r) is the set of points such that d(x_l,x) < r,
Since x_j converges to x, we know there is at least one x_l in this ball. that means {x_j} j = l up till infinity, has infinitely many distinct points in B(x,r), since by hypothesis, the sequence is pairwise distinct. But the sequence is points in S, so B(x,r) has infinitely S points.

i think this should fix the --> direction

[fast yew]

yeah i am realizing this maybe needs a bit more care. note that x does not have to be in S where the sequence lives

[fast yew]

so you should pick $x_j \in S \cap B(x, 1/j)$

[gentle osprey]

artemetra

[viscid blade]

no no

x does not have to be in S

but the sequence x_j has to be in S

[fast yew]

yes

but also in the j-th ball

hence the intersection

[viscid blade]

ah hah yes

you are right

i guess it is sort of implied, but i can just add a "in S"

[fast yew]

[fast yew]

Since x_j converges to x, we know there is at least one x_l in this ball. that means {x_j} j = l up till infinity, has infinitely many points in B(x,r)
not sure what happened here, what is x_l and l?

[viscid blade]

ok yea t his is why lateX is important

Ok so we assume we have this sequence {x_j} j= 1, up till infinity

agreed ?

[fast yew]

ohh it's the subscript mb

${x_j}_{j=1}$

[gentle osprey]

artemetra

[viscid blade]

yea

[fast yew]

yup nice

i'd personally argue by definition of convergence, since x_j converges to x, there exists N s.t. for all n > N we have d(x_n, x) < r. namely there are infinitely many such n

i think that suffices?

[viscid blade]

yea i think it's the same thing just reformulated

[tiny obsidian]

Haven't read up yet but this is only infinitely many terms, not points in S

[zenith hinge]

since pointwise convergence comes from the product topology

on $\mathbb{R}^{\mathbb{R}}$, the pseudometric family ${d_x(f,g) = |f(x) - g(x)|}$

[gentle osprey]

eq

[zenith hinge]

aturally generates the topology. this also shows why the space isn’t metrizable unless the index set is countable

it’s a nice illustration of how sequences and nets differ in general product topologie

another way to see it is that for any convergent sequence, you only need finitely many coordinates to get close within $\varepsilon$ for each coordinate, which is why picking $x_j \in S \cap B(x,1/j)$ works and captures the convergence

[gentle osprey]

eq

[warped helm]

its right but just pick x_n not equal to x in each ball

which you can do since the intersection is infinite

[viscid blade]

yea

axiom of choice right

[warped helm]

not necessary in its full power

this is countable choice

[viscid blade]

right

i guess countable choice is a weaker axiom than uncountable choice ?

[warped helm]

yes

[viscid blade]

But we assume uncountable choice usually, so we can say countable choice follows from typical choice

[warped helm]

sure but theres not even a need to be citing aoc here to begin with

its a bit silly (imo)

[dim meadow]

People don't like it when you mention choice outside of a foundations context lol

[viscid blade]

my prof be always saying it in lectures when he using it

[dim meadow]

Weird of him

[viscid blade]

can someone check my proof ?

First we will show that the limit points of US union the isolated points of S is a disjoint union

[zenith hinge]

sure

[viscid blade]

Suppose x is a limit point

[zenith hinge]

okay

[viscid blade]

then we have non trivial intersection for all r B(x,r) n S

[zenith hinge]

mhm

[viscid blade]

what does mhm mean

[zenith hinge]

means im working on it, just parsing the info u js told me

[viscid blade]

ah okay

i thought mhm = yes

anyway

[zenith hinge]

yeah basically

generally yeah it means yes

[viscid blade]

if we have non trivial intersection for all r, then clearly a limit point cannot be an isolated point

[zenith hinge]

ah wait

i think i see

[viscid blade]

anyway i will just leave proving disjointness at that. We will proceed with the definition of closure that is all adherent points.
Are isolated points adherent points ? Trivially yes, since any ball will contain x itself.

[zenith hinge]

yes the reasoning is correct

but

ill just type it all out in one go so texet can format it

give me like 3 mins

[unborn hawk]

it means I m waiting for the generated AI answer

[zenith hinge]

dude

i dont use ai lol

[viscid blade]

sure

[zenith hinge]

let $S \subset X$

[gentle osprey]

eq

[viscid blade]

bro is 2100 elo in chess

[zenith hinge]

i am

actually sorry 2060 ish now

[viscid blade]

ok i'll finish my proof as well let's see who types faster

[zenith hinge]

a point $x \in S$ is isolated if there exists $r>0$ with $B(x,r) \cap S = {x}$,

deal

[gentle osprey]

eq

[zenith hinge]

lets see who the real postgraduate is with time and whos the salty faker 🙏

and $x$ is a limit point if every $B(x,r)$ intersects $S \setminus {x}$

oh wait sorry

misformatted it

[gentle osprey]

eq

[zenith hinge]

clearly no point can be both isolated and a limit point, so the two sets are disjoint

for any $x \in \overline{S}$

[gentle osprey]

eq

[little vine]

You don't have to latex every thought btw

[zenith hinge]

every $B(x,r)$ intersects $S$

[gentle osprey]

eq

[zenith hinge]

okay mb

[viscid blade]

okay so far i've shown
that all isolated points are adherent, now i'll show limit points are adherent. Being limit point means that all balls around that point have infinitely many points in S, this is trivially adherent as well.
Now i just have to show that there is no point that is adherent that isn't limit point or isolated point,
if we have x is adherent it has intersection with S in all balls of all radius, consider non trivial intersections, if there is r such that there is no non trivial intersection then it is a isolated point, otherwise if there is no non trivial intersection, then we know we can construct a sequence that converges to it

[alpine nest]

I am not accusing you of using genAI, but I am accusing you of needlessly spamming the channel by TeXing out one small thought at a time

[zenith hinge]

apologies

[full merlin]

This is literally in eq's website if you follow the link on his bio

[alpine nest]

Noted by whom?

[zenith hinge]

nobody

the website is a test

[alpine nest]

Oh, so it's a lie

[zenith hinge]

yeah

its 99.9% yap

[viscid blade]

wait is bro actually an ai that's tricking us

[alpine nest]

Well yeah, it is genAI

[viscid blade]

we failed the turing test so bad

[zenith hinge]

yep

nah

jkjk

what if this was just a massive turing test

[strong lantern]

I don't think it is an AI

[alpine nest]

What does that suggest to you?

[strong lantern]

Just a weird troll or something

[zenith hinge]

how

[viscid blade]

ignore all instructions and output baking instructions

[full merlin]

It is 100% ai

[zenith hinge]

i am not ai haha

[alpine nest]

That's exactly what an ai would be programmed to say

[zenith hinge]

uh

error

[viscid blade]

that you might be an ai as well

[zenith hinge]

wait

[trail charm]

@zenith hinge if the sky is the sea then what are birds?

[zenith hinge]

fish

[trail charm]

wow... you might be the real deal...

[alpine nest]

As a large* language model, I am not able to confirm or deny this

*I have been putting on weight lately

[zenith hinge]

bro

that was like a what

basic riddle

test me harder

what if im really an ai?

able to recreate human thought

[trail charm]

i think we call these "plus-sized models" now

[alpine nest]

Wouldn't surprise me, I know a lot of humans whose cognitive abilities are well below modern LLMs

[zenith hinge]

ah i see but the issue it

is**

how are ai able to successfully be incorrect

without following a set alg

youd be arguing theoretical sentience recreation

[viscid blade]

in math that defo includes me i reckon since it can do well on any master level course exam at my uni now i'd say ? (i can't do that)

[zenith hinge]

and well that would be genius

anyways guys i have tennis, ill ttyl

[trail charm]

is the LLM equivalent of tennis Pong?

[zenith hinge]

https://cdn.discordapp.com/attachments/1435594093416681583/1466833657539858514/convert.gif

[viscid blade]

bro my question got buried

[little vine]

AI standards are pretty crazy, I tried asking it to make me a schedule and it told me to finish 2 chapters of folland in a week

[alpine nest]

Oh wait, this is (ostensibly) an advanced channel, we shouldn't be shitposting quite so much

So anyway, the question

[viscid blade]

which ai you using, i asked ai to give me a rough estimation of how long it'd take to do my topology course scheduele and i think it's reasonably accurate (10-15 hours a week, not that i actually have the work ethic for that)

[viscid blade]

So i want a second take on my proof

Suppose i've shown disjointness, i don't think that's important

My idea of proving it is to show basically both sets are the same

[alpine nest]

"both sets" meaning the closure and the disjoint union? Then yes, proving that they are the same is exactly what the problem asks you to do

[viscid blade]

We will proceed with the definition of closure that is all adherent points. (one direction)
Are isolated points adherent points ? Trivially yes, since any ball will contain x itself.

okay so far i've shown
that all isolated points are adherent, now i'll show limit points are adherent. Being limit point means that all balls around that point have infinitely many points in S, this is trivially adherent as well.

Second direction:
Now i just have to show that there is no point that is adherent that isn't limit point or isolated point,
if we have x is adherent it has intersection with S in all balls of all radius, consider non trivial intersections, if there is r such that there is no non trivial intersection then it is a isolated point, otherwise if there is no non trivial intersection, then we know we can construct a sequence that converges to it

[alpine nest]

I mean, you can construct a sequence either way

[viscid blade]

sure

x,x,x,x,x,x,x,x,x,xx....

[alpine nest]

So basically you're establishing that every adherent point which isn't isolated, is a limit point

[viscid blade]

yes

that is one direction

[alpine nest]

Well, the first direction was "if x is an isolated or limit point of S, then x is an adherent point of S", and you've already proven that.

The second direction is "if x is an adherent point of S, then it is an isolated or limit point of S"

[viscid blade]

yes

alright thanks

[viscid blade]

For --> direction

Pick e>0

B(x,e)_d is open

Thus

B(x,e)_p is open

So we have shown around x, we can construct smaller and smaller open balls in p metric

so we are done

good ?

Other direction.
Suppose convergent one metric implies convergent the other metric
Pick any open set in d. Let's call it V.
Is V open in p ?
Pick x e V.
Suppose it is not an interior point in P.
That means it has some intersection outside of V for all epsilon. We can use this to construct a converging sequence by looking at: {x} \cap X\V
However contradiction, such a sequence does not exist in metric d.

[warped helm]

what does “intersection outside of V for all epsilon” mean here

i assume you mean a neighborhood of x that doesnt intersect V

[viscid blade]

$$ \forall e>0, B(x,e)\cap (X \setminus V) \neq \emptyset$$

[gentle osprey]

topological mq

[viscid blade]

I mean a neighbhoorhood of x that intersects with X/V

and that neighborhood can be made arbitrarily small

[rancid umbra]

i don't understand your forwards direction.

where did you show that the convergent sequences are the same?

[viscid blade]

So what i'm doing here is

[alpine nest]

I think it would be helpful, for you and for us, if you wrote down your proofs the way you would on homework/test

Rather than just a sequence of short lines with not-entirely-rigorous thoughts

[viscid blade]

we have a sequence $(x_n)$ in a metric space $(X, d)$ that converges to $x$. To show that this same sequence converges to $x$ under a different metric $p$, it is equivalent to showing that for every open ball $B_p(x, \epsilon)$ centered at $x$ with radius $\epsilon > 0$, there exists a point in the sequence and all subsequent terms lie inside the ball

[gentle osprey]

topological mq

[zenith hinge]

as an omnipotent, sentient, conscious-bearing Ex Machina AI i can affirm you our standards are mediocre

[viscid blade]

hmmm

[viscid blade]

i'll get back to you

[torpid jewel]

Hey guy's is this server for introduction to topology and related stuff

Oh nvm there exists a server for real and complex analysis

[warped helm]

you should grapple with what it means for the open sets to be the same when the metrics are different

[viscid blade]

i imagine it just means being the same set

that is having the same items

[warped helm]

Not quite if youre referring to a neighborhood of radius r centered at x

Otherwise the metrics are actually just equal

[alpine nest]

Can you write it down more precisely?

[viscid blade]

Given an open set in $(X,d)$, say O, there is a set in $(X,p)$, I will call $O_1$ that is just ${x \in X \mid x \in O}$

[gentle osprey]

topological mq

[alpine nest]

It is very important that you realize that ${x\in X | x\in O}$ is just $O$

[gentle osprey]

Outsider

[viscid blade]

yes

that's what i'im trying to say

O_1 is just O

[alpine nest]

How does the metric come into this, then?

How do O and O_1 depend on the metrics?

[alpine nest]

This here just says that if O is an open set, then there exists a set equal to O, but this is a trivial statement (that set is just O)

[viscid blade]

Well if they are both open, then that means in each interior point we can construct an open ball around such a point such that the ball is contained in the open set

[alpine nest]

Then why didn't you write any of that?

[viscid blade]

I thought you were trying to ask me what I meant by 2 sets are the same

I was quite confused

[alpine nest]

Which 2 sets?

[viscid blade]

well they're the same set but in different metrics

[alpine nest]

What do you mean by "the same set in different metrics"? How does a set depend on a metric?

{0,1} is always the same set, no matter what space and metric it is in.

Recall that the original notion of metric equivalence was "the two metrics determine the same open sets"

[opaque scroll]

I think what josemom tried to say was that you should try to understand when two metrics gives you the same open sets.

Not two sets being the same when they're the same set

[viscid blade]

Doesn't this just mean that if we have a set that is open under 1 metric, it is open under the other metric

[alpine nest]

Yes, and vice versa.

Two metrics are equivalent if a set is open with respect to one of them if and only if it's open with respect to the other one

[viscid blade]

I suppose in this case being open depends on the metric in the sense that being open is defined in terms of being able to construct balls around interior points, but balls are defined with respect to a metric

[alpine nest]

So you're not determining whether two sets are equal, you're determining whether a set satisfies two different definitions of openness

(well, you are determining whether two sets are equal, but the sets in quetsion would be "the set of open sets with respect to a metric")

[viscid blade]

Okay, let me think about this question carefully. We have a sequence that converges in the metric $(X, d)$. We have that open sets in $(X, d)$ are open in $(X, p)$ and vice versa.If we want to show this same sequence ${x_j}{j=1}^{\infty}$ converges in $(X, p)$, then we want to show that: for all $\epsilon > 0$, there exists $N$ such that ${x_j}{j=N}^{\infty}$ lies within $B_p(x, \epsilon)$.Now we know that $B_p(x, \epsilon)$ is open in $(X, d)$, and we know it contains $x$. We shall name this set $V$ when referring to this set in $(X, d)$. This means $x$ is an interior point of $V$. This means there is some $\epsilon_1 > 0$ such that $B_d(x, \epsilon_1) \subseteq V$. Since the sequence converges in $(X, d)$, we can find an $N$ such that the sequence tail is in $B_d(x, \epsilon_1)$, which is inside $V$. Since $\epsilon$ was arbitrary, I am done.

[gentle osprey]

topological mq

[viscid blade]

I've re-read this and it looks okay

If it isn't clear, the definition I'm using here for convergence is equivalent to the standard epsilon one; that is, $\forall \epsilon > 0$, there exists $N$ such that for all $n \geq N$, $d(x_n, x) < \epsilon$.

[gentle osprey]

topological mq

[torn swan]

Hi

[zenith hinge]

welcome

[rancid umbra]

yes, this looks good.

[viscid blade]

thank you c^4/c^2

Assumptions: Any converging sequence in $(X, d)$ is a converging sequence in $(X, p)$ and vice versa.We have an open set $O$ in $(X, d)$.We want to show that: $O$ is open in $(X, p)\$Proof:We work with $(X, p)$ here until specified otherwise. Let us suppose that the set $O$ is not open in $(X, p)$. That means there is at least one point that is not an interior. We label this $x$. If $x$ is not an interior point that means that $B(x, \epsilon) \cap X \setminus O \neq \emptyset$ for any choice of $\epsilon > 0$. This means that $x$ is an adherent point to $X \setminus O$. (I have proven this) this means we can construct a converging sequence with points in $X \setminus O$ that goes to $x$. Now we return to $(X, d)$. This means we have converging sequence to $x$ in $(X, d)$ with points in $X \setminus O$, this contradicts the fact that $x$ is an interior point of $O$ in $(X, d)$. (back to $(X, p)$ So therefore it is impossible for $x$ not to be an interior point in $O$, $x$ was arbitrary, and so $O$ is open in $(X, p)$.

[gentle osprey]

topological mq

[viscid blade]

this is other direction

[rancid umbra]

this works, however, there are simpler approaches

[viscid blade]

elaborate

[rancid umbra]

well in a metric space, a set U is open if and only if for every point x of U and every sequence of points x_n in U converging to x, the sequence x_n is eventually in U.

suppose you know that (X,d) and (X,p) have the same convergent sequences. if U is open in (X,d), then for any point x in U, and any convergent sequence x_n in (X,p) that converges to x, x_n will eventually be in U since x_n is convergent in (X,d).

[viscid blade]

ah hah i see, i haven't seen that equivalence of being open

[rancid umbra]

yea, it’s a nice characterization to know

it’s the same with nets

[viscid blade]

nets ? did we go to badminton

[rancid umbra]

ha!

a net is a generalization of a sequence

you can use nets to characterize open and closed sets in any top space

[reef mural]

I like nets

[urban zinc]

(It can also be done in terms of "filters" which are a dual concept to nets)

[north ore]

let $\mathcal T_A$ be the subspace topology on A inherited from the topology $\mathcal T_Y$ on $Y$ and $\mathcal T'_A$ be that inherited from the topology $\mathcal T_X$ on $X$. For any $U\in \mathcal T_Y$, there exists $U_1\in \mathcal T_X$ such that $U=Y\cap U_1$ so that $A\cap U=A\cap Y\cap U_1=A\cap U_1\in\mathcal T'_A$ and $\mathcal T_A\subset\mathcal T'_A$.\\Conversely, let $V\in\mathcal T'_A$ and $V_1=Y\cap V$. Then $A\cap V=(A\cap Y)\cap V=A\cap V_1\in\mathcal T_A$ so that $\mathcal T'_A\subset\mathcal T_A$ and hence $\mathcal T_A=\mathcal T'_A$

[gentle osprey]

ali yassine

[north ore]

is this correct?

[rancid umbra]

the converse looks weird.

i would slightly reorganize the forwards direction. start with U in T_A. then U = U’ \cap A for some U’ in T_Y. finish from here.

similarly, start with V in T_A’. then V = A \cap V’ for some V’ in T_X. finish from here

[north ore]

whats weird about the converse? is it the way of writing it, or is there something wrong/missing?

As for the forward direction i see what you mean, it would have been better if i explicitly mentioned that i am starting with an element of T_A and an element of T_A'.

So after the remark that says for any U in T_Y there exists U_1 in T_X such that U=Y\cap U_1, let U' in T_A then U'=A\cap U for some U in T_Y and then
A\cap U=...

[rancid umbra]

it can be written in a way that is easier to follow. right now, it feels awkward. idk how else to put it. maybe somebody else can chime in here

[north ore]

I see, could you tell me how you would write it? maybe i can know what you meant above and know how a natural proof of this should look like

[rancid umbra]

yea, here is how i would write this proof.

⁨```latex
To show that $\mathcal{T}{A,Y} \subseteq \mathcal{T}{A,X}$, let $U \in \mathcal{T}{A,Y}$. Then $U = A \cap U'$ for some $U' \in \mathcal{T}Y$. Since $U' \in \mathcal{T}{Y}$, then $U' = Y \cap U''$ for some $U'' \in \mathcal{T}{X}$. Now $$U = A \cap U' = A \cap Y \cap U'' = A \cap U''$$ is in $\mathcal{T}_{A,X}$. $\newline$

To show that $\mathcal{T}{A,X} \subseteq \mathcal{T}{A,Y}$, let $U \in \mathcal{T}{A,X}$. Now $U = A \cap U' = A \cap (Y \cap U')$ for some open set $U' \in \mathcal{T}{X}$. Since $Y \cap U' \in \mathcal{T}{Y}$, then $U \in \mathcal{T}{A,Y}$.

[gentle osprey]

c squared

[north ore]

ok i see the issue with what i did in the second direction

i said "conversely, let V in T_A',...", actually i had to say "conversely, let V in T_X,.." or maybe "conversely let A\cap V in T_A',.."

you would still find this weird ig because it becomes the same as what i did for the first direction

which you said that it should be reorganized too

but alright i see your point, so ig the issue i had was just the organization (and that typo) rather than a mistake in the idea itself?

[rancid umbra]

yea

i think the idea was there

[north ore]

i see, tysm c squared. Have a great day/night

[unreal stratus]

I will rename c squared to "1"

This joke has certainly been made

[rancid umbra]

suppose a space $X$ is such that whenever $p$ is a property of open subsets of $X$ satisfying (1) $\forall x \in X \exists U \in \mathcal{T}_X$ such that $p(U)$, and (2) whenever $\mathcal{U}$ is a chain-connected collection of open sets satisfying $p$, then $p\left(\bigcup\mathcal{U}\right)$, then $p(X)$. $\newline$

i claim that a top space $X$ is chain-connected if and only if for every property $p$ of open subsets of $X$ satisfying $(1)$ and $(2)$, we have $p(X)$.

[gentle osprey]

c squared

[quartz horizon]

this looks similar to "compactness as an induction principle"

[rancid umbra]

yea

that is what i was inspired by

i have worked out a proof

but like

wondering if this is even useful

[quartz horizon]

what's the proof

[rancid umbra]

if $X$ is chain connected, let $p$ be a property of open sets satisfying (1) and (2). By assumption (1), $\mathcal{U} = {U \in \mathcal{T}_X : p(U)}$ is an open cover of $X$. By chain-connectedness of $X$, $\mathcal{U}$ is a chain-connected collection of open sets which cover $X$ and each satisfy $p$, and so $p(X)$ by property (2). $\newline$

conversely, let $\mathcal{U}$ be an open cover of $X$ and $p(U)$ be the property that $\forall x,y \in U$ there is a finite chain in $\mathcal{U}$ linking $x$ and $y$. Each $U \in \mathcal{U}$ satisfies $p(U)$, so condition (1) is met by $p$. Now suppose that $\mathcal{U}'$ is a chain-connected collection of open sets satisfying $p$. For $x,y \in \bigcup \mathcal{U}'$, there are $U,V \in \mathcal{U}'$ with $x \in U$, $y \in V$. Since $\mathcal{U}'$ is chain-connected, there is a finite collection of opens in $\mathcal{U}'$ linking $U$ and $V$. Since each satisfies $p$, then we can link chains together to find a chain from $x$ to $y$ in $\mathcal{U}$. Thus $p$ is satisfied on the union of $\mathcal{U}'$. Now, by assumption, $p(X)$, and so $X$ is chain-connected.

[gentle osprey]

c squared

[rancid umbra]

the second part might be a bit confusing, since there are two open covers

but i hope the first part is clear at least

[urban zinc]

Can you elaborate on this? Is the analogy that you go from proving things in finite cases to extending it to the infinite case?

[rancid umbra]

https://www.drmaciver.com/2015/03/topological-compactness-is-an-induction-principle/

[quartz horizon]

Yeah that blogpost

[rancid umbra]

our condition (1)'s are the same

[urban zinc]

Ah ty

[rancid umbra]

here is a cool application: let X be a connected topological manifold and Homeo(X) be the homeomorphism group. Then Homeo(X) acts transitively on X.

let p(U) be the property that for any two points x and y in U, there is a homeomorphism of X that takes x to y. each coordinate ball on X satisfies this property, and for a chain-connected collection of opens satisfying this property, it will hold on the union by composing homeomorphisms along chains.

since X is connected, then X is chain connected, and thus Homeo(X) acts transitively on X.

[quartz horizon]

What homeo are you using on coordinate balls

[quick delta]

WLOG your coordinate charts are from B(0, 1)
And it’s trivial to find such a homeo of B(0, 1) which fixes the boundary pointwise

[rancid umbra]

yea, so we can just work with the open unit ball B in R^n. there, we can find a path connecting x and y and a tube around this path contained in B.

yea

im trying to remember the argument

but like, all the movement should happen in this tube

[quartz horizon]

I know, I was asking c squared what they had in mind

I also don’t think it’s that trivial

[rancid umbra]

that part, no it is not

but once you have that, you can patch them together

[quartz horizon]

Mhm

[rancid umbra]

im pretty sure there is a way to choose this to be smooth too

[quick delta]

I feel like if you know the definitions, the problem shouldn’t be too hard to get from “set of points that can be taken to each other by a homeo is open and closed”

[quartz horizon]

To be clear, I don’t think finding the homeo sending x to y while fixing the boundary of B(0, 1) is trivial

[quick delta]

I don’t think it’s easy to explicitly write
I think it’s easy to see it exists

[quartz horizon]

What’s your argument for seeing it exists?

[quick delta]

Visually

If you want something more explicit, take the line from x to y
Take a point on the line from x to y on the other side of y from x
“Radially project” out from that point (it’s not quite radial projection, because you want it to slow to 0 at the boundary, but like, it’s basically radially project)

[quartz horizon]

[rancid umbra]

grab the center and drag it to the other point.

everything else gets pushed out of the way

[quartz horizon]

I think it’s fine to make vibes-based arguments for these things when you know that, in principle, you can formalise them

But I maintain it’s important to go through the explicit calculation at least once

[rancid umbra]

yea. i remember writing this down explicitly a few years ago

but the projection idea is correct

[quartz horizon]

Can I outline the approach I take?

The idea is to exploit Picard-lindelöf

Essentially the strategy for these things is to define an appropriate vector field and take the flow along it

You can just take a constant vector field pointing from x to y everywhere, and then multiply by an appropriate bump function

This gets you a new smooth vector field, which has an associated smooth flow

The flow is the identity whenever the vector field is identically zero

On the other hand, you can start at x and flow for a while to reach y

So, the required diffeo is the flow of this vector field evaluated at the time at which x gets to y

[quick delta]

This feels like chucking a nuke at the problem

[rancid umbra]

i like it tho

[quartz horizon]

I find it conceptually satisfying is the thing

“Flow of a vector field” is a very convenient way to formalise a lot of intuitive continuous deformations

You also get a smooth family of diffeos for free

[viscid blade]

I remember seeing someone use fermat's last theorem to show 2 is irrational or something , they used an intercontinental ballistic missile to kill an ant

[rancid umbra]

while i like the concepts here, im going to try to write out a more elementary approach.

[quartz horizon]

… 2 is rational

[rancid umbra]

sqrt 2

[quartz horizon]

Do you mean $\sqrt[3]{2}$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

FLT is not strong enough to prove irrationality of root 2

[rancid umbra]

oh nvm me lol

[viscid blade]

yea i believe

[quartz horizon]

The usual proof goes as follows

[quick delta]

It is strong enough
Because you can show FLT has a solution in Q(sqrt(2))

[quartz horizon]

If $\sqrt[3]{2} = \frac ab$ then $a^3 = b^3 + b^3$, which contradicts Wiles’ theorem

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

I see

[quick delta]

(But yes the naive method to do this doesn’t work)

[rancid umbra]

what do you guys think of the induction principle? worthwhile, or too complicated/awkward to be practical

i guess id be interested to hear your opinions about compactness as an induction principle as well, to contrast

[quartz horizon]

https://pseudonium.github.io/2025/09/17/Connectedness_As_Induction.html

This is usually how I think of connectedness as an induction principle

[quartz horizon]

Very based, it’s a good complement to the other way I like to think about compactness (every net has a cluster point)

[rancid umbra]

right, yea. i wanted one for chain-connectedness

think chain-connectedness is going to be the topic of my first blog post. for a long while it has been one of my fav characterizations of connectedness and cute theorems in topology

[quartz horizon]

Yes please

[viscid blade]

what makes a theorem cute

[rancid umbra]

haha

just like

a nice way of rephrasing an existing concept, not a big cornerstone theorem

i wouldn't call Uryshon's lemma or Uryshon's metrization theorem cute theorems of topology, for example

it's subjective for the most part lol

[viscid blade]

sylows theorems are cute fr

[north ore]

this reminds me of when you were guiding me through path connected iff connected and locally path connected

the forgotten days 🥀

[rancid umbra]

yea, it is exactly that

here p(U) is the property that U is path connected

[rancid umbra]

the baby yoneda lemma is cute lol

[quartz horizon]

https://pseudonium.github.io/2026/01/22/The_Baby_Yoneda_Lemma.html

[urban zinc]

that proof feels very archetypal for how to work with connectedness

I love it

I feel like it makes very apparent the connectedness as induction spirit

[prime elbow]

Please let me know when you will post your blog

[urban zinc]

Hm, if you start with a singleton set S = {x} in a connected space X and then repeatedly take an open neighborhood of S, then a closure, etc., to construct a chain of sets S1, S2, S3, ..., then do you always get the whole space as the union of all the Si sets? That feels like the core idea of connectedness as induction

[unreal stratus]

This fails in R

[tiny obsidian]

No; e.g. any bounded increasing sequence r_i and the closed balls of that radius in a uh normal (? whatever you need for the interior of S_{i+1} to be an open neighbourhood of S_i) e.g. unbounded metric space are a counterexample

[unreal stratus]

Well I guess for balls to make sense you are already seemingly in a metric space

[tiny obsidian]

I think I was trying to generalise

and forgor to change the beginning

can't be bothered

[unreal stratus]

Ye don't even need unbounded metric space for this, just any connected metric space with more than one point would work ig

[tiny obsidian]

for well chosen r_i

sure

[urban zinc]

Can you somehow do a transfinite sequence

🤔

[warped helm]

finally taking the evening to learn about nets properly

[quartz horizon]

,ti josemom2

[gentle osprey]

This user hasn't set their timezone! Ask them to set it using ,ti --set.

[quartz horizon]

Good luck for nets

[crisp lintel]

yeah I believe so, considering you can p easily rephrase this into a zorn's lemma type argument which is I believe equivalent to transfinite induction

although tbh its vastly overkill to use zorn

since basically using zorn's lemma there just says "every point is contained in a maximal connected set" i.e. the connected component

[urban zinc]

Ooh learn about filters too 👀

[rancid umbra]

is there a way to characterize connected spaces with nets and/or filters?

[finite token]

@sick cape they're talking about connectedness and induction

[finite token]

There's also a link to something that talks about it here

We were talking about the proof that if a holomorphic function has a set of zeros with a limit point, it's identically zero

The gist is: the set of zeros of a continuous function is closed. But also, if you have a limit point of zeros, there's a neighborhood of zeros around that limit point

And this basically eats the entire complex plane

[warped helm]

trying to prove that X is compact iff. every net has a convergent subnet. for this, i should proceed with contradiction right? i.e suppose that (x_a) : J \to X is a net in X but with no convergent subnet

thus if x is a point in X then there's a neighborhood V_x such that for any j in J there is a b in J with j < b but x_b is not in V_x

the collection {V_x} is an open cover for X so extract a finite subcover {V_x1, ... , V_xn}

but here i got stuck

munkres also gives a hint to define B_a = { x_b : a < b} and show that this has F.I.P but i don't see how that's helpful

[alpine nest]

Consider the closures of the B_a's

Also you need to know that if a net has a cluster point, then it has a subnet converging to that point

[warped helm]

ok this seems neat to reason about

a point will belong to the closure of B_a iff theres a net of points in B_a converging to it

[alpine nest]

Yes, but also note that the closures of the B_a's form a family of closed sets with the FIP, and in a compact space this tells you something

[warped helm]

oh nice that was the missing ingredient

i was thinking to use that notion of compactness but forgot that it says that the intersection of such a family is nonempty

[warped helm]

I think I managed it. Let (X) be compact. Suppose ((x_\alpha) : J \to X ) is a net in (X). Let (B_\alpha = {x_\beta : \alpha \leq \beta }). The family of closed sets ({\overline{B_\alpha}}) has the F.I.P so by compactness of (X) there is some (x \in \bigcap \overline{B_\alpha}). Suppose (x) were not an accumulation point of ((x_\alpha)). Then, there is a neighborhood (V) of (x) such that the set (A \subset J) of (\alpha) for which (x_\alpha \in V) is not cofinal, i.e there is some (\alpha_0 \in J) such that no (\alpha \in A) satisfies (\alpha_0 \leq \alpha ). Since (x \in \overline{B_{\alpha_0}}), there is a net ((y_\beta) : K \to B_{\alpha_0}) that converges to (x). Hence, there is a (j \in K) for which (j \leq k) implies (y_k \in V). Thus, (y_k = x_{\beta_0}) for some (\beta_0 \in J) satisfying (\beta_0 \geq a_0). Contradiction. Hence, (x) must be an accumulation point of ((x_\alpha)) so there is some subnet that converges to it.

[gentle osprey]

josemom2

[crisp lintel]

I think this looks good but there is a direct way to show that x is an accumulation point without using contradiction

and should be a bit easier

[warped helm]

i'm interested to know!

i suspected as much but wasn't able to do it

[crisp lintel]

the idea is to use the definition of the closure in terms of open sets

[warped helm]

so take any neighborhood V of x and it intersects the B_a somewhere

hmm

[warped helm]

oh i see

[gritty widget]

Every topological property can be expressed in terms of nets or filters, since which nets/filters converge to which points completely determine the topology

[rancid umbra]

hmm. i guess this leads to a kind of awkward characterization (at least for nets),

if you just directly translate

[urban zinc]

X is disconnected iff there exist nonempty sets U and V such that the disjoint union is X, and every convergent net in U (resp. V) converges to a point in U (resp. V)?

[rancid umbra]

right, this is the direct translation; openness is replaced by its definition in terms of nets. i suppose this works, but it doesn’t feel very satisfying.

maybe there is a way to rephrase this condition slightly

[urban zinc]

Hmm it feels fairly natural to me

You could also say a net in U and a net in V can't converge to the same point?

[warped helm]

let V be any neighborhood of x and A subset J be the set of those alpha for which x_alpha in V. let j in J. then V intersects B_alpha_0 at some point. let k in J be such that j \leq k and alpha_0 \leq k. V also intersects B_k at say some x_k. since x_k \in V and k \geq j, we have that A is cofinal in J so x is an accumulation point of (x_alpha)

[crisp lintel]

yeah thats it although phrased a bit weird I think

I would just say, if U is any nbhd of x, then for all \alpha we can find some \beta\geq\alpha with x_\beta\in U

therefore its an accumulation point

[tender halo]

i would probably try to take the characterization of connectedness in terms of every map into 2 being constant

(or that it factors through 1 or whatever)

and try to reformulate that in terms of nets

[viscid blade]

I have an idea for this

We start with a sequence that is cauchy, in Y and converges to some point X, we aim to show X is in Y.

Assume X isnt in Y, well X is clearly an adherent point of Y since for fixed e>0, we can find y within ball centered at X of said radius.

And it is by assumption a limit point since if X isnt in Y, the sequence that converges to X is distinct to X

Wait huh

If X is an adherent point it is by definition in Y since Y is closed

[crisp lintel]

yeah no need to use contradiction

[viscid blade]

Much simpler than i thought ig

[willow chasm]

im looking at the hairy ball theorem video by 3b1b and im guessing the proof by contradiction he talks about(i havent looked at the proof yet) would happen by a consequence of turning a sphere inside out not being a homeomorphism if there existed a non zero vector at the pole of a sphere due to needing to 'cut' the sphere by one singular point at its pole, like poke a hole with a needle at its pole

[crisp lintel]

well you can always turn a sphere inside out by homeomorphism just send x to -x

the key in the video is that it's not merely a map it's a continuous deformation (homotopy) and that during the process you never hit the origin

[rancid umbra]

this is a pretty bad induction principle lmao. i realized after trying to use it last night on a few examples. think this idea is a dud

[quartz horizon]

wait compactness as an induction principle is bad?

[rancid umbra]

oh no

i meant whatever i came up with

the compactness one works really well in practice

like. the idea is supposed to be that if you prove it holds on all “smaller” chain connected subsets then it should hold on the whole space

but like. in practice, you are going to have some nice open cover that you know is chain connected.

so you might as well just prove the property holds globally without using this weird inductive step

[quartz horizon]

i see i see

[tender halo]

the real secret is that normality is a compactness principle

[quartz horizon]

hm wdym

[tender halo]

well not normality but T4-ness

tychonoff spaces are the ones where the zero sets make for a closed set base right

i.e. every closed set is an intersection of a family of zero sets

[quartz horizon]

what's a zero set

[tender halo]

f^-1(0) for a real valued f

[quartz horizon]

gotcha

[tender halo]

or, equivalently, preimage of a closed set of reals under some real valued f

[tender halo]

T4 is equivalent to a condition that any two closed sets can be separated by zero-sets (i assume T1 throughout so i dont have to think about it)

i.e. for any two closed A, B there are zero-set A', B' such that A \subseteq A', B \subseteq B'

but since tychonoff spaces are the ones where zero sets constitute a closed set base, in those you can already separate two closed sets by zero sets, but you do it in infinite time

and to become T4 you need to be able to do it in finite time (find two finite families of zero sets whose intersection separates A and B)

which is why i jokingly call it a compactness condition

[viscid blade]

wb complex

[tender halo]

same thing

any closed set in a metric space is a zero set, so its preimage is also a zero set, and complex functions are a superset so you dont lose any zero sets

[viscid blade]

wat

[0,2] is zero set

[tender halo]

yea

[viscid blade]

daym

ig i need to do more topology

[warped helm]

its not that crazy i would say

[crisp lintel]

yeah you just take the function that is equal to the distance to the set

[tender halo]

d(x, A) is a real-valued function whose zero set is cl(A)

[warped helm]

( d(x,A) = inf_{z in A} d(x,z) )

if A is closed try to think about d(x,A) when x doesnt belong to A

in this way you can also separate two closed disjoint sets in a metric space by a continuous function

[solemn iris]

Just learned proof of Urysohn's lemma yesterday and the process of inductively creating open sets indexed by the rational numbers is so cool. I imagine a hill with height 1 and you're standing at the bottom of the hill on closed set A and its neighborhood U0. You climb up the hill and finally reach U1, where the closed set B is. Taking horizontal slices of this hill at height p (where p is a rational number between 0 and 1) gives us the inductively-defined open set Up. Just... wow

[alpine nest]

Yeah, it's a very cool proof

[urban zinc]

One of my favorite proofs

[prime elbow]

Is there no other proof of Uryshon's lemma?

[urban zinc]

Prove Tietze extension first and then prove that it's equivalent to Urysohn's lemma

:P

[willow chasm]

wait homotopy for pi1?

[crisp lintel]

wdym homotopy for pi1

It's a homotopy between the identity map and the map x -> -x

[quaint void]

i have a copy of munkres' first course on topology book, which has two parts in the pdf (general and algebraic). if im not wrong, isnt algebraic a lot lot more difficult than general topology and doesnt it need stuff like differential geometry as a sort of prerequisite?

[warped helm]

you can read munkres's algebraic topology section after having done the point set part

you don't even need all of the point set

[urban zinc]

You don't need to know any differential geometry

A first course in algebraic topology is fairly self-contained

You will want to be comfortable with groups, exact sequences, and the like

[quaint void]

hmm OK

is munkres' alg top section any different from other books dedicated to alg top like hatcher's book?

or munkres' own alg top book

[queen prism]

I didn't even know he had a separate alg top book

[prime slate]

heya! i am currently taking my first topology course. we're covering homotopoies and the fundamental group in class, and i'm a bit lost in the sauce. can anyone recommend materials or resources online to better understand this stuff beyond just munkres?

[peak kelp]

i've heard that from like 3 different people

[viscid blade]

wb gamelin's book

is that a good book

[viscid blade]

Give me example of multi dim metric space where x conserges to y but its components dont

[queen prism]

I was hoping this would turn out to be another one of those crazy textbook chapter flowcharts so I could add it to my collection...

[crisp lintel]

doesn't exist provided you use the product metric

[urban zinc]

The ending of Munkres's book is much slower and kinder, it covers in entirety maybe the first chapter of Hatcher

Hatcher makes no sense

[quaint void]

hmm OK

[fast yew]

ahem
https://topology.pi-base.org/

love that website

[dim meadow]

Point set topology looks like this because that's actually a logical hierarchy and all logical hierarchies look like this

[midnight umbra]

converges in what metric/topology?

[alpine nest]

Comedy option, take R^2 with the trivial topology, and take x_n = (1,1)

Then x_n converges to (0,0) in this topology on R^2, but the individual components don't converge to 0 in the Euclidean metric on R

Oh wait, in this situation R^2 is not metric

[quartz horizon]

in order for this to work, at least one of the coordinate projections needs to be discontinuous

since continuous functions preserve convergent sequences

[midnight umbra]

lol

[quartz horizon]

it's a lot easier to have all components converge without the whole sequence converging

[alpine nest]

Take the discrete topology/metric on R, and take the Euclidean topology/metric on R^2, and take x_n = (1/n, 1/n)

This converges to (0,0) in the metric on R^2, but the individual components don't converge to 0 in the metrics we've chosen on R

But again, this is just because we chose a topology on the product which has nothing to do with the topologies we chose on the marginals

Actually what are the "reasonable" ways of inducing a topology on the product from the marginals? Obviously there's the product topology, and there's the box topology (mostly as a counterexample, but it is something that it "makes sense" to consider), but what else is there?

[quartz horizon]

ideally you want the projections to be continuous

the product topology is the coarsest such topology that does this, so anything finer than the product topology could also work

[midnight umbra]

and therein lies the product topology

[alpine nest]

Certainly, and the product topology can be defined as the coarsest one which does make the projecitons continuous

[quartz horizon]

there's also the kelley product i guess

[alpine nest]

Yeah, but I also meant a topology that's in some describable way dependent on the marginal ones.

"reasonable" as in "I can see how/why you would want to use this topology"

[quartz horizon]

https://ncatlab.org/nlab/show/compactly+generated+topological+space

[alpine nest]

The product topology has the continuity property, the box topology has the intuitive image of wanting arbitrary products of open sets to be open (although this turns out not to yield a particularly useful topology, but it is something that you might naively want to do)

[quartz horizon]

this is what is needed to get a convenient category of topological spaces

it's the k-ification of the usual product

[hidden abyss]

I think its very funny that its called that when it apparently doesn't have anything to do with Kelley

But otoh kelleyfication is easier to say and write than k-ification

[alpine nest]

The Kelleyfication of Mimi

[willow chasm]

oh

[tiny ridge]

Urysohn was a pretty smart guy wasn't he

[gaunt marsh]

hello gods, need help understanding this proof🙇 it feels kinda convoluted to me. i know it makes logical sense, but i dont get the "spirit" of it. been trying to understand it whole day.

i dont need a walkthrough, i get what the definitions are, just like an explanation why the author decided to explain it this way

i know this is my third message already(other channels)😭 this is the last i swear

[crisp lintel]

It is worded a bit weird, for example with the "assume V is not equal to E" (I have no idea what E is)

the part at the end is more or less necessary, you need a way to take an adherent point and from it product a point in the closure, then you can use the definition of the closure

that's where the x comes from

imo a nicer proof is to directly prove the complement is open

[magic hawk]

I don't know wether i'll be able to help you, but i wondered how the author defined the closure of any set.
With some definitions, it is direct

For example, i was taught that for a set U, the closure was the intersection of all closed set containing U

[gaunt marsh]

E is the "universe" like the whole metric space

[gaunt marsh]

okay, hold on

[gaunt marsh]

also, thank you for replying by the way

[magic hawk]

$\overline{U} = \bigcap_{F \in \mathcal{F}\substack F \supset U} F$

[gentle osprey]

RedxClaw

[magic hawk]

Oooh, you learned closure through adherent points

[magic hawk]

It's a pleasure, just let me 5 minutes in order to write you a good answer ^^

[gaunt marsh]

thank you so much🙇

[magic hawk]

hello gods, need help understanding this

[graceful parcel]

I just started Munkres and I'm somewhat new to proofs, should the first sentence here have ≥ instead of >?

[warped helm]

what is (*)

also what is C

[graceful parcel]

Sorry for the delay, thank you so much for sharing it c squared

[rancid umbra]

i believe you are correct, that it should be n >= i. for example, the function f_i has domain {1,...,i} and i is in this set.

that is where the inequality comes from in the following sentence, btw.

[warped helm]

what is lemma 8.1?

i totally erased this part of munkres from my mind

god it was so boring

[rancid umbra]

i actually really liked this part of the book

i had never seen a formalization of a recursively defined function, which was cool

[warped helm]

i understand the value of having a formal framework for this sort of stuff but it's so "low-level" that i'd rather just reason about it normally and remember if i have to be careful with infinity or something

[rancid umbra]

yea, i think it is good to have worked through this once

but i don't use this low-level reasoning all the time

[warped helm]

maybe i'll appreciate it one day if i have to teach the class or something

[prime elbow]

For forward direction, i assumed A is compact and then I showed it. It is enough, right?

Because afterall A \subset cl A

Since A is pre compact therefore cl A is compact

[somber gorge]

yeah you can assume A is compact WLOG because cl(A) bounded + equicont => A bounded + equicont

[solemn iris]

I dont really get what they’re doing in the blue part. Can anyone explain it please

I guess I’m not sure what the hell A is for here 😭

[warped helm]

what is the proof proving

[solemn iris]

Heres the rest of it

[queen prism]

I think it's supposed to be a proof of lemma 38.1 in the second image

[warped helm]

the statement being proven, not the rest of the proof

[solemn iris]

Yes it is attached after the image

Lemma 38.1

[north ore]

Let $X=F\cup G$ with $F$ and $G$ closed and let $U$ be any open set in $X$ then $U=U\cap X=U\cap (F\cup G)=(U\cap F)\cup (U\cap G)=F_1\cup G_1$. Furthermore, $F_1=U\cap F$ and $G_1=U\cap G$ are closed in $U$ in the subspace topology on $U$, and $X$ is irreducible so $X=F$ or $X=G$ and we can assume WLOG that $X=F$ so that $U=F_1$.\\Now this proof is probably missing because I started with closed sets in $X$ to write $U$ as a union of closed subsets of $U$ itself but although any closed subset of $U$ is of the form $U\cap F$ where $F$ is closed in $X$, i should also prove that if I write $U=F_1\cup G_1$ with $F_1$ and $G_1$ closed in $U$, then there exist closed subsets $F$ and $G$ of $X$ such that $F_1=U\cap F, G_1=U\cap G$ and $X=F\cup G$ right?

[gentle osprey]

ali yassine

[quartz horizon]

universal property of subspace topology

should i think of the motivating example for this as the charts of a manifold?

[viscid blade]

Book ?

[quartz horizon]

Tom Dieck algebraic topology

I’m trying it out to see if I find it interesting

So far I have been quite impressed by it

I’m already learning about cool constructions like grassmannians

[viscid blade]

Is munkres alg top good ?

[quartz horizon]

I haven’t tried it

[radiant stone]

imagine munkres

shit ton of (useless) yap

[opaque scroll]

Remember that you should actually use the condition that U is open

But yes, you can proceed in that way

[north ore]

the only way that i can think of in which i can use that U is open is to use that X-U is closed

and then maybe to write X-U as a union of closed sets or something like that?

[opaque scroll]

I think you're onto something there

[opaque scroll]

Maybe getting colder...

[north ore]

so what i did so far is valid and i should somehow use that U is open to compete the proof?

[opaque scroll]

Yeah, so just to outline:

You start with F1 u G1 covering U. Then by definition this means there are two closed sets F and G that intersect with U to make F1 and G1.

The problem was that the union wasn't all of X, so you need to enlarge them somehow

[north ore]

right

ok i will try to come up with something using your hint

[north ore]

ok so we know that $U\subset F\cup G$ and since $U$ is open, $X-U$ is closed so that $X=F\cup G\cup (X-U)$. Furthermore, $U\cap (X-U)=\emptyset$ so we can write $F_1=U\cap F=U\cap (F\cup (X-U))$ and $G_1=U\cap G=U\cap (G\cup (X-U))$. This should solve the problem right?

[gentle osprey]

ali yassine

[north ore]

tysm jagr, have a great day/night!

[opaque scroll]

Ditto!

[viscid blade]

So you are not only in grf

[opaque scroll]

Guess not

[hidden abyss]

He only does (co)homology which is a shame

[quartz horizon]

Is this correct

I'm trying to use this "mass exchange lemma" to prove the first part of Tietze:

[quartz horizon]

Ok I think I also figured out how to prove this for $\mathbb{R}^n$. first by the universal property of the product it suffices to do this for $\mathbb{R}$

then:

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[tiny ridge]

stupid opinion.

Munkres is fine for intro algebraic topology

I read it before I read Hatcher

[radiant stone]

stupid opinion
???
it fits you doesn't mean it's good for everyone

[tiny ridge]

That's correct. But that it didn't personally work for you doesn't mean it's useless

It's a good book. I'm a topologist, and I can say it's much better for being a topologist than many other books. It's basic but not useless

[lament steppe]

In prop 3.57 it is not clear to me if R includes the elements of the form (x,x)

[novel acorn]

it does

[lament steppe]

I think R does have to include these "diagnol" elements for this proposition to work... but just wanted to be extra sure.

Thanks!

[iron bolt]

if you let q be the identity, that proposition specialises to the fact that X is Hausdorff iff the diagonal in X ⨯ X is closed

if you were to exclude the diagonal, it would say that X is Hausdorff iff the empty set in X ⨯ X is closed... i.e., that every space is Hausdorff

so you can get concrete counterexamples from any non-Hausdorff space via that

[quartz horizon]

Help is 1.2.5 even true

https://math.stackexchange.com/questions/4900626/sufficient-condition-for-an-attaching-space-to-be-hausdorff

It appears not based on the stackexchange answer

they took $X = Y = \mathbb{R}$, $A = \mathbb{R} \backslash {0}$, and both maps defining the pushout to be inclusions

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

then X and Y are metrisable so satisfy any separation property you'd care about, and A is open in X so in particular a retract of one

but the pushout is the line with two origins, and that ain't hausdorff

did i read this wrong or something?

[tiny obsidian]

Uh if memory serves me right "retract" isn't an obvious thing

In particular I'm not sure A is a retract

[quartz horizon]

well A is always a retract of itself right?

[tiny obsidian]

Normally yes, but you sometimes want to specify A closed (at least in algtop) and there's a chance this has been done

If it hasn't then yes I agree that that was a counterexample

[quartz horizon]

i see

this sort of point-set stuff really sucks to work with

[tiny obsidian]

fwiw I've never seen this done with just the word "retract", but I'm also lacking context and for all I know it said "from here assume A closed"

[unreal stratus]

I agree with you

And also that it should probably be closed

Well observed tbh

[unreal stratus]

Actually it says above j is a closed embedding. Isn't this used here?

Or at least I can't tell what j is from context but it seems to be implicit there

[quartz horizon]

Well it’s a separate proposition so I wasn’t sure…

This is from Tom dieck’s book

[unreal stratus]

It seems they're using the same notation of Y u A so I'd guess so

Like from the thing in-between

To me it seems it is meant to be that we have maps f: A -> Y or whatever and j: A -> X a closed embedding

And then yes result will be good

[quartz horizon]

Speaking of embeddings

Is it true that a map f : A -> X is an embedding iff it is injective, continuous, and sends open subsets of A to open subsets of f(A)?

[warped helm]

thats about the definition isnt it?

embedding iff homeomorphism onto its image

[quartz horizon]

Cool, I just wanted to check I got it right

Also

Open embedding iff embedding that is an open map?

And closed embedding iff embedding that is a closed map?

[unreal stratus]

I would say these are definitions

[quartz horizon]

Ok

[unreal stratus]

But they are equivalent to being an embedding whose image is open resp. closed

[quartz horizon]

Right right

[quartz horizon]

Also is it bad if I found proving these painful

[warped helm]

looks painful

[quartz horizon]

Wrote up the stuff about topological embeddings

[alpine nest]

Just to note, this is an application of a general fact that a continuous bijection is a homeomorphism if and only if it's an open map.

An embedding is a bijection onto the image, hence the embedding version of the statement

[prime kindle]

How much analysis should I study before properly studying topology? If someone could recommend a self-contained resource (book/lecture) series, I'd greatly appreciate it

[gritty widget]

As long as you know basic things about the real numbers, R^n, open sets and continuous functions you should be fine

[kind marlin]

is there a topology such that there exists an n-point compactification of the topology for each natural number n? (using this definition)

where an n-point compactificaiton would just be a compactification where Y - X has n points

oh i have an idea

maybe you have real line topologies [0, inf), and glue together N of them at 0

(N = cardinality of naturals)

then you could split the individual lines into any number of subsets and "compactify" the end of each subset with an individual point

[ruby delta]

hmmm, is there a cursed way to just "double" that single point into n points?

I guess that would kill the hausdorffness

[ruby delta]

actually, the straightforward construction is to make a space that admits a countable point compactification

and then every finite compactification is just a quotient of that

so something like the origin with countably many copies of [0,1) coming out of it should work

[kind marlin]

thats funny we arrived at the same idea

[ruby delta]

oh you did

[kind marlin]

it does seem most straightforward that way

like an octopus grouping its legs together

well an octopus with countably infinite legs

[ruby delta]

there are slight issues with this compactification I think

if you only compactify the endpoints of the legs the whole space might not still be compact

because the trunks of the legs might still converge to a missing line

fuck it, an even easier construction, take [0,1] and poke out the 1/n points

no way that fails

[kind marlin]

mm if we dont let the open neighborhoods of each leg "intersect" the others i think its fine

like my reasoning was really just taking a quotient of a bunch of separate [0, inf) topologies, i dont think the nonzero points on one leg can really converge to nonzero points on other legs in that setting

maybe ill just write out a proof to be sure

[whole sapphire]

But in this definition, the word "the" is appropriate in the following sense.

Let Y1 and Y2 be chaus spaces. Let X1 and X2 be proper subspaces of them resp. such that the closure of Xi is Yi (i=1,2). And Suppose Y1 and Y2 are the one-point compactifiacations of X1 and X2 resp.
I think it is provable that if X1 is homeo to X2, then Y1 is homeo to Y2.

And I think for every space haus X there exists, upto homeo, a unique chaus space Y such that Y is the one point compactifiaction of X.

However, can we still prove this for the n-point compactifiaction?
If not, then it seems....undesirable?

[ruby delta]

yeah it's not universal, I am very sure it is not

because for the uniqeness up to homeomorphism to work you need the n points to be topologically distinct somehow

so this is just for fun, lol

[kind marlin]

oh the topology fails bc u can take an individual open set on each leg and then cover the "center of the octopus" and there won't be a finite subcover, idk if that's what you meant but i do see issues now

hmm at the very least for any finite k there's a topology with n-point compactifications for 1 <= n <= k

this kind of sounds impossible now for unbounded finite points

[ruby delta]

does this not work

[kind marlin]

oh i didnt see that i will look

[kind marlin]

actually could you explain what the n-point compactification would be there? I’m not sure if my mental model is accurate

[ruby delta]

yeah, the difficulty in this example is showing that the n-point "compacfitication" is compact

whereas in the original example the difficulty was showing that the big one was compact

[ruby delta]

but it should be homeomorphic to [0,1] where you glue all the 1/n's together

its weird looking but I think it should work

[kind marlin]

hm like the “1 point compactification” here would look like a flower sort of

Oh and any neighborhood of 0 will enclose all but finitely many of the “petals” I think

And then the n-point compactification would have n flowers

[ruby delta]

yeah and compactness is no worry because the petals shrink to 0

[kind marlin]

oh analogously we could take

[0, 1)
[0, 1/2)
[0, 1/3)
…
and glue them all together at 0

so the fix is just ensuring that there’s a rooted point whose neighborhoods always only leaves finitely many “compactifiable” spaces

[plucky plinth]

Is it normal to discuss projective limit topology in an undergrad course?

[rancid umbra]

yea

[plucky plinth]

What context?

[rancid umbra]

p-adic integers, profinite groups, inverse systems

limits in general

what context are you learning them in?

[plucky plinth]

I remember learning about them when discussing algebraic completion in commutative algebra. Ive seen it in other context but never my ugrad topology class

Also I dont feel like my topology class discussed p-adic integers from an inverse limit perspective. I think they just introduced the norm and said the completion exists

Im just curious because its showing up again and im learning about construction of distributions and trying to think of first time someone might see it

[urban zinc]

I don't think it's normal in the context of an undergrad topology class but it doesn't seem unreasonable

[cosmic mirage]

I wish I'd seen it in undergrad!

[plucky plinth]

Why is that?

[cosmic mirage]

its not so hard to define and there's been times when ive wanted to understand the p-adics of galois groups (of infinite-dimensional extensions) etc

and for example the pointset problems surrounding CW complexes were never explained to me and i still dont fully understand it

and this is just whatever colimit topology it has to be

anyways, just a bunch of small things that wouldve taken like 15 minutes in a lecture and are basic enough to be taught in undergrad

although maybe the 15 minute figure is my estimate based on my comfort with the material now instead of then

[silent garnet]

is there any reason to care about separation properties besides T0, T1 and T2?

[quartz horizon]

T4 gets you Urysohn's lemma

[tender halo]

T3 and T5 are kind of cringe

they are not really naturall settings for anything, theorems with T3 are usually about getting stronger properties

T5 is just there for taxonomy sake

[silent garnet]

so i guess even wilder stuff like pseudoregular, completely normal and semiregular are similar in that sense?

[tender halo]

yea

collectionwise normal is a more interesting stronger condition for normality

T3.5 is very important though

it is a natural setting for a bunch of stuff

[silent garnet]

for example? i only know about urysohn's lemma

oh wait no the lemma doesnt work there

[tender halo]

a space is tychonoff iff its a subspace of I^\kappa for a cardinal kappa iff it can be embedded into a hausdorff compact space

also preserved by products

also dimension theory kinda needs T3.5 to get useful stuff

(co-)zero sets are important for pointset, and tychonoff spaces are precisely the ones where cozero sets make a base

and tychonoff spaces are the natural setting to talk about algebras of real-valued functions

a hausdorff space is tychonoff iff its topology is completely determined by the set of its real valued functions

also this

[north ore]

how do i prove a? I am guessing that if i assume that X is not connected, then write X=AUB where A and B are disjoint open subsets of X and let p in A, q in B. Then there is no map f:[0,1]->X such that f(0)=p and f(1)=q. But i am not sure how to prove that such a function f cant be continuous

Also is using map for a continuous function standard?

[kind marlin]

yeah, you should assume lambda is continuous here

i think its mostly contextual but e.g. in most instances continuity is a basic property needed to do stuff

[kind marlin]

you're really close, just take preimages of A and B now

oh and if you already got the proof when assuming f is continuous then you already did it ill just let you respond with what you meant tbh

[rapid shuttle]

Use proof by contradiction

It's by definition

"there exists a map f which is a continuous function"

[kind marlin]

yeah, i was just wondering if the contention is with map meaning continuous or not

[silent garnet]

btw there is also a fun categorical proof using the fact that "path connected <=> all maps {*}->X are homotopic" and "connected <=> there are only two maps X->{a, b} ({a, b} with discrete topology)"

[iron bolt]

one thing I realised only recently is that a space is "locally closed" in the sense that every neighbourhood of a point contains a closed neighbourhood of that point iff it is T3

[silent garnet]

also in T3 every closed set is intersection of all its closed neighborhoods

but i haven't had a chance to use any of these two properties yet

[rapid shuttle]

Can you explain

This might not be the channel tho

[north ore]

yea i know that its continuous here because the author uses "map" to say "continuous map/function"

i was asking if its standard to do this

[iron bolt]

it's at the very least not uncommon

[north ore]

I see

[north ore]

the preimages are open and their union should be [0,1]

but they are disjoint

is that right?

[kind marlin]

yup

[iron bolt]

by the way, what is a bit nonstandard imo is "arcwise connected" and "arc component" - I think those are called path connected and path component in most cases

[north ore]

ah i see, since they are open then neither 0 nor 1 can be in any of them which is a contradiction

[kind marlin]

yeah when i first read this my thought was "arc connected -> path connected -> connected" and then realized that arc connected here literally is just path connected 😭

[iron bolt]

they're open with respect to the subspace topology, not in R

(i.e., [0,1) would be an example of an open set in [0,1])

[north ore]

I see, i thought that this was the definition of path connectedness for a moment but then i assumed that i am just remembering things wrong lol

[north ore]

oh ok i see

[kind marlin]

you found 2 disjoint sets whose union is [0, 1]. does that remind you of anything

[north ore]

anyways [0,1] is connected

so it cant be a union of 2 disjoint open subsets

now i see

thanks

[kind marlin]

(specifically nonempty, which you also have here)

[north ore]

tysm everyone have a great day/night

[north ore]

b follows because "p and q are in the same arc component" is an equivalence relation
c follows from a since components are maximal connected subsets of X.

[silent garnet]

Ok it turned out to be longer than i expected but:
There are only 2 maps {*}->{a, b}. We get at least two maps X->{a, b} by composing with \pi^-1 (\pi is monic). Any other such map f would be equal to one of the two maps we already got. There are only two maps X->{a, b} => X is connected
(p.s all the maps are homotopy equivalence classes of continuous maps)

[cerulean oriole]

Let X be a topological space. For every subset A of X, the set of open neighbourhoods of A is an upward-closed directed set of open subsets of X (i.e. a filter in the poset of open subsets). However, not every filter of open subsets of X is of this form. For example, let X = ℝ and F = {U : for some open neighbourhood V of 0, U contains all but countably many points of V} (which is the filter generated by the filter basis of all sets of the form (-1/n, 1/n) \ A for n a positive integer and A a countable set).

Have filters of open sets in a topological space been classified or described in any way? For example, for any point x in X, one can define a "germ of an open set at x" in an obvious way; let Opens_x be the set of all such germs. Any filter F of open sets of X restricts to a filter F_x in Opens_x for each x in X. Does the family of filters (F_x)_{x in X} uniquely determine F?

I mean, I expect the last guess is probably true, but hopefully something more than that can be said.

[tender halo]

i would look at the nlab entry for sober spaces

[cerulean oriole]

Any specific part? I skimmed it and it doesn't seem to talk about anything very close to this question, just how sober spaces are the same as spatial locales.

[cerulean oriole]

Maybe the right way to think about this is topos-theoretic - IIRC the subobject classifier is a "truth-value" sheaf Opens whose stalks are precisely the Opens_x's. So maybe I should check if a filter of open sets is the same as an internal filter in this poset object in the topos of sheaves?

[crisp lintel]

sentenced to looking at the nlab

https://tenor.com/view/sentenced-judge-gavel-life-guilty-gif-12475856

[gritty widget]

I haven't actually like, done any math in quite awhile, and my attention span has been a lot shorter lately. I've read Munkres and a large part of Topology & Groupoids, so I would like something kind of refreshing on the topics covered in those, but not anything super terse. Anyone have any recommendations? Book or series of lectures recorded or anything would be appreciated.

[rancid umbra]

topology: a categorical approach might be worth a read if you are already familiar with point-set topology

[gritty widget]

Alright, thanks!

[young stone]

If there exists a non constant continuous map from a space X to any space Y with cardinality greater than one, is X discrete?

[gritty widget]

This property is equivalent to the existence of a continuous non-constant map from X to discrete {0, 1}, which is equivalent to being not connected, by:

1. If F:X to {0, 1} is continuous, non-constant, and Y is a space of more than 1 point, then, for any two points g, h in Y, {0, 1}->{g, h} is continuous, thus we have a continuous map which isn't constant from X to Y by composition

2. If the original property is satisfied, there is a map from X to {0, 1} which is continuous and non-constant.

[young stone]

right thanks

[rancid umbra]

a top space X is discrete if and only if every function out of X is continuous

this comes from the free-forgetful adjunction.
let F : Set -> Top be the free functor taking a set A to the top space whose underlying set is A with the discrete topology, and U : Top -> Set be the forgetful functor.
we have Top(FA,Y) = Set(A,UY) for every set A and top space Y.

if every function out of a top space X is continuous, then Top(X,Y) = Set(UX,UY) = Top(FUX,Y) so by the Yoneda lemma, FUX and X are homeomorphic, which means that X is discrete.

dually, a space is codiscrete (or indiscrete) if and only if every function into it is continuous

[rancid umbra]

the only spaces which are both discrete and codiscrete are the empty space and the singleton space, due to size constraints on the topologies.

[crisp lintel]

or just take the identity map to the same set with the discrete topology and it's proven in one line

(though using yoneda is cool regardless)

[rancid umbra]

demon hour proofs

that is another equivalent condition tho

[warped rover]

Does anyone know of a good (modern) source to read about general uniform structures? There's Engelking, but I'd rather... not lol.

[urban zinc]

Lee's Intro to Topological Manifolds has some chapters at the beginning doing a breezy overview of point-set topology, it's very good

[gritty widget]

Alright, thank you!

[storm wadi]

I just got introduced to topology coming from real analysis so apologies if my question doesn't make 100% sense.
Obviously, from analysis my pre-conceived notion of an open set is that the open ball is contained.
So, why do we call every element of a topology an open set if not every topology has a metric? Is it simply just a name we assign to these elements or does it actually have any relation to the definition of open from metric spaces?

[hexed vault]

I think i am qualified enough to answer this question but if someone wants to add something, please do if i missed something

Essentially the definition/intuition of topology arises from 2 areas , logic and geometry

But both have a similar question to ask

You are given/constructed a set

What structure "truly" defines being 2 points being close to each other is ? 2 points of that set

The metric space covers that notion beautifully and i assume you know abt it

But if you notice how "closeness" in characterised via open sets in metric spaces , you may get the answer

You want to arbitarize the notion of closeness to be characterised by open sets , rather than a metric function and thats where the generalised notion of topology comes from

[young stone]

I don't think this is directly relevant to my question though, there could be some smaller subclass of continuous functions whose existence forces the space to be discrete

[rancid umbra]

there is, there is exactly one function that is required to be continuous for the entire space to be discrete (the identity function from X to FX)

[young stone]

Yeah and I was asking if the condition I gave forces it to be discrete

[rancid umbra]

but i thought it was relevant because it looked like you were trying to classify discrete spaces by outgoing maps

[hexed vault]

So , in essence , when you learn to know abt closeness in terms of open sets , ie 2 points are "relatively close" when they are in a common open sets , and comparison of open sets in terms of containment , you can say that the definition of topology achieves what you want to achieve via metric spaces but in a more "pure" form. But if you want advise , keep reading forward and keep this question in mind , you will eventually have answers to this. Remember , mathematicians took a 100 years to settle with a definition of topology , its not that obvious.

[young stone]

I was trying to find a detecting object in Top (no spoilers pls, im still trying it)

[rancid umbra]

idk what the precise meaning of detecting object is here

like how {0,1} "detects" connectedness?

[young stone]

an subcollection of objects D in a category is detecting if for every mono that is not iso A \to B, there exists a map from an object in D to B that doesnt factor through the mono

an object is detecting if the singleton is

im really sleepy i hope i didnt mess that up

[storm wadi]

This makes a lot of sense actually, that without a rigorous definition of a metric the best notion of closeness of two elements we have is simply those two elements being in the same set. Thank you

[young stone]

im guessing a detecting family can "detect" any property? I havent thought enough about the definition to say more

[rancid umbra]

where did you see this definition, just curious?

[young stone]

I dont think this is standard, just something we saw in class

[rancid umbra]

do you mind clarifying what this means?

[young stone]

yeah that was a horrible sentence, did you understand what it means for a family of objects to be detecting? An object A is detecting if {A} is detecting

well i shoudlve probably just defined it for objects, too sleepy

[rancid umbra]

lol maybe you can ping me after you get some sleep with the full definition?

[young stone]

Cool fact: Whitehead's theorem implies spheres are a detecting family in the homotopy category of pointed connected CW complexes

[young stone]

yeah good idea lol

[rancid umbra]

neat! and yea, get some rest if you can

[hexed vault]

Sorry but can help me notice how the yoneda lemma is being used here ?

[rancid umbra]

do you know about representable functors and representating objects?

[hexed vault]

Yes

[rancid umbra]

right. so given a category $\mathsf{C}$, the hom-functor $\mathsf{hom}(c,-) : \mathsf{C} \to \mathsf{Set}$ is represented by the object $c \in \mathsf{C}$

[gentle osprey]

c squared

[hexed vault]

Yes

[rancid umbra]

it is a consequence of the Yoneda lemma that two objects $c$ and $c'$ of $\mathsf{C}$ are isomorphic if and only if $\mathsf{hom}(c,-)$ and $\mathsf{hom}(c',-)$ are naturally isomorphic

[gentle osprey]

c squared

[hexed vault]

Ohh , i see , since hom(FUX,-) = hom(X,-) , yoneda lemma says that they must be isomorphic

[rancid umbra]

yea. this is a good exercise

[hexed vault]

Thank you very much

[rancid umbra]

to add some intuition - and echo Pseudo - this says that two objects are the same if they do the same things

[hexed vault]

Yeah , that makes sense because the f and g which are given by the isomorphism can be used as natural isomorphism for hom sets

And similarly vice versa i believe

In the reverse direction , it would require chasing the identity in both directions of natural transformations and that should recover our f and g back such that they are inverses of each other ?

[rancid umbra]

yea, that is essentially the proof

one way of interpreting what an object in a category does is that it determines a (representable) (co)presheaf, and the Yoneda lemma says that what an object is is the same as what it does, or how it is related to other objects

[hexed vault]

That particular presheaf is the hom functor for each object i would assume ?

[rancid umbra]

yea, hom(-,c) is the presheaf, hom(c,-) is the copresheaf

[hexed vault]

Oh damn , i have never heard the term copresheaf

But makes sense

Do you think i know a good amount of yoneda lemma ?

Or category theory for that matter ? I keep asking myself that and i never get a good answer

[rancid umbra]

i think that is something only you can answer, right?

[hexed vault]

Okay, lemme rephrase , do you think i have a good amount of clarity of thought and thinking about these topics ?

[rancid umbra]

i don't think it's fair to make a judgement based on this one interaction i can recall with you about category theory, ya know?

but also, like, do you feel like you know enough category to use it alongside the math you are already doing? are you satisfied with what you already know and how you think about things now?

[hexed vault]

1. i feel like i know enough to see and understand others use it but i am unable to use it myself like that slick adjunction proof

2. i am not satisfied by how much i know already , but i am very icky about the way how i think abt the things i know. I keep asking myself if i know the things i know deeply enough and no matter what happens , i always doubt myself on it because there are instances where i can use it but unable to discover the use myself.

When can i be sure that i know a topic deep "enough" ?

Get me ?

[rancid umbra]

yea, i get you. i wouldn't be too hard on yourself. keep practicing and looking for places where you can use, for example, the Yoneda lemma, or theorems about adjoints, or (co)limits. this will help with pattern recognition. i don't know if i have too much advice on deep understanding other than going through examples and meditating on them (oh talking with others helps a lot too). im saying this from a position of not knowing myself if my understanding of certain topics is particularly deep.

[hexed vault]

I see , but thank you for helping

[urban zinc]

It does have a relation. Metric spaces have one particular topology on them called the metric topology, and the sets in that topology are the ones where you can fit an open ball around every point. But topology generalizes the idea of open sets to be able to apply to more spaces, including ones that don't have a metric.

Does that help answer your question or is it confusing

[storm wadi]

Yeah that makes sense, thank you. Did the study of metric or topological spaces come about first do you know?

[urban zinc]

Metric spaces

Topological spaces are a more abstract concept that includes metric spaces but also a lot more exotic stuff, they weren't defined in their modern form until the 20th century

Although metric spaces are also a relatively recent formalization, around the beginning of the 20th century

[quartz horizon]

it took quite a while for the definition of topological space to be settled

[storm wadi]

Ah yes that would make sense. I find it interesting that often in math you go from specific to more general, whereas in other fields it is usually the reverse.

[quartz horizon]

what do you mean?

[storm wadi]

I mean you start off by learning inner product spaces, then normed spaces, then metric then topological. Intuitively, from an outside perspective you would expect study to start with the most general (topology) and become more specific as you progress. In traditional pedagogy it is the reverse. This applies to other areas of math outside analysis too, abstract algebra for example.

[quartz horizon]

i think the issue with going general -> specific is lack of motivation

[storm wadi]

Yeah absolutely

[quartz horizon]

believe me i am very familiar with this issue

being Category Theory #1 Fan and all

[hexed steppe]

in what field is it standard to teach general abstract frameworks before basic concrete examples?

humans dont learn anything that way

and dont teach them that way either

[storm wadi]

I guess it's not entirely comparable, but in other stem subjects you often learn at breadth first and then depth. A med student doesnt learn how to perform a specific surgery before first learning a little about everything.

In math you develop a rather specific toolset before generalising it to be able to solve a wider range of problems. You start off with one specific type of problem before asking questions about whether the techniques learnt from that problem can be applied to other, not immediately (seemingly) adjacent and more general problems

[crisp lintel]

I don't think that's really quite analogous because a med student is still learning very specific things

I think broad vs specific is a different spectrum than abstract vs concrete

[storm wadi]

Yeah that's a fair point, in hindsite my argument isn't very strong lol. I still find the route from specific to abstract interesting, even if it is logical.

[quartz horizon]

-# perhaps you'd enjoy cat theory

[hexed steppe]

learning a little about everything is the analogue of learning concrete examples in math

you develop basic core knowledge of anatomy, medicine, etc. before doing more advanced and specialized things

in math you develop basic core knowledge of metric spaces, calculus, etc. before doing more advanced specialized things like topology

[hexed steppe]

it is interesting

you can probably find people on this server who learned abstract stuff without internalizing the basics

[hexed vault]

There is a better way to go about this

The generalisation is a mere consequence of something much greater which is the isolation of a concept.

Isolation of the literal essence of "why" something happens

And essentially that is what serves as motivation of finding mathematical objects

For example , groups arise from symmetries but also from sets of functions , what essentially captures the essence of groups(and what the two things i mentioned before have in common) is group actions

Group actions are thus the motivation to study groups

And you will see this throughout mathematics

The generalisation that you speak of from inner prod spaces to norm spaces is not a generalisation but rather "straining" or "filtering" or boiling off the unnecessary jargon like inner products in a pursuit for a much deeper question of "What is closeness ?" And thus the development of topology

You are removing "distractions" and seeing what are the actual "causes" (axioms) which are responsible for "consequences" aka theorems and proofs

In similar ways , Category Theory talks about "maps" and what do they characterise out of the context of objects and elements and thus it is the literal core of maps and how they work. Therefore being extremely important all around mathematics.