point-set-topology

[drowsy iron]

I see, thank you very much

[sonic crane]

I am trying to show product of two connected spaces is connected. I was able to show it by contradiction when X x Y = (U1 x V1) union (U2 x V2) is a disjoint union of opens in X x Y that make up X x Y, but I am having trouble when going to general case of some arbitrary open set in X x Y

When X x Y = (unions of some Ui x Vi) union (unions of some Uj x Vj)

[tiny obsidian]

X x Y = (U1 x V1) union (U2 x V2) is a highly unusual way to write the product, since purely as sets you need very specific U1, U2, V1, V2 to have a decomposition like this

Either U1 = X and V1 = Y, or U1=U2=X and the V1 u V2 = Y

Or the equivalent statements for the other way around

[tiny obsidian]

I think a better place to start whether you have any intuition for why the result is even true

Alternatively, if you have any intuition for why the product of path connected spaces is path connected, and perhaps see if that argument can be generalised in any way

[sonic crane]

I was just trying to be like: if X x Y was not connected then it is a disjoint union of open sets

And I just took U1 x V1 and U2 x V2 to be the disjoint open sets

But yeah, youre right i dont really have much intuition rn for why its true. I feel like whenever i see product topology im not really sure how to intuitively think of the topology of it

[tiny obsidian]

I mean, the obvious place to start with intuition is R^2

specifically the open balls induced by the 1-norm |(x,y)| = |x|+|y| are squares, which are basic opens in the product topology on RxR

[sonic crane]

Yeah

I guess I just get uncomfortable with how not every open set in the product can be like “open cross open”

But yes even earlier today i realized not even every subset can be written like “subset cross subset” anyway

[tiny obsidian]

Try prove (or at least justify) that a square is connected (taking for granted that an interval is connected). In particular pay attention to what subsets of the square you know are connected

And then see what you can generalise to a general product of topological spaces

[sonic crane]

A square is connected considered in the subspace topology from R^2?

[quick crane]

Actually it's easy to prove that the product of path connected spaces is path connected, though this might not help you generalize.

[plush folio]

Isn't that the converse of what you're trying to show? For the contrapositive I think you should assume either X or Y is disconnected, then show that X x Y is disconnected. In this case I find it easier to write the separation as a clopen set U instead of two disjoint open sets U and V

[fair idol]

Is there an easy example of a continuous bijective function that isn't a homomorphism?

[tiny obsidian]

yes, although I now realise I shouldprobaby clarify I mean the inside as well (like a disc, not a circle), so that this is the same topology as the product topology of I x I where I is the interval [0,1] (or really any interval will do)

another example if you choose the correct domain:

yet another example could be seen from the fact that any function into a set with the indiscrete / trivial topology is continuous, but very few functions out of it are continuous. So the function f(x)=x considered as going from a set with a nontrivial topology to the same set with the indiscrete topology would be continuous, bijective, but with discontinuous inverse (this generalises to whenever the codomain has a coarser topology on the same set as the domain)

[sonic crane]

Proving this square is connected using the fact that [0,1] is connected?

[bronze wadi]

if Y is the one point compactification of X, is the subspace topology of X the same as the topology of X?

[rancid umbra]

yes

X should get embedded into its one point compactification

e.g., open unit interval gets embedded into S^1, the open unit square gets embedded into S^2

[bronze wadi]

alright thx

[potent prism]

Hello, I'm looking for references describing topologies on natural numbers? I'm wondering if these even have any interesting properties?

[craggy cedar]

Topologies on countable sets can be interesting, sure. There's the discrete topology, cofinite topology, the topology of the rationals (fun fact, by the way, metrizable, countable metric spaces without isolated points are all homeomorphic to Q!), and I'm sure various others that don't come to mind immediately

[grave solstice]

The p-adic topologies are very interesting

If you don't impose conditions you will have many uninteresting topologies, so you should narrow it a bit, imo

[prime elbow]

Furstenberg topology

[tender halo]

p adic topology is just a cantor space topologically though

and definitely not countable

[chrome dew]

I guess it's worth mentioning Ostrowski's theorem limits what properties your topologies can have to not being induced by an absolute value

[low flame]

is this essentially saying that (X, Sigma 1) has less open sets, meaning its coraser?

Therefore it may not seperate points from the interior vs the boundry as finly as (X, Sigma 2)?

Thus whats in the closure of A in (X, Sigma 1) >= closure A in (X, Sigma 2) since the open sets in (X, sigma 2) can seperate points further???

theres less open sets, so theres less closed sets?

so you can take an intersection further in (X, sigma 2) so its always gonna be equal or smaller

[unreal stratus]

Yeah exactly

Your description is essentially the proof lol

You can also write it quite cleanly: note it is enough to show that Cl_1 A is closed in (X,omega_2) (why?)

[low flame]

because theres an open set thats the compliemnt of it in Omega_2?

[unreal stratus]

Yes correct, I meant like why does that mean we're done? :)

[low flame]

ah i get it

i think hah

because CL2 is the smallest set containing A in omega 2

thus its has to be a subset of a bigger set IE CL1

[unreal stratus]

Yes exactly

[unreal stratus]

Smallest closed but ye

[low flame]

i like this problem

very nice

[unreal stratus]

It's helpful to think of closures as the smallest closed thing containing it

Bit easier to work with

[low flame]

alrighty yeah i get it.

How do you determine which problems are important? im doing self study but i feel i get bogged down on trival problems sometimes.

how do you figure it out? im doing viros problem book

for example i dont know if this is important or not in the greater scope of topology

[unreal stratus]

To be honest it is hard to know I'd say

unless you have already learnt the subject lol

[low flame]

True, True. oh well. ill keep pickin

[unreal stratus]

But for the record like

This question is imo not very important for most users of pointset topology (including me) but is a fun thing to consider

The other you jusy did is good to know and a good exercise, like just helping develop intuition

[low flame]

good, i dont wanna think about how many different combinations of closure and interior i can do on a set hahaha

[tiny obsidian]

Yes

[prime elbow]

It is interesting one

[prime elbow]

Viro is good

[low flame]

yeah its a great book, its why im doing half a page to a page a day 🙂

[prime elbow]

If you want, I have one related article to it

[low flame]

go ahead

[sonic crane]

Yeah im having trouble with showing product of connected spaces is connected

Can I use this somehow?

[prime elbow]

And I am not doing 🥲

[prime elbow]

Yes I think

[sonic crane]

Any hint on how? Ive been strugglin

[prime elbow]

Let me see Munkres 🥲😂

[sonic crane]

Can I use the fact that union of connected is connected if they have nonempty intersection?

[prime elbow]

Actually this is important

[sonic crane]

Yeah can I write X x Y as union of Xy Yx kind of guys

[prime elbow]

Can you show that X × Y is a union of connected space such that they have a common point ?

[sonic crane]

I think so

[prime elbow]

If you can then you done it

[prime elbow]

This exercise shows you that Y_x is connected

[sonic crane]

Ya

For some x and y fixed, Xy and Yx share something so their union is connected

Then X x Y is just union of those guys over x in X and y in Y

[prime elbow]

Yes

[prime elbow]

But you have to show that every set has a common point, in case of fixed x and y you have different common point

X_a and X_b have no common point

Which book is that ?

[sonic crane]

This is my homework from class actually

[prime elbow]

Interesting

You can refer to Munkres

[sonic crane]

But (Xa union Yb) and (Xc union Yd) have a common point right

So union of all (Xi union Yj) is X x Y

[prime elbow]

Which point ?

[sonic crane]

(Xi union Yj) unioned over all i in Y j in X?

[prime elbow]

But ( Xa union Yb) and (Xc union Yd) have no common point

[sonic crane]

Really?

[prime elbow]

Sorry I don't see, which one ?

[sonic crane]

(d, a) is in Xa and thats in Yd

[sonic crane]

Well they have a common point for the same reason that any Xi and Yj have a common point ..?

[prime elbow]

yes (j,i)

[prime elbow]

https://math.osu.edu/sites/math.osu.edu/files/Kuratowski14Sets.pdf

[prime elbow]

i think you are correct, now you can look at Munkres, there are same approach if you look at that you will understand why i am refer to Munkres

[sonic crane]

🫶

[plush folio]

Stupid question: isn't (X \ U) x Y union U x Y equal to X x Y?

[prime elbow]

Yes it covers all X × Y

[sonic crane]

Showing R and R^n are not homeomorphic: my only thought is something to do with if u remove a point from R you get a disconnected space but if you remove a point from R^n its still connected:

so to prove it, can I suppose there is a homeomorphism, and then show that the image of R remove 0 must be disjoint opens that make up R^n{point removed} but thats not possible since R^n{point removed} is comnected

[plush folio]

Okay, then I think it should be straightforward to prove your proposition by contrapositive @sonic crane. Assume WLOG that X is disconnected and prove that X x Y is disconnected

[prime elbow]

Yes idea is correct

[sonic crane]

The contrapositive is “if X x Y is not connected then X or Y is not connected”

Is that the same thing

[plush folio]

omg, sorry, I had a brainfart

[sonic crane]

I thinj its not the same thing

Np

Ya i was gonna say yesterday i dont think thats the contrapositive lol

[prime elbow]

@sonic crane please verify that because there is a chance that I am wrong but I think you already verified it.

I am not talking about set theory stuff

[sonic crane]

Oh?

Hmm

Oh

Well we are using that union of connected spaces are connected if they all have some common point

[prime elbow]

Yes

Now I am sharing the idea which used in Munkres

So let T_x = X × {b} union {x} × Y then T_x is connected.

Then T_x over x in X is X × Y and they have a common point a × b.

Same idea

[sonic crane]

7b) is easy isnt it? f is continuous and {c} is closed in R so f^-1({c}) is closed ?

Is it that easy or am i missing something? Also i wouldve thought it connected to 7 a) or else why are the questions linked like that

[prime elbow]

Yes it is correct

[prime elbow]

You know 7 a, is pasting lemma.

[sonic crane]

Yea i saw it in munkres before too ha

[prime elbow]

I think there is a link

[heady skiff]

can i have a hint for showing that the closure of a connected subset A is connected

i'm assuming for contradiction that it's not connected so that I can write Abar = U u V for disjoint open sets

(with subspace topology on Abar)

[tender halo]

if you have a family of connected subsets such that there is a subset A such that none of the others are separated with A, then their union is connected

in particular, take A and the set of singletons consisting of points that lie in the closure of A but not in A

[heady skiff]

wdym by separated with A

[tender halo]

A and B are separated means neither contain points from the closure of the other

[sonic crane]

Dude i was just looking at that in my notes

And i was confused on my profs proof

I was literally just looking at this and also thinking about it lol

[heady skiff]

lol nice

we didn't even go over connectedness in this class

mf just put all theorems on a problem set

[prime elbow]

It is equivalent to disconnected space definition

@sonic crane X is disconnected if and only if there exists a continuous map from X onto {0,1}, two-point discrete space.

[tender halo]

its not

[prime elbow]

If A is connected then every continuous mapping from A to {0,1} is constant

[prime elbow]

In metric space?

[tender halo]

its a definition about a pair of sets, not about a space

[prime elbow]

But if A and B are separated then A U B is disconnected

[tender halo]

not true

consider sets of positive and negative reals

[heady skiff]

Let $(X, \mathcal{T})$ be a topological space. We first show that if $A$ is a connected subspace of $X$ and $A \subset B \subseteq \overline{A}$, then $B$ is connected. From this, it will follow that $\overline{A}$ is connected, since $A \subset \overline{A} \subseteq \overline{A}$. Suppose that $B = U \cup V$ for disjoint open sets $U, V$. Then either $A \subset U$ or $A \subset V$ (otherwise $A = (U \cap A) \cup (V \cap A)$, which constitute a separation of $A$) so suppose that $A \subset U$. If $x \in V$, then $x \notin A$, since $A \subset U$ and $U$ and $V$ are disjoint. Hence $A \cap V = \varnothing$. Let $x \in V$, so that $x \in \overline{A}$. Then $V$ is an open set containing $x$, so $A \cap V \neq \varnothing$ by the definition of $\overline{A}$; since this a contradiction, we conclude that $B$ is connected.

how this look

[gentle osprey]

okeyokay

[tender halo]

they are separated but their union is connected

[prime elbow]

How it is union is connected there is no 0, means it is not interval

[tender halo]

uhh sorry i meant the opposite

they are not separated but their union is disconnected

[prime elbow]

How are they not separated?

[tender halo]

uhh ok im tripping up

https://tenor.com/view/cat-cute-fluffy-stare-awkward-gif-27184320

ah ok yeah ur right

sorry brain fart

[prime elbow]

Okay

@sonic crane So if you want to prove that A and B are connected then A × B is connected.

Let f: A × B -> {0,1} be continuous. We have to show that f is constant.

f(a,•): B -> {0,1} and f:(•, b' ) : A -> {0,1} is continuous.

Since A and B are connected so these maps are constant.

Thus, f(a,b) = f(a',b') because f(a,•) and f(•,b') is constant and agree at (a,b').

Hence, f is constant.

From Carothers

[unreal stratus]

This is a great proof lol

[prime elbow]

I think it doesn't matter which topology we are let on {0,1}

[plush folio]

It has to be the discrete topology, because any map to the trivial topology is continuous

[prime elbow]

Yes

In Janich, he introduced a disjoint union of Topological space.

Disjoint union of sets, if X and Y are sets, their disjoint union or sum is defined by
X + y = X × {0} union Y × {1}.

So topology on X + Y is given by { U + V | U is open in X and V is open in Y }.

So now I want to verify this, so an empty set belongs to this topology and also X + Y belongs to this topology.

Now let U_i + V_i be open sets over index set I, then their union is union U_i + union V_i, since U_i and V_i are open we get it is U + V.

Similarly for finite intersection.

Is it correct?

[sonic crane]

I was stuck showing R^2\Q^2 is path connected, but then i realize elements of this set dont have to be (irrational, irrational)

[plush folio]

took me a second products are confusing

[sonic crane]

My prof is following Janich. I do not like that book

[sonic crane]

Right?

They always trip me up

So much struggles in topology come from set theory

[plush folio]

Agree! And the fact that UxV is open in XxY doesn't mean that U and V are open in X and Y

[sonic crane]

Aww bruh yeah 😭

Why it gotta be like that lmao

[prime elbow]

Any reason?

[sonic crane]

Well i feel like its more of a casual discussion than a textbook

I feel like im missing so many details when reading it

[prime elbow]

Okay

[heady skiff]

is this a mistake? d(f(y), s_i) < e/2 by the triangle inequality

so e/2 is too strict of an upper bound

also I don't see how this shows that f is in F_{\psi}

[unreal stratus]

Yeah lol

It's quite cute actually uh

You can prove that R^2 \ X is path-connected where X is any countable subset of R^2

[sonic crane]

Oooh

Poopology

[lunar linden]

Hey do you guys have an example of an open map that discontinuous from IR to IR

where IR is endowed with the standard euclidean topology

[plush folio]

Here's an interesting one: https://math.stackexchange.com/a/2933144

[paper wedge]

is S^2 removed from R^3 homotopy equivalent to S^2?

[paper wedge]

a sphere

but ig i got the answer

also similary how does R^3 complement the solid torus look like

[robust drum]

Wait, what? Aren’t the projections open maps?

pi_1(UxV) = U, which is open because U x V is open by assumption and because pi_1 is an open map

[white oxide]

Sphere wedge sum a circle.
Retract to the ball which contains the complement. This retracts to a sphere with a line segment connecting its poles which is then the wedge sum of a sphere with a circle.

[sonic crane]

Hmm @plush folio did u mean to say something different?

[iron bolt]

they are, but pi_1(UxV) = U only holds when V is non-empty

so there's a bit of an edge case there

[robust drum]

Hmm that’s insanely cursed

Well I mean actually

Not really that cursed

Cause that’s basically just saying “everything times emptyset is emptyset and thus everything times emptyset is open”

[plush folio]

Oops, yeah that's wrong I was thinking about how open sets (or any set really) in XxY are not necessarily of the form UxV

[heady skiff]

are the connected components not immediately connected due to the definition of the equivalence relation?

[paper wedge]

yeah

its asking u to show they are closed tho

[heady skiff]

ye i know i was just asking about the connected part

[paper wedge]

yeah

[heady skiff]

singletons are connected right

[tiny obsidian]

yes

two disjoint nonempty sets contain at least two points between them, so a singleton can't possibly be disconnected

[heady skiff]

assuming that i've shown that the closure of a connected set is also connected, could I just say that if y is in the closure of a connected component, then y ~ x for some x in that connected component and not in the closure, and hence y belongs to the connected component

Since $x \sim y$ if and only if $x$ and $y$ belong to the same connected subset, we see that $[x]_{\sim}$ consists of all points in $X$ which are contained in the same connected subset as $x$. Thus each connected component is connected by the definition of $\sim$. We now show that each connected component $C$ is closed. Clearly $C \subseteq \overline{C}$. Conversely, let $y \in \overline{C} \setminus C$. By problem 2, $\overline{C}$ is connected; in particular, $x \sim y$ for some $x \in C$. It follows that $y \in C$ by the definition of $C$, so $\overline{C} \subseteq C$.

[gentle osprey]

okeyokay

[hoary sphinx]

Yeah this is good

There's another way to think of [x]_~

It is the union of all conected subsets that contain x

If you had that already, then by the fact that the closure is also a connected set containing x you could conclude that the closure is contained in [x] directly

[hoary sphinx]

Is there a relatively simple example of a quotient map that isn't open or closed?

The map f:R²->R given by f(x,y)=x+y is open but not closed

And if X=[0,1]² with lexicographic order topology and we take
(x,y)~(x',y')
then the canonical projection is a quotient map and is not open

I think it's closed though

But I'm not completely sure

Anyway, is there some example where it's neither open or closed?

I have an idea maybe partitioning the ordered square into
[(x,0),(x,1)] if x is not 1/2
[(1/2,0),(1/2,1))
and {(1/2,1)}

The image of ((0,0),(0,1)) is {[(0,0),(0,1)]}

But that isn't open since the preimage of that is [(0,0),(0,1)] which isn't open

And I thinkkk that the image of the closed set [(1/2,0),(1/2,1)] isn't closed

Nvm it is lmao

[gentle osprey]

Shaloming Home

[vocal jay]

I think I see what I was missing: a local homeomorphism only guarantees there exists some open neighbourhood of a point such that those properties in the first picture hold, it doesn’t say you can pick any neighbourhood you would like.

[low flame]

1 is true, since you take the union of two sets that contain all their limit points

2 is false. For example take Closure(Q) intersect Closure(R). You would end up losing limit points right? so its false

or could i take the intersection of Q with R-Q, which is empty and then thats not everywhere dense right

[chrome dew]

yeah your original answer for 2 doesn't work but that does

[low flame]

cool cool yeah i see, this is fun thank you 🙂

[chrome dew]

yeah you're welcome

[sharp panther]

I want to show that the 2-sphere and Torus are not homeomorphic. My intuition is that if we remove a closed surve from torus such that the torus is path connected (and hence connected), the image of that curve under a homeomorphism would be again a closed curve on 2-sphere. Now I want to apply Jordan curve theorem but I am not sure if it needs subtle approach since the result is for R^2. Any help is much appreciated

[opaque scroll]

The result translates immediately to the 2-sphere if you know that R^2 is homeomorphic to the sphere minus a point

[unreal stratus]

There is

I am trying to remember one lol

Pretty sure there is an example in Munkres tho

[sharp panther]

Oh, you refer to the one-point compactification of R^2. I am sorry but how does it translate to 2-sphere exactly?

[tender halo]

one-point compactification of R^2 is the sphere

in general, Alexandroff compactification of R^n is S^n

[opaque scroll]

It is the one point compactification yes, but the important thing is that if you remove any point from the sphere you get the plane. So if you have a simple curve on the sphere, then if you just remove a point not in the curve you get a simple curve in the plane.

[heady skiff]

is this closed since it's the preimage of {0} x [-1, 1], which is closed in the product topology of R^2? also, how do we know that closed set => largest element?

[tender halo]

yes; compactness

[heady skiff]

oh ya

thanks

[heady skiff]

when showing something is connected, is contradiction usually the way to go

[heady skiff]

can somebody tell me if this proof is correct pls

[low flame]

7.D. In the set N of positive integers, the relation a|b (a divides b) is a
nonstrict partial order.
7.1. Is the relation a|b a nonstrict partial order in the set Z of integers?

this is a partial order right? since like 3 doesnt divide 4, but 2 does?

[plush folio]

yeah, I think that works 👍 I think you should write U = union of pi^{-1}(y) though, since U = {pi^{-1}(y) | y in Y} is a set of sets

[heady skiff]

oh ye ur right

thanks

[low flame]

ag wait i see it isnt for example -1/1 = 1/-1 but -1 != 1

[woeful coral]

Suppose I have a metrizable space $X$ and a continuous function (wrt. metric) on $X \setminus A$ where $A$ is compact. Is there a way I can extend this continuous function to X?

[gentle osprey]

brayden

[woeful coral]

to being continuous on $X$ of course.

[gentle osprey]

brayden

[quick delta]

Not necessarily

Consider $\bR \setminus 0$ and the function which is 0 on $x < 0$, 1 on $x > 0$

[gentle osprey]

Micose

[woeful coral]

Okay, maybe ill just be a bit more specific.

I am working on the sphere S^{n-1} here.

And

The compact set A is just going to be the closure of a neighbourhood

[quick delta]

You can easily construct something that blows up at some point of the boundary of the neighbourhood

Something like 1/d(x, A) where d(x, A) = inf_{y \in A} d(x, y)

[woeful coral]

if the neighbourhood is open then maybe?

i.e. the function is uniformly continuous

[quick delta]

Are we still assuming A to be compact or no?

[woeful coral]

No. A is just open and simply connected.

say, A is an open ball.

on the sphere.

[quick delta]

I think the answer is finally yes
If A is an open ball, then extending the function from $\partial A$ to A continuously is equivalent to giving a homotopy of the restriction to $\partial A \cong S^{n-2}$ to the constant function, which you can do as $\bR$ is contractible

[gentle osprey]

Micose

[woeful coral]

Nice

And then if A is a neighbourhood an argument can be made using this

not nessecarily connected

[heady skiff]

why can we replace U with U n U'?

moreover, why can we assume U bar is compact

[tiny obsidian]

Instead of spending the rest of the proof talking about U n U'

They're saying we're going to refer to it as U since it has nicer properties

Clearly if we find U_x subseteq U n U' with the desired property, it will also work for our original U so we will have still proved the lemma

And (U n U') bar is compact, hence why we can assume U bar is compact (because we've renamed U n U' to U)

[heady skiff]

got it thanks

why is the highlighted part in the last line of this screenshot true

[tiny obsidian]

Could you send the remainder of that argument so we can see what the "hence" is for

[heady skiff]

sure one sec

[tiny obsidian]

ngl

I have absolutely no idea

And this whole proposition and proof is just a mess

Why is there a vector space V in the set up of the proposition that isn't ever referred to again

"open closed W subseteq V" is another disaster given that I'm pretty sure that's supposed to be "closed W subseteq R" (just because nothing else would make sense in that context)

[heady skiff]

yeah i think these notes were typed up during class and the prof tends to sometimes make errors or maybe he made a typo

maybe i can find a proof online

[sonic crane]

Im stuck on this

[hoary sphinx]

What's your problem?

What have you tried?

Remember that to prove a function is continuous you can also prove that preimage of a closed set is closed

[sonic crane]

i dont know like

Take C closed in Y, h_inv(C) = f_inv(C) union g_inv(C)

so uh they are both closed

ok done? like i dont get it

[plush folio]

Hmm, I don't think that can be the complete solution, because you haven't used the fact that A and B are closed. I think an implicit assumption here is that they're continuous in the subspace topology of X

but h_inv(C) = f_inv(C) union g_inv(C) should be correct

[hoary sphinx]

Yes lmao

Like yeah f_inv(C) is closed

Given that f is continuous

Same with g

And that's it lmao

Oh wait

Oh yeah there's the detail that f_inv(C) is closed in AuB given that it's a closed subset if A, which is closed in AuB

[sonic crane]

ok cause right now we know finv(C) is closed in A but maybe not closed in X?

[hoary sphinx]

Yep

[sonic crane]

its closed in A because

[hoary sphinx]

But closed subsets of A look like A intersected with a closed subset of AuB

[sonic crane]

and A is closed

[hoary sphinx]

Yep

[sonic crane]

When are we using the fact that f = g on A intersect B

[hoary sphinx]

Ans intersection of closed sets is closed

[sonic crane]

is that just so the function is well defined

[hoary sphinx]

So that h is a function

[hoary sphinx]

Yes

[sonic crane]

my problem is we didnt even prove this in class

[hoary sphinx]

Hmm

[sonic crane]

so annoying like we didnt even prove that basis for subspace top is basis element intersect A

i dont know why we are not proving these helpful results

i feel like i gotta do topology handicapped

[hoary sphinx]

You didn't cover subspace topology?

[sonic crane]

We did

but we didnt cover that basis elements for it are basis of X intersected

for example

[hoary sphinx]

That's stupid ngl

[sonic crane]

Bro ik

lmao

like its making me confused

on the topics

cause i dont have those basic results in mind

[hoary sphinx]

Yeah one needs the whole picture

[sonic crane]

Yeah

[hoary sphinx]

Like although it seems kind of backwards

The more theorems you prove the more you understand the thing

[sonic crane]

Yeah exactly

[hoary sphinx]

Like you see how things relate much better

Intuition gets much much better

What book are you using?

[sonic crane]

So again for example we never said that closed sets in subspace topology are closed sets in X intersected with the subspace set

[sonic crane]

Janich. worst book ever

I hate it

[hoary sphinx]

I love munkres

[sonic crane]

I asked prof about munkres and he said he hates it

[hoary sphinx]

That sucks

[sonic crane]

Munkres was what i liked when i was studying a bit before classes started

He says its too dry

it is a bit dry but at least it really gives u the full picture

[hoary sphinx]

Yes Munkres is very thorough

[sonic crane]

When im reading janich im like bro is not explaining anything

[hoary sphinx]

Also my class is very good

Im taking topology rn too

[sonic crane]

lucky

[hoary sphinx]

Yeah that sucks

Well that's life i guess

[sonic crane]

Yup

[hoary sphinx]

Sometimes you get a good book

Most of the time you don't

[sonic crane]

lol

yea and my prof for algebra is following hungerford

like bruh y not dummit and foote

hungerford scares me

lol

[hoary sphinx]

I dont like dummit and foote

[sonic crane]

thats fair tbh

[hoary sphinx]

But this is getting off-topic lmao

[sonic crane]

lame

[plush folio]

wow, I looked up the subspace topology in Janich, and he literally devotes only three paragraphs to it

[hoary sphinx]

Oh

Lmao

[sonic crane]

😂

its like coffee-table topology

put it on the coffee table for like a casual whatever read

[plush folio]

I really like Lee's ITM for topology, I think he explains a bit better than munkres, and he covers the universal property for subspaces, products and quotients

[sonic crane]

ITM?

[plush folio]

introduction to topological manifolds

[sonic crane]

Oh i see yea i wanted to get Lees books on diff geo

[hoary sphinx]

You know you study math

When you start throwing acronyms around

And they mean introduction to topological manifolds lmao

[sonic crane]

for textbooks lol

yall like a/m?

i do fuck with atiyah macdonald

I like how its thorough but then at the same time doesnt painstakingly explain each detail

forces the reader to figure it out more properly themselves

[dire warren]

Do you need any more than 3 lines for subspace topology tho

[rich cliff]

I'm trying to understand why the Bit

``$X \cup_f Y$ consists of the disjoint union of $X$ and $(Y-A)$"

[gentle osprey]

plexcty

[rich cliff]

is true

So $X \sqcup Y$ has multiple definitions, say I choose to define it as ${(x. 0)~|~x \in X} \cup {(y, 1)~|~y \in Y}$

[gentle osprey]

plexcty

[rich cliff]

then $X \cup_f Y = (X \sqcup Y)/\sim$ will be a particular subset of the power set of $X \sqcup Y$

[gentle osprey]

plexcty

[rich cliff]

This is clearly not the same "type" of object as $X \sqcup (Y-A)$

[gentle osprey]

plexcty

[rich cliff]

What am I missing?

[alpine nest]

coffee-table-top-ology

[plush folio]

well yeah, I would atleast expect a mention of the fact that C is closed in A iff it is the intersection of A and a closed set in X. And ideally a mention of the universal property, how the inclusion map is continuous, and maybe an example or two

[dire warren]

Fair enough universal property is important

[iron bolt]

they're not the same set, but there's a canonical bijection between them

taking each element of the disjoint union to the equivalence class it belongs to

[sonic crane]

For projection map X x Y onto X, i am trying to convince myself that opens in X x Y map to opens in X

Ok yea i thinj its good

Opens in X x Y are unions of (open X) x (open Y) so when you collect all the x values that will still be open in X

[iron bolt]

yep. specifically, the key lemma you figured out here is that in order for a map to be open, it suffices to check whether the generators of the topology on the domain map to open sets

there's a few useful lemmas like that; for example, for checking continuity it suffices to check that the preimages of the generators are open etc

I'm sure there's more, but I forgot 💀

[sonic crane]

Yea! Thanks!

[wind bronze]

Unfortunately I do not have a copy of Steve Vickers' Topology via Logic. However, hearing about it made me curious what the "logical" approach to locales / topological spaces is. How do they arise from logic ? I'm really curious.

[rich cliff]

Thanks, I was able to prove that it is a bijection

[sonic crane]

The circle is homeomorphic to a closed interval?

or like half open interval?

its not

when considering as subspace topologies at least i guess

oh but we can still argue the circle is connected from the interval being connected which is what i was trying to say anyway

because u can still have a cts function

[shrewd night]

Exercise 12

[sonic crane]

I think this is not complete

Now Im thinking i need to show a common point for ALL Xy union Yx simultaneously

each pairwise thing i can find a common point but now im not sure how to show there is a common point for everything at once

[low flame]

A, B ⊂ X and any f : X → Y
f(X-A) = Y \ f(A)

Well if x is in X, and x is not in A then x is in X-A and F(X-A)

let X be {0,1,2} and A be {2}.

Now lets say we have the constant function. that gives us lets say {3}

Then F(X-A) = {3}
while X-A is {0,1}

then Y - F(A) is {3] - {3} = \empty

So theyre not equal? is this right?

Also i cant seem to find any kind of inclusion here

if you got a response please ping me im going to lay down to sleep 🙂

[prime elbow]

Okay but have you seen other proof?

[cerulean oriole]

Are there any more well-known constructions of the Stone–Čech compactification of a topological space X than the following?

• Take the closure of the image of X in ∏_{f: X → C, C compact Hausdorff, |C| <= 2^2^|X|} C.
• Take the closure of the image of X in ∏_{f: X → [0, 1]} [0, 1].
• When X is discrete, take the set of ultrafilters in the powerset P(X) of X (with an appropriately constructed topology).

[cerulean oriole]

That's correct: it's not true in general that f(X \ A) = Y \ f(A). Neither inclusion holds in general either.
To be more precise:

• Y \ f(A) ⊂ f(X \ A) if and only if Y ⊂ f(A) U f(X \ A) = f(X) if and only if f is surjective.
• f(X \ A) ⊂ Y \ f(A) if f is injective, since then f(x) for x not in A cannot be equal to f(x') for any x' in A.

[cerulean oriole]

Instead you can use the following lemma:

Let X be a topological space and { U_i : i in I } (with I non-empty) connected subspaces of X such that for any i, j in I, we can find i = i_0, i_1, ..., i_n = j in I such that U_{i_0} ∩ U_{i_1}, U_{i_1} ∩ U_{i_2}, ..., U_{i_{n-1}} ∩ U_{i_n} are all non-empty (in other words, for any two of the U_i's there is a chain of other U_i's joining them with consecutive U_i's intersecting). Then X is connected.
(Hint to prove the lemma: ||suppose Y is a clopen subset of X, so that we want to show that Y is either empty or all of X. Consider what the intersection of Y with each U_i can be. Then consider whether it is possible for Y to intersect one U_i but not another.||)

[cerulean oriole]

bumping since I buried my own question

[low flame]

Thank you!

[rancid umbra]

what have you tried?

we aren't just going to do this for you

[shrewd night]

I'm not so sure about how to start. I'm just asking for hints

Never said I wanted you to do it for me

[rancid umbra]

you never said anything

anyways, start with the definition of compactness

if K is compact, then every open cover of K should have a finite subcover

[paper wedge]

does f(x,y) = (-x,-y) on the Torus serve as a counterexample for browuers fixed theorem?

on the torus.

also for the borsak-ulam?

[north dust]

Why is null set clopen?

[paper wedge]

it is open by the definition of the topology

and it is closed since its complement is open

[north dust]

so in real number set for ex because null set has no elements therefore it has no boundary points so it is open but because R/(empty set) is still R meaning it is open,we can conclude its closed too?

[candid wyvern]

in an ultrametric space an open ball is both closed and open. whats the intuition behind that. i got the proof via some mse but whats the intuition?

[prime elbow]

Yes if you have shown that the empty set is open then also it is closed. But in general, we define topology on a set X, as collection of subsets, in that collection we assumed that empty and X must belong, and follow some properties, closure under union and closure under finite intersection

[shrewd night]

Yeah I know that part but I'm still not sure about how to go from there.

[heady skiff]

does anybody know what 1_X is here?

is it just the function which sends everything in X to 1

[heady skiff]

I get that the map is continuous, but why is its inverse continuous? (example 5.6 at the bottom)

nvm

ok wait so X is compact

but how is A* hausdorff?

[plush folio]

I don't have an answer, sorry, but I'm curious, what subject is this? Looks like a mix of topology and algebra

[heady skiff]

these are course notes from a graduate class called "intro to topology and analysis"

sometimes the professor marc rieffel goes on rants about algebra and stuff

so it's included in the course notes

[plush folio]

I see I guess you're not following a particular book?

[heady skiff]

he recommended lang real and functional analysis

but yeah basically he just teaches as he wants

[plush folio]

Btw, is it possible they meant that A* has the weakest topology such that ev_a : A* -> k is continuous? Because a |-> ev_a is a map from A to Hom(A*, k), I'm struggling to understand what it means for this map to be continuous

[heady skiff]

yeah i talked to the guy who made these notes and he said that it doesn't make sense

and that he would revise it

[tiny obsidian]

yes, judging by the fact that it later sends x to 1

[heady skiff]

lol okay thanks

[grave solstice]

Let $\varphi : S^1\times S^1\to \R^3$ be defined by
$$
\varphi(e^{i\alpha}, e^{i\beta})=((a+b\cos\alpha)\cos\beta, (a+b\cos\alpha)\sin\beta, b\sin\alpha)
$$
where $a>b>0$ are constants. What is a simple way of proving that $\varphi$ gives a homeomorphism $S^1\times S^1\to \varphi(U)$, where $\varphi(U)$ has the subspace topology? That it's continuous and bijective is clear, it's just showing that it's open. I mean I can do it but I don't see a way to do it without writing too much

[gentle osprey]

croqueta3385

[plush folio]

I think you can use the closed map lemma, since S1 x S1 is compact and R^3 is Hausdorff

[grave solstice]

thanks!

[heady skiff]

the line "$x_0 \in S \text{ with } a \in B(x_0, \frac{\epsilon}{2})$" is a mistake right since we only know balls of radius epsilon cover $X$

[gentle osprey]

okeyokay

[heady skiff]

i guess it doesn't matter since we can just take epsilon = epsilon/2 in all parts prior to that sentence

[worldly star]

I’ve recently started studying topologies, and I learned about topology inheritance. In the lecture I’ve watched, however, said that if there is a continuous map $f:M\rightarrow N$, and we create a topological structure from a subset $(S,T); S \subseteq M$ then there is a continuous map $f|_S: S \rightarrow N$
In an attempt to prove this, I tried to show that there exists a continuous “identity map” so to speak, that: $id: S \rightarrow M; id(s) = s$, but I could only show it was continuous iff S was an element of the topology with respect to M. And as far as I’m concerned this is a special case of topological inheritance. If anyone can help me with this, I’ll be grateful.

[gentle osprey]

Schiza

[tiny obsidian]

you've made a mistake somewhere in your proof, since f|S is definitely continuous for all subsets S

you should post your attempted proof so that people can figure out what that mistake is

[worldly star]

I’ll try and formalize it once I’m home

[sonic crane]

If Y is a subspace of X and A is a subset of Y but is a connected subspace of X, is A a connected subspace of Y?

I said that weird and also i guess obviously its still a connected subspace

I said that so confusingly wtf 😂

Like the open sets of A as a subspace of Y is the same as A as a subspace of X right

[sonic crane]

Yes: my question is,

A is a subspace of Y which is a subspace of X. Is A considered as a subspace of X the same as it being under Y?

[quartz horizon]

This is true

[sonic crane]

Ty

For this, could we instead conclude by bar(A) subset bar(C), but C is closed in B bc it is part of a separation so bar(A) subset C

So B has to be contained entirely in C

Basically im asking could we have concluded like that instead of saying “since C bar and D are disjoint blah blah”

[plush folio]

Yeah, I prefer your argument

[plush folio]

If f : X -> Y and g : Y -> X are both continuous and surjective, is that enough to conclude that X and Y are homeomorphic?

I see, thanks for the examples 👍 I had an idea that if f : (X, T_1) -> (X, T_2) is continuous then T_2 is a subset of T_1, and vice versa for g : (X, T2) -> (X, T1), but I realize that argument doesn't quite work (I think it only shows that T_1 and T_2 are equinumerous)

[iron bolt]

yep. I'd say the reason for this is that the subspace topology on A w.r.t. X can be defined as the pullback of the topology on X along the inclusion A → X, this inclusion is the composition of the inclusions A → Y → X, and there is a lemma that pullbacks behave well under composition

i.e. whether you pull back the topology to Y or first and then to A or directly to A doesn't matter

of course there's many ways to prove simple things like that though, this lemma is just one of them (but a useful one I think)

[quartz horizon]

Hehe, taking the universal property route i see

I get that but honestly in this case it may be more straightforward to just look at the actual topology

[iron bolt]

that is true, but: pushforwards and pullbacks are my new best friends

the reason is that I've finally understood that they actually occur all over the place, and can be used to get the same results in many categories at once

topological spaces, measurable spaces, diffeological spaces, bornological spaces, converge spaces etc all have the property that the set of structures (i.e. topologies/sigma-algebras/...) on any fixed set forms a complete lattice, and that you can push and pull those along functions in a way that makes sense

same thing with topological/diffeological/measurable monoids/groups/modules and other things like that, just that you're looking at base categories other than Set there

[sonic crane]

Is this a category theory thing

[iron bolt]

what I wrote after that somewhat is, but pushforwards and pullbacks of topologies are just a topology thing

idk, maybe you even know it already but under a different name

what I mean by pushforward is that for any topological space X, set Y and function f : X → Y, there is a finest topology on Y such that f is continuous; it consists of all sets whose preimages are open

and in the other direction, for every set X, topological space Y and function f : X → Y, there is a coarsest topology on X such that f is continuous, consisting of all preimages of open sets in Y

so you can sort of pull back or push forward topologies along arbitrary functions. there's a lot of good properties this process has

[sonic crane]

“Every closed subspace of a compact space is compact”

What is a “closed” subspace? Is it a subspace that is closed in the ambient top space?

I guess there is really nothing else it could be lol

[quick crane]

I think subspace is the same as subset here

maybe they used the word "space" to indicate that the subset got the subspace topology

[iron bolt]

my guess is that they so far only told you what a compact space is, not what a compact set is, and so are using the word subspace to hint that the set can be considered as a space with the subset topology

[sonic crane]

Yea this was just from munkres

Does it even necessarily matter if we throw out the X-Y set ?

Ohh yea cause we just want finite subcollection from A

[worldly star]

Regarding this, I think I proved it, here’s my work (a bit messy)

bruh

I dind't respond to the original post

hold on

[worldly star]

this

[fierce lily]

Can someone tell me what does the first sentence agree on a dense open of X that is not the same on all of X mean? And if possible, can someone give me some hints for this one?

[paper wedge]

you want two maps f and g

such that f(D) = g(D) where D is a dense open subset

but f is not equal to g

[fierce lily]

for a dense open subset, I can think of the complement of cantor set, denote [0,1]/C, but I can not come up with a space non-hausdorff, any suggestions here?

[paper wedge]

the cofinite topology is non-hausdorff

try that

so maybe an example would be take some discrete space and map it into a cofinite space

so that the continuity is easy to check

[fierce lily]

i consider map from {1,2,3}->{ empty, {1,2,3}, {1,2},{1,3},{2,3}, {1},{2},{3} }, do you think this one will work?

[paper wedge]

you need two functions

that agree on a dense set but do not agree over the whole space

[fierce lily]

so another set still from {1,2,3} but onto different space {empty, {1,2,3}, {1}}, is that correct?

[fierce lily]

I have a question: Is discrete space dense?

[paper wedge]

wdym dense

a set is dense iff its closure is the whole space

a space is seperable if it has a countable dense subset

are u asking if the discrete space is seperbale?

seperable*

[paper wedge]

wdym another set

do u mean another function

u need maps

[fierce lily]

yes, I seperately list f and g maps

[fierce lily]

the question writes agree on a dense open of X

[paper wedge]

what is ur dense subset

[fierce lily]

{1,2,3}, but I think it can not be dense subset so Z is correct?

[paper wedge]

do u know what the closure of a set is

[fierce lily]

i think closure of Z is Z, because all elements in this set is both open and closed, so they are all dense because it equals closure

[paper wedge]

what is z

[fierce lily]

i mean integer

[paper wedge]

im sorry im not following

what is X and Y in ur problem

and what are the two maps

[fierce lily]

f:{1,2,3}->{ empty, {1,2,3}, {1,2},{1,3},{2,3}, {1},{2},{3} } g:{1,2,3}->{empty, {1,2,3}, {1}}

[paper wedge]

.

also u have to define f

im not following

at all

[fierce lily]

what does define f mean?

[paper wedge]

i gtg continue doing hatcher

u should first define

whta is X and Y and what are the topologies

and then define two functions from X to Y such that they agree on a dense open subset but do not agree on the whole space

i think u should try doing it with more knonw topologies instead of like trying to define the topology on ur own

it is easy to prove the claim for when Y is hausdorff btw u can try htat

remember that a dense subset intersects every open

good luck

[fierce lily]

You said I can try a map from a discrete space to cofinite space, I know about cofinite topology but I can only come up with set like {1,2,3}(but it is hausdorff and I think it is cofinite space?),are you willing to share how you will construct these maps?

[paper wedge]

the cofinite topology on an infinite set is never hausdorff so u can take for example N

[fierce lily]

so you set Y=N with cofinite topology on N? And how will you set D such that f(D) =N?

[wise mist]

i have proved this, any hint for the converse to be false ?

[gentle osprey]

yeshua

[wise mist]

i should have used L_1 and L_inf to represent the norm

[slender glen]

if two metrics generate the same topology, what can we say about them?

[novel acorn]

https://en.m.wikipedia.org/wiki/Equivalence_of_metrics

[iron bolt]

not much aside from all topological properties being the same, I think

the uniform structure and bornologies on the two spaces can both be different, so pretty much anything related to uniform continuity or boundedness doesn't have to be preserved, I think

[wise mist]

will it be worthy to spend time with hilbert cube ?

[cerulean oriole]

Can a Hausdorff completely regular/Tychonoff/T3 space X be recovered from the R-algebra C(X) of continuous functions from X to R, possibly equipped with some further "purely R-algebra" structure on it like the topology of uniform convergence on compact sets?

As some motivation, this is possible when X is compact and Hausdorff because X = {maximal ideals of C(X)}, with the closed sets of X precisely the intersections of the zero-sets of families of elements of C(X) (in other words, with the weakest topology making every element of C(X) continuous).

[chrome trout]

Does the Cantor-Bendixson derivative respect countable union of nested sets?

for example, is $\left(\bigcup_{\beta<\alpha}U_\beta\right)^{(\gamma)}= \bigcup_{\beta<\alpha}U_\beta^{(\gamma)}$ if $U_1\subseteq U_2\dots$?

[gentle osprey]

qwerty

[chrome trout]

I seems like It should be true, but I can't quite prove t

[iron bolt]

by that derivative you mean just the set of accumulation points, right?

[chrome trout]

yes

[iron bolt]

do you place any condition on the U_β, like compactness? the source I found when looking Cantor-Bendixson derivatives up requires them to be closed and compact

also, what is α? you do mean to take the union over all countably many U_β, right?

[chrome trout]

The U_\beta sets are not necessarily closed or compact

yes take the union over countably many U_b

[iron bolt]

then your answer is no

I mean, that one of the two is a subset of the other is easy to see

but if you try to find a counterexample to the inclusion in the other direction, I think you'll find a simple one

[chrome trout]

Would the result be different if the U_b were compact?

[iron bolt]

the funny bit is that if the U_b are all compact Hausdorff and their union is too, the sequence actually becomes constant after finitely many U_b, so you actually just have the finite case

wait, no. I messed up somewhere

[chrome trout]

Hmm I see, I will work on it a bit and see what I come up with

[iron bolt]

there actually is a counterexample where the U_b and their union are all compact Hausdorff spaces

[iron bolt]

(just in case you're curious where I previously messed up: what I said here is true only if you also require each U_b to be a subset of the interior of the next)

[chrome trout]

Can I get a hint to the counterexample?

[iron bolt]

what I have in mind are some really simple subsets of [0,1]

you can also think about it more analytically though. every accumulation point of an U_b is also an accumulation point of their union; so if you want the equality to not be true, you're looking for an accumulation point of the union which is not the accumulation point of some U_b

||that point being an accumulation point could for example mean there's a sequence of points in the union of the U_b converging to it. what does it not being an accumulation point of any U_b mean for the elements of that sequence? can you construct such a scenario?||

[chrome trout]

Ah yes, the sets [1/n,1] are counterexamples, how dumb of me

Wait, not quite

[rancid umbra]

0 is a limit point of the left hand side while it is not in the right hand side (= (0,1])

[chrome trout]

No, the definition Im using is that the derivative of a set is the limit point of the set that is in the set

[rancid umbra]

0 is not a limit point of any of the [1/n,1] sets

[iron bolt]

it's also not in the union of the [1/n,1] though, which I think is what he means

it's not hard to modify the example though

[chrome trout]

Ok, I think I am misunderstanding the definition. I will read up some more. Thanks for the help

[wise mist]

the open unit ball in $(C[0,1],L_\infty)$ is not open in $(C[0,1],L_1)$

[gentle osprey]

yeshua

[wise mist]

help

or hint

i did 57.
i cant even fathom the hint 58(maybe the last line), so

[mellow basalt]

Draw a 'spike' function. It can be as high as you want and the area as small as you want. Then argue by contradiction.

[fierce lily]

Can someone tell me what does the - above intersection represent and why we can get these two intersection in this diagram?

it is about irreducible space

[tribal palm]

my best guess is that’s a (incredibly bad) notation for the closure

[visual rock]

\bar and not \overline

[gentle osprey]

emme

[limpid mural]

no help anymore, I am stupid bc C is a cross lol

[shrewd night]

If a set is connected, is the closure and interior also connected?

I also want to prove this so any ideas on how to start would be nice

[shrewd night]

I got the proof now

[fierce lily]

I have proven part a, b, c. I need help for part(d): I suppose cl(Z) is irreducible and cl(Z)=A union B with A,B proper closed of cl(Z), and it is also closed in X. I have shown if Z is open, then it works, and it is trivial if Z is closed. But if Z is neither open nor closed, I am stuck

summarize: I want to show if Z is neither open nor closed, cl(Z) is irreducible as well

or if you have alternative approach, please share

[novel acorn]

well if \overline Z is reducible it can be written as A \cup B
What happens when you intersect A and B with Z?

[fierce lily]

you mean (A union B) intersect Z?

[novel acorn]

intersect both individually

[fierce lily]

or A intersect Z and B intersect Z?

[novel acorn]

well that's fine

it doesn't change anything

[fierce lily]

so do you come up with the solution to this one?

[novel acorn]

A intersect Z is a closed subset of Z
B intersect Z is a closed subset of Z
Their union is Z
But since Z is irreducible Z has to be contained in one of them

now I'll let you do the rest

[fierce lily]

Z=A intersect Z or Z=B intersect Z. .hence Z contained in either A or B. since cl(Z) is smallest closed set containg Z but we can find either A or B be that smaller closed set containing Z, then we see the contradiction

[novel acorn]

Yeah basically

[heady skiff]

can i have a hint for finding a homeomorphism between (-oo, a) and (b, oo)

(explicitly and not just using some theorem for existence)

[unreal stratus]

First reduce to a = b = 0

Then imo it is pretty clear

[opaque scroll]

A hint could be that the number -1 exists

[heady skiff]

LOL okay

[heady skiff]

this would just be -1/x i hope

and then for the general case we just translate by (b - a)...

so something like -1/x + (b - a)

b - 1/(x - a) mayhaps

[tiny obsidian]

yes though you can pick an even easier map

[heady skiff]

oh just -x lol

[iron bolt]

yup

[heady skiff]

so something like -x + b - a

for the general case

wait

[iron bolt]

got an infinite ray pointing to the left and want it pointing to the right? just rotate it around

rigid body transformations already work in this case

[tiny obsidian]

almost, your signs aren't quite right

[heady skiff]

bro im trippin

i know -x won't work because what if a > 0 and b > 0

maybe -x + b + a

that would work if they're both positive

uhhh

ok yeah this would map into (b, oo)

[tacit basin]

For an example of a space $X$ which is not connected but $X/\sim$ with the quotient topology is, does $X=\R$ with the discrete topology and $a\sim b$ for all $a, b\in\R$ work?

[gentle osprey]

Sara

[quartz horizon]

Seems like it, yes

[low flame]

is this realted to the subspace topology at all?

[quartz horizon]

Yes in that the subspace topology is a special case of 9.5

And is also the same as 9.D

[paper wedge]

if i have a covering space E --> X

if x is in X; U being the nbd of x evenly covered

must the sheets cover E?

[solemn temple]

i don't think so

[arctic edge]

nop

helix cover circle

take any arc

containing x

that isn’t the whole circle

[paper wedge]

yeah makes sense

i saw this proof of something and it said that so ig im misunderstanding

ty every1

[shrewd night]

21

Did a and c

[iron bolt]

do you know that the interval is connected?

if so, that's your hint

[sonic crane]

How do we know exactly why c = sup A0 must be in [a,b]?

Is it just because b is an upper bound of A0 and b is in [a,b] so the least upper bound of A0 surely will be in [a,b]

[shrewd night]

Which one?

(0,1)??

[iron bolt]

oh, I only just saw that the map p defined there has domain R. I thought it was defined on some interval

R then

[shrewd night]

R is definitely connected

[iron bolt]

yep. then what can you say about A_0 and B_0, using a)?

[shrewd night]

It says they're separated which I managed to provr

Prove*

Hold on a sec

[iron bolt]

a) says they're seperated, yes

[shrewd night]

OMG of course

I think I got it

If t_0 doesn't exist then R would not be connected because it would be a union of separated subsets

As in if there was no t_0 not in A union B @iron bolt

[iron bolt]

the exercise is phrased a little weird, because it defines the map on R and but then asks you to find a t in (0,1)

so it is actually connectedness of (0,1) you need

but either way, you got the idea

if case you're curious, I would actually prove that convex subsets are connected with a slightly different sequence of steps than the exercise suggests:

1. [0,1] is connected
2. images of connected sets under continuous maps are connected, so γ([0,1]) is connected for any path γ
3. a set is connected if every two points in it are contained in a connected subset of the set, so in particular path-connected sets are connected
4. star-shaped sets are path-connected
5. convex sets are star-shaped

each if those should be doable if you think about it for a bit

[sonic crane]

Wow, im like severely confused by this proof

[iron bolt]

and it does tell you a lot more about connected sets than just proving the result for convex sets directly

[sonic crane]

Why is there some open interval (d,c] contained in B0

Im confused why its closed at that endpoint

Oh

(d,c] is not an open interval

Ok

Ok im slowly understanding this but its more confusing than i first thought it was gonna be tbh

[sonic crane]

Verifying my understanding of a proof of the claim: If X is connected and f: X -> Y is locally constant for each x in X, then f is constant on all of Y.

Choose a y in f(X) so the set finv({y}) is nonempty. finv({y}) is open, since each x in there has an open set around it by hypothesis (then just union them all up). finv(Y-{y}) is open by the same argument, yet this set must be empty otherwise it would make a separation of X, and we know X is connected

So finv({y}) = X

[iron bolt]

"locally constant for each x in X" does not make sense to me

do you just mean locally constant?

also, I'd f say is constant on all of X then, since that is the domain

[tiny obsidian]

I mean replace 'for' with 'at' and these are equivalent

[iron bolt]

oh, I see

I've never seen somebody talk about a map being locally constant at some point

[junior granite]

Let $f: X \to Y$ be a continuous surjection of topological spaces. Suppose $Y$ is Hausdorff, and that for all $y \in Y$ the subspace $f^{-1}({y})$ of $X$ is open and Hausdorff. Show that $X$ is Hausdorff.

[gentle osprey]

Vic

[junior granite]

How would one go about proving something is Hausdorff? I am trying to view examples in my textbook but am struggling to understand them as I'm not sure I understand the definition of Hausdorff.

[plush folio]

Pick two points in X and show that they have disjoint open neighbourhoods

[iron bolt]

unrelated to what the exercise wants you to show, but: all fibers of a continuous map being open is a really strong condition

actually, I think it's equivalent to being locally constant?

yep

[junior granite]

would this suffice as a proof?

oops that should be U_1 \cap U_2 in the first line of the third paragraph

[iron bolt]

one thing I'd change is that you don't actually need a surjection

like, at this point instead of saying "such y exist" you could just say "let y := f(x)"

or not even name them, and just talk about neighbourhoods of f(x_1) and f(x_2)

the proof is correct though, those are just stylistic things

[junior granite]

i mainly included it because it was in the problem statement so i figured i'd include it in my response.

but stylistically i see your point.

[iron bolt]

hehe. in my eyes it's always cool when you can just drop unnecessary hypotheses from the problem statement and prove it anyways

[plush folio]

It's not 100% correct, you haven't used the assumption that the preimages are open and Hausdorff. The reason you need those assumptions is that x1 and x2 might map to the same element in Y

but your proof would work if f was injective

[iron bolt]

oh, right

well, if a proof is what convinces other mathematicians, it did work as a proof on me 😆

[plush folio]

Mochizuki would be proud

[iron bolt]

I even thought about the f(x_1)=f(x_2) case earlier, but then didn't think of it when reading your proof

[junior granite]

wait are these line not using that assumption?

[iron bolt]

nope, preimages are always open and preimages of disjoint sets always disjoint

[plush folio]

oops, I meant that the fibers are open

[iron bolt]

"we can find disjoint nhds U_1 and U_2" is the problem

[plush folio]

fibers are the preimages of singletons, which are usually not open, so continuity of f doesn't necessarily mean that the fibers are open

[iron bolt]

that's why openness of the fibers is assumed in the statement

I guess another way to prove it is actually just that openness of the fibers means that X is the disjoint union of its fibers, and so a disjoint union of Hausdorff spaces if they're Hausdorff

you don't need Y to be Hausdorff

uh, sorry if that was a spoiler - I figured it was fine since you already have a proof that almost works, it just needs that one additional case

[junior granite]

i think i see what you are trying to explain. let me try to execute that into the proof hold on.

[plush folio]

Nice proof you're right that that is a pretty strong condition

it would also mean that X is disconnected unless f is the constant map, right?

[iron bolt]

yep

locally constant maps are always constant on connected components - and that condition is just equivalent to being locally constant if you think about it for a bit

the one with the fibers I mean. being constant on connected components is not strong enough to imply local constancy

[plush folio]

ah, I see it now, it is almost the definition of locally constant; just take the fiber as the neighbourhood, which is constant

[iron bolt]

yep

took me a bit to see too

I wonder if the book the exercise is from is doing that intentionally - not phrasing the statement in the cleanest and most general way possible, to give the students some room to explore and make improvements

it is what doing actual research feels like too after all - often you seek understanding, not proof of something you already know the precise and most general statement of

at least from my limited experience so far

[plush folio]

yeah, might be a good idea that some exercises are like that - usually when doing an exercise, the first thing you do is to gather up all the assumptions and see how you can use them. You never get to practice the art of knowing which assumptions to use

[junior granite]

sorry to interrupt, but i'm having trouble seeing where this would fit into the proof. could you perhaps give a hint at which paragraph of my proof i should look at further?

[iron bolt]

^

[junior granite]

oh oh i missed that entirely lol

[iron bolt]

no worries, lol

[plush folio]

Do case analysis on whether f(x1) = f(x2). You have done the case where they are different already. If they are equal you can't use the Hausdorffness of Y to find disjoint neighbourhoods

Concrete Mathematics has some interesting paragraphs about different levels of problems in textbooks

I rarely see level 3, and I don't think I've ever seen level 4

[iron bolt]

I've already seen too many mistakes in papers to not see many lemmas as exercises of level 3, lol

got myself convinced to have found a counterexample to a lemma quite a few times already; a few times it was true, most others not so much

working through the proof with the supposed counterexample in mind usually clears it up quite quickly

the problem with level 4 is that there's no longer a unique correct answer, but many answers that are true to the spirit of the exercise to varying degrees

Q := P always works for a start

[plush folio]

True, it's hard to grade such a question

and of course, there's level 5 💀

[junior granite]

is this better? or did i make it worse D:

[iron bolt]

if you change 4 to an "interesting sufficient and necessary condition" this is actually just level 4 applied to P := true :p

nope! this bit is actually false

[plush folio]

true 😆

[junior granite]

sadness so i didn't understand what you meant.

[iron bolt]

Hausdorffness guarantees you disjoint neighbourhoods for every two distinct points. otherwise only the empty space would be Hausdorff, since x never has a neighbourhood that's disjoint from x

if f(x_1) = f(x_2), that doesn't give you a contradiction. it's just its own case you have to deal with

hint: ||x_1 and x_2 live in the same fiber then, and you have some assumptions on those...||

[junior granite]

maybe i'm confused by this - what do you mean by fiber? I think i'm confusing terminology.

[iron bolt]

oh, sorry about that

just any preimage of a single point

the terminology is most often used in the context of fibre bundles and in particular coverings I think, but it's also often just used regardless of what map you're dealing with

[junior granite]

ah okay okay

let me

think

lol

[tiny obsidian]

Out of curiosity, what were 1 and 2? "Verify that P(x) holds for specific x", and "Prove P(x)"?

[plush folio]

Nice extrapolation, that's correct Prove P(x) for a specific object x, and prove P(x) for all x in some set X

[plush folio]

Just to clarify: you seem to have picked y1 and y2 in Y first, and did a case analysis on whether x1 is equal to x2. That's the wrong way around; we pick distinct x1 and x2 in X first, then we map them over to Y. So we know that x1 and x2 are distinct by definition. And remember that surjectiveness doesn't really matter

[junior granite]

i have a feeling i'm straying farther from the answer the more i think about this.

[iron bolt]

oh. this is indeed even more wrong now

sorry for putting it so bluntly, lol

one thing I'd recommend is to never introduce the shortcuts y_i := f(x_i) - just call them f(x_i) directly

so then, what you're saying at the bottom is that you can find a neighbourhood of x_i that does not intersect f^-1(f(x_1))

can you see the problem with that?

[junior granite]

no no i appreciate the help haha. bluntness is what i need.

[iron bolt]

it's also still true that you need to take care of two cases, that where f(x_1) = f(x_2) and that where they're different

so basically, your proof structure after "we want to show that X is Hausdorff" should be something like this:

To do that, let x_1, x_2 be two distinct points in X; we now want to show that they have disjoint neighbourhoods. If f(x_1) ≠ f(x_2), we can do that by...
If however f(x_1) = f(x_2), we can...

[junior granite]

was what i originally wrote many moons ago sufficient for the case where f(x_1) \neq f(x_2)?

actually let me just rewrite everything and resend it.

[iron bolt]

it was, yup

it was a bit more complicated than it needs to be, because it used the unnecessary condition that Y is Hausdorff

but it did work, so you only had to also account for the case f(x_1) = f(x_2) to get a valid proof

[junior granite]

okay

i am super tired i think i need to put this proof away for now and revisit in the morning. thank you all for your help though!!

[iron bolt]

sounds like a good idea, yeah

[chrome trout]

Am I allowed to post stackexchange link here and ask question about one of the solutions?

[fierce lily]

for the first sentence of the hint, is it correct to write there is a map q: S^n->S^n/C2 defined by: x->-x for all x in S^n? Second, can someone tell me what does Cw complex with number of cells in each dimension from 0 to n mean, namely what does cell refer to?

third, what does the symbol reverse cartesian product represent? I remembered that for disjoint union topology, it should be of form like {0}*A union {1} * B?

[iron bolt]

cells are just the things you glue together to get a CW complex

so the interval [0,1] for example would with its canonical cell structure have one 1-cell and 2 0-cells

the antipodal map x → -1 is relevant, but it isn't q; it's the map that C_2 acts on S^n with

q maps every point x to the equivalence class {x,-x}

[iron bolt]

it's the disjoint union. the set of all pairs (i,x) where i ∈ I and x ∈ X_i is one possible representation, but really it's anything that functions as a disjoint union of all X_i in some canonical way

[chrome trout]

Here the question is about the Cantor-Bendixson rank of a tree

https://math.stackexchange.com/questions/2596284/cantor-bendixson-rank-of-a-tree-over-omega-possible-values-and-more

At the very last part of the answer, they say that $T^\alpha_\alpha={\langle\rangle}$. I just can't see why.

[gentle osprey]

qwerty

[fierce lily]

and here, x is in S^n and C_2 acts on S^n maps x to equivalence class {x, -x}?

[fierce lily]

may I ask what does canonical cell structure have one 1-cell and 2 0-cells look like?

[iron bolt]

your 0-skeleton is just two points {0,1}

the 1-cell is D^1 = [0,1] glued to them by the attaching map that takes 0 to 0 and 1 to 1

it's all really close to the definition of a CW-complex - you probably want to look at that a bit more

[fierce lily]

so 0 cells means singleton right? and if possible can you tell me what is CW-complex, when I search google it tells it obtained from gluing copies of k-cells, but i have no idea how to measure number of cells.

or I can just think n-cells is object in n dimension? like [0,1] is 1-cell by gluing singleton boundary point. and circle is 2 cells by gluing all its boundary in 2-dimension and keep going?

[iron bolt]

it's a bit like asking "given a graph, how can I tell how many edges and vertices it has?"

there's not some clever way of doing it, it's just part of the definition

a graph is defined as a set of vertices with a set of edges between them

similarly, a CW-complex is defined as what you get from gluing disks together in a certain way

cells is just what those disks are then called, similar to how you call the points of a graph vertices just because that's the convention there

one slight nuance I guess: when people say cell, they usually just mean the interiors (i.e. open disks), not the full closed disks that are glued together to get the space

but of course for the number of cells that doesn't matter

[fierce lily]

so you describe one-cell as an interval [a,b], and attach more than one intervals by attaching its boundary. How you will describe like say 2-cell and 3-cell?

namely, for 2-cells or 3-cells or more cells, how you will glue these disks, any examples here?

[iron bolt]

like the definition says - an n-cell is a disk D^n, glued to the (n-1)-skeleton via its attaching map S^(n-1) → X^(n-1)

i.e., you specify how the boundary ∂D^n = S^(n-1) of D^n gets glued to the gluing of lower-dimensional cells you already have

you could for example describe S^n as a CW-complex by giving it a single 0-cell *, no 1-cells, and a single 2-cell that you attach via the constant map S^1 → {*}

or RP^2 as a CW-complex by giving S^1 the cell structure like above and gluing a single 2-cell to it via the attaching map S^1 → S^1, ω ↦ ω^2

[uneven bronze]

I have a question about the product topology of a set $X$ which is the Cartesian product of a finite number of sets $X_j$. I know a base for the product topology are unions of sets of the form $U_1\times\cdots\times U_n$ where $U_j\subset X_j$ is open. This description of open sets is...descriptive. Now comes my naïve question. Is there a similar thing for closed sets? Can we say that closed sets are unions of the $U_1\times\cdots\times U_n$ where $U_j\subset X_j$ is closed?

[gentle osprey]

psie

[paper wedge]

look at say the product topology on like X and Y

u know how open sets look like

look at their complements

and to answer ur question

no

not unions

you can think of an easy example

or counterexample

to what ur saying @uneven bronze

in R^2

[uneven bronze]

ok, thanks. Well, the reason I'm asking is I'm trying to prove $\overline{A \times B}= \overline{A} \times \overline{B}$ for $A\subset X, B\subset Y$. The argument for one inclusion goes like this: \

First, $A\times B\subset\overline{A}\times\overline{B}$ and $\overline{A}\times\overline{B}=\left(\overline{A}\times Y\right)\cap\left(X\times\overline{B}\right)$ is, as intersection of closed sets, closed in the product topology. Thus $\overline{A\times B}\subset\overline{A}\times\overline{B}$.\

How can I know that $\overline{A}\times Y$ and $X\times\overline{B}$ are closed in the product topology? I have no description of the closed sets in the product topology.

[gentle osprey]

psie

[paper wedge]

closed sets are the complements of open sets

also tho

in ur case

okay

[paper wedge]

use this to prove that the cartesian product of closed sets is closed

now u probably already know that the closure of a set is (by def) closed

so that should clear up ur like misunderstanding

[uneven bronze]

hmm, ok

[paper wedge]

yeah]

[paper wedge]

hint: this is just set theory

[uneven bronze]

ok 🙂

[paper wedge]

good luck

[uneven bronze]

but... (U times V)^c is not necessarily equal to U^c times V^c

[paper wedge]

what is it equal to

[uneven bronze]

If U,V are open, I know their cartesian product is in the base of the product topology I mentioned earlier

[paper wedge]

suppose U and V are closed

figure out what (U x V)^c is

U subset of X and V subset of Y

[sonic crane]

Unfortunately!

I have my topology midterm tomorrow !

[sonic crane]

In my homework we showed that the “slice” {x} x Y as a subspace of product topology is homeomorphic with Y

It should be also true that some finite {x1,x2,…xn} x Y is also homeomorphic with Y right?

Also, the projection map is continuous and an open map, but its definitely not injective so its not a homeomorphism?

Oh wait

[sonic crane]

Hm not really

Because now u have choice of xi , y

That singleton cross Y was homeomorphic with Y cause u just had that same x always tacked on

[fierce lily]

are you willing to share your solutions to this question？I think I can understand much better after I see the solution

[fierce lily]

and also, can I say give a single 1 cells and a single 2-cell here? but it seems like 1 cells should glued to 0 skeleton that map to nothing(am I correct?)

[iron bolt]

I actually did in the case n=2 already

[iron bolt]

right here

so RP^1 is just one 1-cell glued to a 0-cell in the obvious way (it's just the circle S^1 after all). RP^2 can be described as a 2-cell glued to RP^1 via the quotient projection S^1 → RP^1. can you see how that generalised to RP^3 and above?

[iron bolt]

sorry, I mean a single n-cell attached via the constant map S^(n-1) → {*} here. in the case n = 2 it's what I wrote though

that cell decomposition doesn't have any k-cells for 0 < k < n; that's sort of the point of it, it's super simple

you can put different cell structures on S^1, like one consisting of one 0-cell, one 1-cell and two 2-cells (the 0- and 1-cell form the equator, then you glue one disk to it for each hemisphere)

exactly one 0-cell, one 1-cell and one 2-cell isn't possible though

[sonic crane]

On my homework, i said x^2+y^2=1 is connected , but prof said “i see it visually, but whats the quick way to show this?”

In my homework i said its the continuous image of an interval which we know is connected

So then, what IS the easy quick way to show the circle is connected ?

[plush folio]

Show path connectedness maybe? Which is even easier if you define the circle as { exp(2pi*i*t) | t in [0, 1] }

[sonic crane]

Oh ok, yea defining circle in that way is like already stating what the continuous path function is

[robust drum]

I think this argument is fine imo

[fierce lily]

Rp^3 can be descirbed as 3-cell glued to RP^2 via quotient projection S^2-RP^2. but i remembered RP^1 is a line in 2 dimensional space through the orign, so why we can say it is just 1-cell glued to a 0-cell?

[iron bolt]

it's homeomorphic to it

there's many different ways RP^n can be constructed, that are all homeomorphic

one is like you said the space of all lines through the origin in R^(n+1) - but another is as the quotient of D^n where antipodal boundary points are identified

and that's basically what this cell complex gives you

[fierce lily]

for the another part, you refer to S^n/C_2 right?

[iron bolt]

it's related, but not quite

RP^2 is homeomorphic to S^n/C_2

but it's also homeomorphic to a quotient of D^n where only the boundary points are glued in that way

it's basically what you get when restricting the quotient map S^n → RP^n to a single hemisphere

[prime elbow]

In definition of sequential compactness, a metric space (X,d) is said to be compact iff every sequence in (X,d) has at least one convergent subsequence. A subset Y of a metric space X is said to be compact if the subspace (Y,d) is compact.

So if a subset Y is compact that means any sequence in Y has a convergent subsequence which is converge in Y, right?

Let X be Hausdorff space and Y_i are a compact subspace of X so they are closed.

Now their intersection is also compact because they are closed and since intersection of all Y_i is closed subset of Y_j, for some fixed j.

A closed subset of a compact space is compact, right?
Thus the intersection is compact.

[rancid umbra]

closed subsets of compact spaces are compact, yes. what are u trying to prove?

[rancid umbra]

yes

[prime elbow]

In Hausdorff space arbitrary intersection of compact space is compact

[plush folio]

You show that the intersection is compact in Y_j, right? Does this imply that it is compact in X?

[prime elbow]

I think yes

[plush folio]

ah, it's because the subspace topology of the intersection Y in Y_j is the same as the topology of Y in X?

[prime elbow]

I am not sure but yes if Z \subset Y \subset X then subspace topology Z of Y is same as subspace topology Z of X

[heady skiff]

can I have a hint for showing that connected components in Y contain only one element

[rancid umbra]

take a connected component C of Y. suppose C contains c. if C - {c} is non-empty, then what can you say about C?

[heady skiff]

idk, maybe something like C - {c} is open and if I can show that {c} is open then C is not connected which is a contradiction

wait maybe not, bc that would imply that singletons are open

[rancid umbra]

thats okay. take a singleton set {c}. let c_k be the kth component of c in S. then {c} = prod {c_k}

since each of the singletons in S is open, then the above product is open

assuming that your basis is the collection of all products of opens in S

[heady skiff]

i thought that the product topology requires that you only have a finite number of elements in the product which are not equal to the entire space

or is that only for the basis elements

and how would we write a singleton as a union of those basis elements, because all of those basis elements will contain the whole space in one of the elements of the product and thus "cover" the kth component for each k

or wait maybe i'm mixing up union

[rancid umbra]

ah, okay. so i was working with the box topology. all good

[heady skiff]

got it

[heady skiff]

in showing that every element of B is actually closed, why do we know that every element of B is open (which is implied in the problem statement?)

i was thinking about writing {x} = {x} n B and since B is open in Y, {x} is open in Y, but this argument wouldn't work since this would generalize to any topological space to show that singletons are open

[rancid umbra]

basis elements are open since they are the unions of basis elements

[heady skiff]

can you elaborate pls

he's asking to show that every element of basis element are closed

not that every basis element is closed

or is that a typo

at first i assumed it was a typo so I showed that every basis element is closed but now i'm not so sure

[rancid umbra]

open sets generated by a basis are unions of basis elements. a basis element b is the union of just b

hence b is open

[heady skiff]

yeah i know B is open

but i'm confused as to why every point of B is open/closed

[rancid umbra]

points of the B in this picture are basis elements

since that B is the collection of all basis elements

[heady skiff]

how do we know that the points are basis elements?

[rancid umbra]

it says, "Let B denote the usual base for the product topology"
so B is the collection of all basis elements

points, or elements, of B are basis elements for the product topology

[heady skiff]

i don't understand, the product topology has basis elements which are of the form \prod U_i where U_i = X_i for all but finitely many of the i; hence a point cannot be a basis element for the product topology, since a basis element must have one of the open sets in its product equal to the entire space

[rancid umbra]

B is the collection of all sets of the form you have just specified

the elements of B are those sets

[heady skiff]

OHHHHHH

i'm a dumbass

[rancid umbra]

lol

[heady skiff]

thank you

[rancid umbra]

still thinking of a hint for you about the connectedness part

[heady skiff]

yeah, i'll come back to it probably

been banging my head against the wall for like 2 hours on it

[rancid umbra]

might not be able to respond until later today tho

gotta go soon

[heady skiff]

all good, i'll come back to it probably later too

i appreciate the help

[mighty hull]

Hint: There's a reason they want you to first show that the basis elements are clopen.