point-set-topology

[mighty hull]

If you glue two edges together, you'll get a hollow cylinder, like a pretend telescope.

[sonic crane]

Hmm, in munkres he also identified the 4 corners as its own equivalence class too

Why is that?

I see in that gif identifying the lengths as the same and the widths as the same folds it into a torus, but where does the four corners as one equiv class comes in?

[fringe thorn]

ah, I didn’t know Munkres did it like that

but isn’t that already accounted for during the fold?

you can see in the GIF that during the cylinder forming process, the 4 corners were collapsed into 2 points

and then during the torus forming step, those two points were jammed together

[sonic crane]

Any ideas on whats happening here?

[fringe thorn]

I mean, I stand by the comments I made back here

[sonic crane]

Im not sure what u mean

[sonic crane]

Oh bruh ok

[fringe thorn]

yeah, all four corners are already glued together by identifying the edges as we did

[sonic crane]

Yeah I see that in the visualization, i guess im trying to convince myself on why mathematically it means we have/need a whole separate equivalence class

Just for the corners

Ohhh wait

[fringe thorn]

well, we don't

[sonic crane]

I think i misunderstood a big thing

[sonic crane]

The four point set he says

Isnt that its own equiv class

[fringe thorn]

in Lee, he doesn't make a separate eq. class

this is what I'm used to

I'm not sure why Munkres chose to do what he did

[sonic crane]

I get what munkres did now I misunderstood at first

I missed that the two point sets for the edges are all seperate classes as well

He’s like manually identifying each grouping of points that will be eventually pasted together

[plush folio]

But if you do it like Lee, wouldn't the corners be part of both the horizontal and vertical lines, forcing the whole boundary into the same equivalence class?

[sonic crane]

I actually like how he did it now

[fringe thorn]

edit: nvm lol

[plush folio]

oh wait, nvm, I misunderstood

[fringe thorn]

wait, maybe I'm being silly lol

[plush folio]

you don't identify the line with itself, you identify each point on a line with a point on the opposite side, right?

[fringe thorn]

yes

[sonic crane]

Yea this is what i missed in munkres description

[plush folio]

so this doesn't make sense, what Lee did is correct 🙃

[sonic crane]

Lee and munkres did the same thing munkres just explicitly wrote it all out more

Lee’s description also has an equivalence class {(0,0),(1,0),(0,1),(1,1)}

[plush folio]

yeah, makes sense

[sonic crane]

So i guess technically speaking we should be able to prove that the quotient topology on that subspace of R^2 is homeomorphic to the torus as a subspace of R^3?

In the sense of like, we just kind of said that when we make this identification we can think of the new space as the torus but we didnt really prove it

[mighty hull]

I doubt most folks have another characterization of the torus in hand!

Like, I couldn't give you an explicit parameterization of the torus in R^3 offhand.

So, yes, you have this space and you have some parameterized thing in R^3 and you'd want to prove they're homeomorphic.

Which one you call "the" torus then becomes a matter of preference

[plush folio]

there's a third characterization of the torus: the product of circles S^1 x S^1, which is sometimes more convenient

[mighty hull]

This paper by Mazur is great and relevant here: "When is one thing equal to another thing?"

https://people.math.osu.edu/cogdell.1/6112-Mazur-www.pdf

[sonic crane]

A homework problem i have rn is showing that a quotient topology on some equivalence relation on R2 is homeomorphic to that

It was p ~ q if p-q in Z2

R2 \ ~ is homeomorphic to S1xS1

[plush folio]

What does p-q in Z2 mean?

[sonic crane]

Sorry I mean Z x Z not Z mod 2

The equivalence relation on R2 being p ~ q if p-q is in ZxZ

The quotient topology on the set of equivalence classes is homeomorphic to S1xS1

[sonic crane]

Oh, coooool

[plush folio]

Interesting I guess it's similar to the result that R/~ is isomorphic to S^1 where ~ identifies the integers to a point

[sonic crane]

Yeah! I guess the way to show this would be just extendinf that method

Im trying to do it rn

[mighty hull]

Define $f: \mathbb{R}^2 \to S^1 \times S^1$ by $$f(x,y) = \left(e^{2 \pi i x}, e^{2 \pi i y}\right)$$

[gentle osprey]

Cufflink

[fierce lily]

here is what I confused, for each pair of closed set A and B, X-A and X-B is clearly disjoint open sets containing B and A respectively, then it seems that every space is normal??

[iron bolt]

are you sure about that "disjoint"?

[fierce lily]

ok, it can be not that disjoint, ex:if I have three singleton {x}, [y},{z}, the intersection can be not disjoint

I mean singleton in R

[iron bolt]

X\{x} and X\{y} are never disjoint, unless X only consists of those two points

generally, for disjoint subsets A and B X\A and X\B are only ever disjoint if A and B form a partition of X, i.e. X = A ⨆ B

if you want the neighbourhoods to be disjoint you have to choose them to be small, in a way: for example, for points x < y in R you could choose the intervals (-∞,(x+y)/2) and ((x+y)/2,∞) as disjoint neighbourhoods of {x} and {y}, but not R\{x} and R\{y}

[sonic crane]

If you have a quotient map p and you have a basis for the saturated open sets, is mapping that basis under p gives u a basis for the quotient topology?

[swift fjord]

I'm blanking here, what standard result is he referring to?

[paper wedge]

diagonal being closed in hausdorff space

[swift fjord]

Oh yea ofc, A = (\tilde f x f')^-1(\Delta)

ty

[paper wedge]

yw

is this rotman?

[swift fjord]

yea i'm refreshing algtop for my oral exam

[paper wedge]

good luck

[swift fjord]

ty

[fierce lily]

Can someone here explain what does minimal uncountable well-ordered mean?

I mean the "minimal"

[paper wedge]

minmal set with property P probably means it contains no set that has property P

that is it is smallest with ordering being the inclusion

[fierce lily]

How can we show this set has the least upper bound?

The thing I am confused is: if X is unbounded, how can I let it have the least upper bound property here?

[iron bolt]

unbounded sets can have the least upper bound property

[mighty hull]

The real numbers have LUB

[iron bolt]

it only says that subsets with upper bounds also have least upper bounds, not that all subsets do

[fierce lily]

so if a subsets have upper bound, is it possible that it does not have least upper bound?

[fierce lily]

The whole R must have least upper bound?

[iron bolt]

in general orders yes. orders with the least upper bound property are precisely those where that isn't possible

[iron bolt]

does R have an upper bound?

[fierce lily]

where that isn't possible? what does that mean?

[fierce lily]

I think no

[iron bolt]

yep, it doesn't

so the least upper bound property doesn't say anything about it

[iron bolt]

what you wrote here #point-set-topology message - in general orders it is possible that a set has an upper bound but no least upper bound, but the least upper bound property forbids exactly that

[fierce lily]

so we want to prove that the set X *[0,1), every subsets having upper bound has a least upper bound in this set?

[mighty hull]

Yes

[fierce lily]

Then i want to write for every subset A * B having upper bound x * y here, because y is in [0,1), then B must have least upper bound. but I am not sure how to prove it correctly

I am not sure if I am on the right track here

[iron bolt]

one immediate problem is that subsets of X ⨯ [0,1) are not necessarily of the form A ⨯ B

[mighty hull]

What is * here?

[iron bolt]

cartesian product of sets and the notation for a pair probably - they're usually denoted A ⨯ B and (x,y) though

fyi, you are right that a good first step is to take some subset S of X ⨯ [0,1) with an upper bound (x,y), and that you'll eventually need to use both the least upper bound property of X and the least upper bound property of [0,1). the proof won't be pretty though

as far as I can see, you will need to go through a case distinction to write down the least upper bound of S explicitly

[fierce lily]

The question here is, it seems that I don't know how to argue that subsets of X having upper bound must have the least upper bound

[iron bolt]

you're given that X is well-ordered - what does that mean? anything you maybe can apply that property too?

well-ordered sets have the least upper bound property, generally

[fierce lily]

The thing I know about well-ordered set is this set with existence of order relation must have smallest element

[fierce lily]

so I can say both X and [0,1) have the least upper bound property, then the subset of X * [0,1) having (x,y) as upper bound must have the lub on both X and [0,1), it seems correct to me but how do we need to go throguh a case distinction here?

[iron bolt]

because it turns out that "seems correct" is not actually enough

you can take the least upper bounds of π_1(S) and π_2(S) seperately, where π_i are the two projections, and that does indeed give you another upper bound of S

but what happens when you try to prove that it's a least upper bound?

[fierce lily]

let lub(pi_1(S))=a', lub(pi_2(S))=b', then a' * b' seems to be the lub for S, is that wrong?

[iron bolt]

it is wrong

it might be instructive to look at an example - take for example ||X := N ∪ {∞} and S := N ⨯ {1/2}||, or ||X := {0,1} and S := {0} ⨯ [0,1)||

in both cases something goes wrong, but what exactly goes wrong is different

[gritty widget]

Guys

This is a PSA made with the interest of the health of the general public in mind. Terms & Conditions apply

[fierce lily]

so in second case, it seems that lub of S is {0} *1? and for the first case, I can not see th lub for that.

[iron bolt]

1 is not an element of [0,1)

[fierce lily]

can you tell me the lub for first and second case respectively

[iron bolt]

in the first case it's (∞,0), in the second (1,0)

I think it's clear once you understand the lexicographic order on products as a picture rather than just an abstract concept. X ⨯ [0,1) is X copies of [0,1) laid in a row, based on the order of X

so {0,1} ⨯ [0,1) for example is two copies of [0,1) right after each other, and S is just the first half of that

[uneven bronze]

I have a silly doubt and don’t know really where to ask. Consider the following shape in R^2. The rectangles are meant to be the same and enclose a cube, but the set consisting of the rectangles does not contain the cube. Is this set closed?

[mighty hull]

The finite union of closed sets is closed

[kindred cairn]

I want to show that a polygonal line (curve consisting of finitely many straight segments) in R^n \ {0} is always contained in some starshaped region. obviously this only works for n\geq3

I can easily argue that there exists a point x, that connects to all the corners of the polygonal line, without meeting 0. I have trouble upgrading this to the whole polygonal line

[dire dove]

For each segment take the 2-dimensional plane containing the segment and 0, if you can show R^n cannot be the union of finitely many linear proper subspaces then you can find a point which is not on any of these planes and this point does the job

Now it's a linear algebra problem, and it's true in general that a vector space over an infinite field is not the union of a finite number of proper subspaces

[kindred cairn]

right, I see. I couldn't come up with an argument that used that n\geq3

now it makes sense, thanks

[dire dove]

Idk if you need a proof for this fact but it's nothing too crazy

[kindred cairn]

should be fine

[grave solstice]

What does this mean? That U subset C^infty(M, N) is open iff there exists some r and open V subseteq C^r_W(M, N) such that U=V cap C^infty(M, N)?

[fierce lily]

When S= {0} ⨯ [0,1), so is it because 1 seems to be the sup on Y, which does not belong to S, so we just consider the sup(S)=(1,0)?

I think I can understand it is just a copy of [0,1) here

[fierce lily]

and for the first example, is it because N is infinite, seems to be infinte copy of [0,1), hence we just consider sup to be (inf, 0)

I mean, it seems we can have a picture of dictionary order, and it can be visualize clearer

[ocean canyon]

How do we know that there is a homotopy between $k$ and $f$, when we assume that there is one between $k$ and $\tilde{f}$, in the \textbf{second paragraph in the proof in the picture, above}; I mean in one sense Lee is only saying that "assume one exists" (but I think we want an existence of such a homotopy since we want to find an extension of $\tilde{f}$).

At first I thought that this was a typo and it should be $H_1 = \tilde{f}$, but now I am not so sure, since we want $\tilde{H}$ to restrict on $S^1$ to $\tilde{f}$, presumably we want to use that $\tilde{f} \circ \omega = f$ where $\omega:I \to S^1$ is explicitly defined by $s \mapsto e^{2\pi i s}$.

[gentle osprey]

Benjamin

[ocean canyon]

This seems like a typo: Errata J. Lee:s "Topological Manifolds" (see p. 4)

[red folio]

Does anybody have a hint for this?
Let X be a second countable and $f:X \rightarrow f(X)$ be open, then f(X) is second countable

[gentle osprey]

snus

[cedar jungle]

How can I prove that Int(A) U Int(B) is a subset of Int(A U B)?

[iron bolt]

I mean - you know that there is a countable basis of X, that f maps open sets to open sets, and you want a countable basis of f(X)

sounds like taking the images of the basis elements to be your basis might be a good approach, no?

then you'll just have to check that that is indeed a basis

[iron bolt]

as always with relatively basic questions - expand out the definitions, and see where you could go from there

[mighty hull]

write 'em out, all the way out

[mighty hull]

Have you written out the definitions?

[cedar jungle]

well, definition of interior is the largest open subset. And with this the formula is very intuitive. But is that enough to prove it?

[mighty hull]

Write it outttt

[mighty hull]

Let x ∈ Int(A) ∪ Int(B). You want to prove that x ∈ Int(A ∪ B).

If x ∈ Int(A) ∪ Int(B) then x ∈ Int(A) or x ∈ Int(B).

Say x ∈ Int(A). That means x is in....

And therefore....

[cedar jungle]

x is in Int( A U B)

[mighty hull]

You have to make the argument.

x ∈ Int(A) means what? Write it out.

"x is in the largest open set contained in A."

Ok, what does that mean, for a set to be the largest open subset? Write it out.

[cedar jungle]

that means that x is contained in an open subset of A

[mighty hull]

There are a few equivalent definitions of the interior of a set. You gave "Int(A) is the largest open subset of A".

That's fine. It works.

But what does it mean for a set S to be the largest open subset of A? What makes it largest?

[cedar jungle]

Maybe that it contains all open subsets of A?

oh, Int(A) is a subset of Int(A U B)

[mighty hull]

eyy

[sonic crane]

is the projection onto the quotient an open map?

pi: X -> X / ~

[mighty hull]

Don't say "maybe" — there's a definition.

In general, if you have a set X then a set S ⊆ X is the largest subset of X with property XYZ if any set in X with property XYZ is a subset of S.

Here, you can prove that's equivalent to Int(A) being the union of all open sets in A, but it's not the same definition.

[mighty hull]

But if x ∈ Int(A) then Int(A) is open and a subset of A.

A is a subset of A∪B, so Int(A) is an open subset of A∪B.

Since Int(A∪B) is the largest open subset of A∪B and Int(A) is an open subset of A∪B, it is a subset of Int(A∪B) by definition of largest.

[cedar jungle]

Thank you

[mighty hull]

You can also think of Int(A) as the collection of all "interior points".

x ∈ A is an interior point if there's an open set U ⊆ A with x ∈ U.

These three definitions are equivalent:

1. Largest open subset of A
2. Union of all open subsets of A
3. Set of all interior points of A

But then x ∈ U ⊆ A ⊆ A∪B, so x is also an interior point of A∪B.

[fierce lily]

I am trying to prove that a long line is sequentially compact, and trying to first prove that every monotonic sequence in this set converges in long line, how you will prove this question?

[sonic crane]

For (R,+) as a topological group, so the operation map on RxR->R is continuous, how can I think of what the inverse image of like (-1,1) looks like?

I guess any pair (x,y) that lies in the inverse image has an open ball around it

So the inverse image is open

Im trying to think of how exactly the inverse image looks but its a union of many open sets right

First time encountering topological groups so im a bit unsure on how they work

[iron bolt]

completely ignoring the topological group context - the operation map you're talking about is addition, right? so (x,y) ↦ x + y

you can just write out the preimage for that, should give you a nice and simple shape

[sonic crane]

Yea

[iron bolt]

i.e., {(x,y) ∈ R^2 | ||x + y ∈ (-1,1)||}

[ocean canyon]

Am I mistaken, or is it not the case that a path-connected topological space X is simply connected iff any two paths with the same initial and terminal points are path-homotopic. But then if one wants to show that R^n is simply connected, is it not enough to observe that we have the straight-line homotopy between any two paths f,g:I ---> R^n where f(0) = g(0) = p and f(1) = p(1) = q?

[iron bolt]

yeah, showing that vector spaces are simply connected is easy like that

[ocean canyon]

sorry, I should add, for convex subsets of R^n

[iron bolt]

just linear interpolation works

[sonic crane]

Right yeah duh, thanks. I guess since the shape is like that it makes sense why i was getting confused by trying to write it in the form “unions of open set x open set”

[iron bolt]

yup

[ocean canyon]

right, I think (correct me if I'm wrong) we are both thinking about H(s,t) = f(s)+t(g(s)-f(s))

[iron bolt]

looks good to me

there are some even more general statements that can be proven btw: convex sets are star-shaped, star-shaped sets are contractible, and contractible spaces are simply connected

just depends on whether you prefer working with general statements like that or concrete homotopies that are easy to visualise

[ocean canyon]

well ok, one should then really conclude that applying this to $\pi_1(\mathbb{R}^n,p) = \pi_1(\mathbb{R}^n)$ shows that any path $f$ is path-homotopic to $c_p$ (the constant map), so the fundamental group is trivial, for each $p \in \mathbb{R}^n$, and therefore, $\mathbb{R}^n$ is simply connected.

[gentle osprey]

Benjamin

[ocean canyon]

oh

[iron bolt]

doesn't invalidate your proof though ofc

[ocean canyon]

yes, I just meant that I was not really finished 🙂

[iron bolt]

if it proves what was to be proven it's complete - everything else is just a bonus 🤷

I was just mentioning it because I think a catalogue of examples and their properties is always good to keep in mind

[ocean canyon]

I am not sure I follow

[iron bolt]

now I'm not so sure either I understood you correctly 😅

but anyway, your proof is correct, that's all that matters

[viscid rover]

Can we define a norm on R^infty as the final topology on R^n's

[heady skiff]

would a sub-base element just look like I_f n (a, b) for some (a, b) in R

[woven trail]

so, I'm trying to prove kuratowski, but I accidentally proved the number is 10, which is obviously not true since there are examples with max 14

$ $please, DO NOT GIVE ME ANY HINTS, just tell me where I'm wrong here:

$\overline{A}^{c} = \cup U{\alpha}$ for all open sets $U{\alpha}$ such that $A \cap U{\alpha} = \emptyset$, which means that $\overline{\overline{A}^{c}}^{c} = \cup U{\alpha}$ for all open sets $U{\alpha}$ such that $\overline{A}^{c} \cap U{\alpha} = \emptyset$

now, $\overline{A}^{c} \cap U{\alpha} = \emptyset$ is equivalent to $U{\alpha} \subset \overline{A}$, which implies $\overline{\overline{A}^{c}}^{c} = int(\overline{A}) = int(A)$
applying the complement to that gives us the previous set and applying the closure gives us $\overline{A}$, so we end up with 5 different sets
if we start by the $A^{c}$, we have another only 5 different sets, which totalizes 10 sets, what has gone wrong?

[gentle osprey]

gabi the ancient

[ocean canyon]

am I correct that to tie this proof together in the end, we are using homotopy invariance of path-multiplication? I.e. we "stitch together" $f_{[a_0,a_1]} \cdot \ldots \cdot f|{[a{n-1},a_n]}$ and then we exchange (where needed) $f|{[a{\ell-1},a_{\ell}]}$ to a path not containing $q$?

So e.g. if there is some interval $[a_{i-1},a_i]$ where $q$ is in the image of $f|{[a{i-1},a_i]}$ then we find a parametrized path $g_{a_{i-1},a_i}:[a_{i-1},a_i] \to M$ homotopic to $f|{[a{i-1},a_i]}$, and then we see that $f_0 \cdot \ldots \cdot f|{[a{i-1},a_i]} \sim f_0 \cdot \ldots \cdot g_{a_{i-1},a_i}$ and then we just continue like this, using transitivity of $\sim$?

I am a little confused since I've mainly seen the definition for path used with domain $[0,1]$ and not $[a_{i-1},a_i] \subset [0,1]$.

[gentle osprey]

Benjamin

[woven trail]

oh, this first fact is because $\overline{A} = \cap C_{\alpha}$ for all closed sets $C_{\alpha}$ such that $A \subset C_{\alpha}$, taking the complement gives that

[gentle osprey]

gabi the ancient

[mighty hull]

https://www.desmos.com/calculator/lwlpbzhdnz

[fierce lily]

for proving that every increasing sequence of ordinal numbers has a limit point, why we can say omega can never be a limit point of a sequence of countable ordinals, is it because it is first uncountable?

Q2: why (an) must converge to countable ordinals and therefore an ordinals represented in real line?

[woven trail]

on #help-8 I managed to come to the conclusion that the maxiumim is somehow 6

pls I have no clue what's wrong with it

[rancid umbra]

given a sequence of increasing ordinals, its supremum is the union. intuitively, if Omega were a limit point of this set, it would certainly have to be within [0, S] where S is the union of the sequence of ordinals. but since each ordinal is countable, and the union of countably many countable sets is countable, then S must be countable

[woven trail]

found out! the closure of the interior is not, in fact, the set!

<333

[woven trail]

is the interior of a set the interior of its closure?

[iron bolt]

nope; you can find that it isn't by looking for example at ||the rational numbers Q in R||

[woven trail]

*dies

but thank you

[iron bolt]

alternatively, since the interior of a set is precisely the complement of the closure of its complement, you can also conclude it directly from the closure of a set being not necessarily the closure of its interior

it's just dual to that

[woven trail]

amazing

[fierce lily]

so is it because S is countable, but omega is uncountable, so omega limit point of a seqence of countable ordinals contradicts that omega is uncountable?

[rancid umbra]

yes, basically

[woven trail]

btw trying to prove kuratawski has sedimented already 2 kinds of topology counter-examples in my head: isolated points and dense sets

these mfers can refute most of my wrong assumptions

[iron bolt]

yeah, Q subset R in particular is a counterexample to a lot of things

it's one of the simplest examples of a space that is not locally compact (at least that I know of); that makes it break a lot of things

[woven trail]

proved kuratowski!!! just need to find the maximal example now, which I suppose will have something to do with all my errors along the way lmao

[fierce lily]

I have another question here: the last two sentence, it says we take the point( a'+1,0), can I change it into like (a'+2,0), and also like (a'+0.2, 0)?

[woven trail]

good example of a problem that actually made me like, objectively better at topology while trying to solve it

I think I have a way better understanding of closures now

and interiors

[rancid umbra]

no

[fierce lily]

why only (a' +1 ,0) is allowed here?

[rancid umbra]

because in the case that dm converges to 1, then (a' + 1,0) is the limit point

[iron bolt]

yep. think of it as a picture: if you exhaust one collection of intervals completely, for your least upper bound you want to pick the first point from the next interval after that

that next interval is the one with index a'+1 - a'+2 is the one after it

and a'+0.2 doesn't make sense when a' is an ordinal

to be clear btw: when I suggest to look at a picture, I don't mean that the correct way to prove the result is to draw one. but often, looking at one makes it clearer why the formal proof needs to be written the way it does

[fierce lily]

I am confused because in minimal first uncountable well-ordered set, 0.2 is reasonable to appear here, so why we can only say +1, +2 etc

[rancid umbra]

no, that is incorrect

0.2 is not an element of Omega, when Omega is defined as the supremum of all countable ordinals

the only elements in Omega are the countable ordinals

[iron bolt]

I think if you want to understand the long line and those proofs related to it, you'd probably benefit a lot from first revisiting ordinals a bit tbh

they are kind of essential to it

or, if this is your first topology course and you have no reason to care about the long line other than it being used as a counterexample here, you could also just ignore it for now I guess - it really comes up pretty much exclusively as a counterexample, not in the proofs of any important theorems or anything like that

[fierce lily]

so you mean countable ordinals only contain element like 1, 2,3...etc?

[fierce lily]

where I can see the material related to it

[iron bolt]

wikipedia, set theory books, even youtube if you like

[rancid umbra]

the only elements that an ordinal contains are other ordinals

that is how they are constructed

[ocean canyon]

I've seen some other answers on mathstackexchange, but is it not enough to note that $L_{g'g^{-1}}:G \to G$ by left-translation is continuous if $G$ is a top. group, so that we have an induced map $$(L_{g'g^{-1}}){*}:\pi_1(G,g) \to \pi_1(G,g'),$$ which is such that if the image of this map is $c{g'}$, the constant map, then $[f]$ must be the map $[f] \equiv [c_g]$, so it is injective. For surjectivity, assume we have $[\ell] \in \pi_1(G,g')$. Then we need a path-class $[\gamma] \in \pi_1(G,g)$ such that $g'g^{-1} (\gamma(t)) = \ell(t) \iff gg'^{-1} \ell(t) = \gamma(t)$. Something with this construction feels off, but I can't pinpoint exactly what.

[gentle osprey]

Benjamin

[rare creek]

does it follow that if $U \in \mathcal T_1$ is open, that $f(U) \in \mathcal T_2$ will be open as well?

[gentle osprey]

BlackBeard

[rancid umbra]

show that if G is a topological group, then the path-components of G are all homeomorphic.

for simplicity, show that the path-component of g in G is homeomorphic to the path-component of the identity using the fact that left translation is a continuous map

since the fundamental group only depends on the path-component of the basepoint, then you can conclude that the fundamental group is independent of the base point for a topological group

[rancid umbra]

you said yourself it was clear in the previous paragraph.
maybe its not so clear to you...

[rare creek]

wiat

[rancid umbra]

not trying to be mean, but i am suggesting that maybe you should investigate why what you said is true

[rare creek]

im cooked in the head

hold on i wrote this a bit ago

[ocean canyon]

this sounds like the answer I read about; can you pinpoint what is wrong with the argument? Is there no way to explicitly give an isomorphism between fundamental groups with different basepoints in G, without involving path-components?

(And ofc, ty for the input :)).

[rare creek]

is it becaue f(T_1) = T_2

[rancid umbra]

this argument is fine after a more careful inspection. you can actually do this a lot cleaner by noting that (L_g)^-1 = L_g^-1, i.e., L_g is a homeomorphism of G.

now L_g o L_g^-1 = Id_G means that the pushforwards are inverses (due to functoriality) and ur done

[ocean canyon]

right! I think I see what you mean. Thanks for checking 🙂

[rancid umbra]

i prefer the connected component argument since it proves something stronger

[ocean canyon]

In Lee:s notation I think you are saying that we see that $(L_{g}){*}:\pi_1(G,g') \to \pi_1(G,gg')$ has an inverse $(L{g}^{-1}){*}:\pi_1(G,gg') \to \pi_1(G,g')$ and we are done (since we know that if $\varphi$ is a continuous map, then $\varphi*$ is a group-homomorphism).

[ocean canyon]

I see; baby steps (for me) 🙂

[gentle osprey]

Benjamin

[rancid umbra]

right

[faint gyro]

is R, {z: Im(z)>0} and {z: Im(z)<0} a valid set of domains for this question

im thinking that the boundary of the latter two would just be the real line as well

actually any separation of C would work right, e.g., the unit circle, its interior, and its exterior

[rancid umbra]

no

remember you need the domains to be simply connnected

also

do you take domain to mean an connected open subset of C?

because if you do, then neither of those work, as the circle and the line are not open in C

[faint gyro]

i'm not sure if domain needs to be open here, and putting that aside, why aren't the examples i gave simply connected?

for any pair of points x,y i can draw a path between them

[rancid umbra]

that's not what simply-connected means

[faint gyro]

oh, is simply connected not path connected?

[rancid umbra]

im fairly certain that the domains need to be open as well. that is the definition of domain, after all

simply-connected sets are required to be path-connected, but being simply-connected is a stronger condition than just being path-connected

you should review your definition of simply-connected

[faint gyro]

i see, simply connected is defined as: D is simply connected if every closed curve in D can be deformed (in D) to a point in D, and through some propositions:

• D simply connected iff its complement is connected
• D simply connected iff its boundary bD is connected

where connected is not defined, so i assume connected just means path connected?

[rancid umbra]

no, those are not correct:

• if D is the closed unit disk, C - D is connected has connected complement equal to D but C - D is not simply-connected
• C - D has connected boundary equal to S^1 but C - D is not simply-connected

[faint gyro]

ahhhh ok there is some misunderstanding here

i'm working on C_infty, so for the first point C-D would still be simply-connected

[rancid umbra]

what do you mean by C_oo?

[faint gyro]

riemann sphere

this is the context if it helps

[rancid umbra]

okay, then yes, that clears it up

[faint gyro]

so the mental picture in my head for prop 5.1.4 is that if D is simply connected, i can shrink it to a point, which means that the boundary of D will also get shrunk down to a point so the boundary is connected too

[faint gyro]

but sadly this is on C, not C_infty so the same tricks don't quite apply i think

[rancid umbra]

i think you should be able to draw them on S^2 tho

maybe there is no need. i actually am not sure how to answer your initial question atm

[faint gyro]

so the book did provide a hint, which was "Hocking and Young Topology, pg143" which lead me to discover https://en.wikipedia.org/wiki/Lakes_of_Wada

[rancid umbra]

yup. no way i was coming up with that

very cool tho

[faint gyro]

it turns out that the answer is a malicious evil construction that no sane human being would ever think of

thanks for the help tho

[rancid umbra]

there may be different constructions, but i imagine even the simplest are of the same flavor

[faint gyro]

yea, i don't think there are any elementary constructions that would satisfy this

[alpine nest]

Welcome to mathematics, yes

And I definitely think it's mean to give this as an exercise without any hint.

[craggy cedar]

As outsider says, welcome to math. More specifically though, you tend to have this kind of moment more in pointset than many other places. There's a reason we have Counterexamples in Topology.

[fierce lily]

What does this maps are stable under composition here mean, and how you will get started with this question?

[jolly sedge]

closed maps when the map is bijection and it's revers continus

stable it's means for any elemant maps you can compose them and get a elemant wiche it's revers it's another elemant in the original set

let's say map is X to Y

[gritty widget]

let F be the set of all x in l_\infty such that x_n = 0 for all but finitely many n. Is F closed? Is F open.

take x =(1, 1/2, 1/3,....) in M\F then for any open ball B(x,r) there exist y in F such that y in B(x,r).
since r>0, there exist k in N such that 1/k < r, now take y = (1, 1/2,...,1/(k-1),0,...). Now, d(x,y) = 1/k < r.

And for x = (1,0,0,.....) there is no open ball B(x,r) contained in F. For B(x,r) take y = (1,s,s,....) , where 0 <s < r.
Now find d(x,y) = s < r.

can anyone verify this ?

[red yoke]

Yes

[gritty widget]

if ( V, | | ) is any normed space, prove that the closed ball S = { x in V : | x | <= 1 } is always the closure of the open ball { x in V : |x| < 1 }.
Proof : Basically we have to show that cl(B(0,1) ) = S. It is clear that B(0,1) \subset S.
observe, we need to only show that x, which is | x | = 1 is in cl ( B(0,1) ).

Now we will show that take any y with | y | = 1, and any open ball B(y, r) there exist x in B(y,r) such that |x|<1.

Since r>0, there are three cases;
Case I : r = 1, in this case take x = y/2.
Case II : r > 1, in this case take x = 0.
Case III : r < 1, in this case take x = y/(1+r).

if there are any mistakes then let me know please.

[alpine nest]

You've only show that S is a subset of the closure of the unit ball (and I didn't verify whether in case 3 the distance of x and y is indeed less than r); how do you know the two sets are equal?

[gritty widget]

in Case III : | y - y/(1+r) | = | ry/ (1 + r ) | = r/(1+r) < r
B(0,1) \ subset S follows that cl ( B(0,1) ) \subset cl S = S

[alpine nest]

Yep, that's fine then.

[gritty widget]

Show that every open set in R is the union of countably many open intervals with rational endpoints. Use this to show that the collection U of all open subsets of R has a same cardinality as R itself.

I done with one but i have no idea how to use one to prove second.
any idea?

[gritty widget]

can i extend this result? cl(B(x,r) ) = { y in M | |x-y| <= r } in Normed linear vector space M ?

[alpine nest]

Yes, the closure of an open ball in a vector space is always the corresponding closed ball.

You can either prove this directly, or take the result for unit ball and use the fact that scaling and translation are homeomorphisms.

(and in fact translation is an isometry)

[prime elbow]

Yes, that's what I thought but I was not sure

B(x,r) = x + rB(0,1)

[gentle osprey]

Gapi

[covert ruin]

Can anyone help to confirm if the relation between them is equivalent ? I’m kind of lost in this. TIA

[real star]

they are the same

you have to show an open set of one is an open set of another

intuitively in every disc you can inscribe a square and the other way around

you can try to just do this for open balls in both topologies

then take a ball of T1 around every element in an open set, inscribe a ball of T2, take the union...

[fierce lily]

for this question, I construct the map f:Z->A, Z with discrete topology and A={0,1} here, then define x->{1} as x>=0, and x->{0} as x<0. Can someone help me check if my construction here is ok?

[prime elbow]

So you have to show an inverse image of a compact set that is not compact

[prime elbow]

{0} is compact but yes its inverse image is not compact

[mighty hull]

Take the constant map from any non-compact space to a single-point space.

[prime elbow]

But here there are four possible topologies exist on A, so specifically define it, but in every case {0} is compact

[fierce lily]

Then if I say set a map from Z->A as setting A={0} while Z with discrete topology, then A is compact and its inverse image is clearly not compact, it seems to me that the map has satisfied the condition?

[real star]

(infact there's only one map in that case)

[fierce lily]

I think the condition I list should work, because it seems that every open covering of A can have finite subcover->A={0,1} is compact as well, but Z is clearly not compact seems it is far too unbounded

[mighty hull]

There's only one map from any space to the single-point space (which is the constant map).

The preimage of that point is the whole domain.

[prime elbow]

Okay

[grave solstice]

If you remove two antipodal points from S^2 and identify antipodal points, do you still get S^2 with two points removed?

[real star]

isn't this like making a hole in the projective plane

[grave solstice]

yeah what I said is not true

okay it is the Mobius strip

if instead you remove disks it's much clearer lol

[real star]

ohh i see it too now

[snow gull]

Hey. I'm looking at a proof that if a fuction f:X->Y is continuous then the following is true.
Where Bf(x) and Bx are neighbourhood bases.

The proof begins by saying:

Given x in X, take a neighbourhood V of f(x) belonging to Bf(x). And they say that since f is continuous then f^-1 (V) belongs to the neighbourhoods of x.

Can someone explain me that? I know being continuous means that f^-1(V) is open but why is the second part also true?

[prime elbow]

f^-1(V) is an open set containing x, right?

[snow gull]

yes sir

I guess that's it then. But does it necessarily contain x?

[prime elbow]

Then by definition of open set there exists an open basis element of containing x, B_x \subset f^-1(V)

[prime elbow]

f(x) in V implies x in f^-1(V)

Inverse image of V under f

[snow gull]

Ok that makes sense. thx

[prime elbow]

Now B_x \subset f^-1(V) implies if y in B_x then f(y) in V

[snow gull]

Yeah I understand now. Thank you!

[runic sinew]

is it supposed to say Let $D^*$ denote the set of all limit points of the set $D$.

whoops

[gentle osprey]

Dilly

[runic sinew]

@cedar pier

[cedar pier]

yeah LOL

typo in the textbook

[runic sinew]

what was the definition for a limit point you were given

[cedar pier]

a is a limit point if there exists an open ball such that a point from D is in the open ball but not a itself

something like that

[runic sinew]

Is it, for every open ball surrounding $a$, there exists a point from $D$ contained within the open ball?

[gentle osprey]

Dilly

[runic sinew]

i didnt need latex for that

[cedar pier]

yeah

[cedar pier]

how do we show every limit point tho

[runic sinew]

Well you would probably start by letting a be a limit point of D union D*

and show that its in the set

[cedar pier]

ok i had that

how do i show that though

so basically

[runic sinew]

what can you say about a being a limit point of D union D* based on the definition of a limit point

[cedar pier]

let a be a limit point in D U D*

if a is a limit point, there exists an open ball that contains points from D

and D* is the set of all limit points

so it's D U D* is the set of all limit points + D

so a has to be inside?

[runic sinew]

well it contains points from either D or D*

[cedar pier]

oh yeah

ur right

@runic sinew

let a be a limit point, if a is a limit point, there exists an open ball with a nonnegative radius that contains points from D or D*

[runic sinew]

should just be positive radius

[cedar pier]

what now

[runic sinew]

nonnegative means you could have a radius of 0

[cedar pier]

ur right

[runic sinew]

which would be a single point

[cedar pier]

what do i have to show now

im still kinda lost

[runic sinew]

also it should be every open ball with positive radius contains points from either D or D*

[cedar pier]

let a be a limit point, if a is a limit point, every open ball with a positive radius contains points from either D or D*

[runic sinew]

yeah

[cedar pier]

what now

[runic sinew]

I'm still thinking tbh

Ok this might be overcomplicating it, but I would split it into a couple cases:

• Every open ball around a contains a point from D
• Every open ball around a contains a point from D* but not from D (this should result in a contradiction)

[real star]

what is D*?

complement?

[runic sinew]

the set of limit points of D

[cedar pier]

set of all limit points of D

[cedar pier]

well pt1 is true from defn of limit point

how is pt2 true?

@buoyant badger

[runic sinew]

yeah if every ball around a contains a point from D then a is in D*

the second case isnt true

[cedar pier]

bc D* is the set of all limit points

[runic sinew]

you will arrive at a contradiction

[runic sinew]

ye

[cedar pier]

why can we say a false statement and say it's a contradiction?

[runic sinew]

so you could also just use the second case to show that every open ball around a must contain a point of D

[runic sinew]

If you assume something is true and it leads you to something thats impossible, then you know it cant be true

[cedar pier]

so if we show that every open ball around a contains a point from D* but not from D is false, does that imply:
every open ball around a contains a point from either D* or D?

how does it imply that

[runic sinew]

it implies that if an open ball around a contains a point from D*, then it also contains a point from D

[real star]

^you can use this to directly prove the statement

[cedar pier]

i dont get how

why does it imply that statement

[real star]

If every open ball around $a$ contains a point from $D*$ or $D$, then there's 2 cases.
Either there is already a point from D in an open ball or a point $x$ from $D*$. But then take a smaller open ball around $x$ and you find a point from $D$ inside that ball. This means $a$ is a limit point of $D$.

[gentle osprey]

sonihi

[cedar pier]

what about D* though

will a point from D* be in the smaller open ball?

[real star]

it does by definition because x is in D*, but that doesn't matter

[cedar pier]

why not?

[real star]

it just has to be a point from D or D*

[cedar pier]

why does this show that it contains all limit points?

[real star]

ok lets go through one step at a time

say $a$ is a limit point of D union D*

written out: every open ball around $a$ contains a point from $D$ or $D*$,

[cedar pier]

yes, i get that so far

[real star]

Now we have to show every open ball around $a$ contains an element from D, which simply means showing $a$ is a limit point of D

so take an open ball $B$ around $a$. If $B$ contains a point from $D$ we have nothing to show. If it contains a point $x$ from $D* $, we take a smaller open ball around $x$ and since $x$ is a limit point that smaller open ball contains a point from D.

this means $a$ is a limit point

[cedar pier]

and what makes us able to make a smaller open ball?

why can we keep making smaller open balls

[real star]

cuz an open ball is open

[cedar pier]

if a is a limit point, how is it every limit point?

sorry im really slow 😭

basically

[real star]

so $a$ is just assumed to be an arbitrary limit point of D union D*

the argument just uses that fact

[cedar pier]

so if a is an arbitrary point it's true for all a right

[real star]

yes

[cedar pier]

ahhhh

lemme write this up and make more senses of it

*sense

[woven trail]

one thing I'm really enjoying about topology is basically doing analysis with no real number algebra

pretty satisfying I'd say

doing analysis with only set theory

that's the vibe

[cedar pier]

shouldnt it be taking a smaller open ball around a not $x$?

[gentle osprey]

emerso2000

[real star]

nah, we want to use a ball around x, because that ball will contain a point from D

[cedar pier]

so from what im understanding

x is the point in D*

[real star]

yea

[cedar pier]

if it's in D* isnt it limit point already

since D* is the set of all limit points of D

[real star]

it is, but that doesn't yet mean "a" is

[cedar pier]

i dont see the end of the argument

if x is a limit point, why is a?

x being a limit point just means there's a limit point in the open ball B

and if u shrink the open ball

[real star]

right but you can take an open ball B' around x that is fully contained in B (since B is open). B' will contain a point from D since x is a limit point

[cedar pier]

wait why will B' contain a point from D since x is a limit point?

bc the definition of limit point right

[real star]

yes

[cedar pier]

x has to contain points from D but not x itself

is that correct

[real star]

wdym?

x is a point not a set

[cedar pier]

sorry

open ball around x, which is B' will contain points from D because x is a limit point

@real star

[real star]

yes

[cedar pier]

and why does this apply to every x again?

bc we picked a to be arbitrary?

[real star]

yea

either a ball B around a limit point "a" contains an "x" in D* or a point in D

just by definition of "a" being a limit point

[cedar pier]

can you check my work

NTS is need to show

[real star]

maybe think this through yourself lol
it seems fine though
I guess a cleaner way to write this is to just say "pick any element x in B" and then say if x is in D or if x is in D*

[cedar pier]

i get all of it until the end

why is a a limit point?

[real star]

you showed any open ball around "a" contains a point in D

i.e. a is a limit point

[cedar pier]

ok i need to think on this a little more

how did you know how to do this?

i was super lost

i wanna be able to do these on my own

[real star]

how long have u been doing topology?

i am somewhat familiar with it

so just took a guess as to how it works

[cedar pier]

uhhhh 1 week

😭

this is my lin alg hw

[real star]

loool good luck

you are doing this for linear algebra?

[cedar pier]

we went over open ball stuff

the course is lin alg + calc 3

i still dont get theorems like this

it seems trivial but idk how this ties into topology

[real star]

oh

[cedar pier]

how do i get better at proofs?

my last midterm i got a 40% and i dont want to repeat that

but i just cant see proofs on my own

my prof told me to prove stuff in the textbook but it's all so hard

[real star]

the one direction is clear, i guess the other direction you have to combine deltas from every individual function

or topologically product topology or smth idk

[wise mist]

u will get better magically

[cedar pier]

u do combine deltas, how do i get the sense to do that?

[real star]

it took me hours to get a proof right at first

[wise mist]

just keep showing up

[cedar pier]

ill keep working at it then

[real star]

in this case you don't have a lot of info other than the definitions

[wise mist]

i think they are talking bout taking min of those deltas ?

[cedar pier]

yep

[wise mist]

but if if is delta its analysis and not topology

[cedar pier]

u take the min of the deltas

that's the proof

😭

im cooked

[wise mist]

u ain't

it's just about getting used yo it

i don't know why min, i felt good to say it and i told u that

[real star]

u are fine, https://math.stackexchange.com/questions/2343261/sequentially-continuous-implies-continuous i tried this at the beginning of my analysis course and literally couldn't do it. the contraposition proof is 2 lines

[wise mist]

yeah don't sabotage yourself anymore

everyone suffers

[civic verge]

Ladies and gentlemen, I passed general topology at the undergraduate level, I thank all those who have helped me to clarify my doubts, thank you very much.

❤️

[wise mist]

yeah, crazy as hell

[heady skiff]

can somebody explain to me why the boundary of H^0 is the empty set? is that just by definition

[civic verge]

A point is closed and is insolated, there is no boundary in this case.(?)

[heady skiff]

i see

thanks

[civic verge]

Are you reading Loring's variety book?

[heady skiff]

i just started on lee's topological manifolds

skipped until the manifolds part since i alr know point set topology

so i'm looking for more of an alg top perspective

[fierce lily]

i have done question 1, and now want to prove (a)->(b) here. I know about the def of closed map right now, let f:X->Y and every cont map g:W->Y here. WTS: X product_Y W->W is a closed map. But the question is , I have no idea about how to show this map is a closed map

[gritty widget]

Suppose that every countable, closed subset of M is complete. Prove that M is complete.

now we have to show that if (x_n) is cauchy sequence then it will be converge.
Now take E = {x_1,x_2,...} it is countable subset of M.
If cl(E) = E then E is closed then it follows by given hypothesis.
Now if there exists x in cl(E) such that x not in E. Then x is a limit point of E implies there exists subsequence of x_n which converges to x, then x_n is convergent.

If there are any mistake please let me know

[quaint spruce]

Looks good

[gritty widget]

Let M be metric space. Now define l(M) is the set of all bounded real-valued function on M with sup norm.

since B(X, Y) is complete space if Y is complete.
Hence l(M) is complete.

Now fixed a in M, and for x in M define f_x in l(M) such that f_x(t) = d(x,t) - d(a,t).
It is bounded and also M is isometric to {f_x } over M.
Now take M' is closure of { f_x, x in M }.
M' is closed follows that M' is complete and {f_x} over M is dense in M.

I want to confirmation only.

[mighty hull]

confirmation ONLY

[gritty widget]

if there is any mistake please let me know

[mighty hull]

confirmation OR DISCONFORMATION ONLY

[gritty widget]

confirmation

[mighty hull]

error contradictory instructions

🤖 bzzt 🤖

Never mind, two different proofs.

[quick crane]

confirmation on what? Is this a proof of something?

[gritty widget]

Yes completion of metric space M

[mighty hull]

How many limit points can E have, as a Cauchy sequence?

[gritty widget]

Sorry I have no idea because they didn't use cauchy- sequence

[mighty hull]

Ah, oops.

A metric space is complete if every Cauchy sequence converges. Every Cauchy sequence is a countable subset.

If (a_n) is a Cauchy sequence then E = {a_1, a_2, ..., a_k, ...} is countable. If cl(E) is countable then you're done.

[gritty widget]

Yes I know the definition

[mighty hull]

"The familiar is not understood precisely because it's familiar." — Hegel

[mighty hull]

How many limit points can a Cauchy sequence have?

[gritty widget]

At most one

[mighty hull]

So |cl(E)\E| <= 1

[gritty widget]

I don't understand why we consider this question?

[mighty hull]

If E is countable then E plus its limit points is still countable.

[gritty widget]

Yes

[mighty hull]

Ok...that's cl(E). Is cl(E) closed?

[gritty widget]

Yes

[mighty hull]

So cl(E) is complete by hypothesis.

Whatever E converges to in cl(E), it will converge to in M

[gritty widget]

You mean this question?

[mighty hull]

Oh, apologies, I doubly confused myself. I thought they were two different proofs and then your comment made me think I was wrong about that, but I see you were verifying you were asking if your construction of the completion was correct.

[gritty widget]

No problem

[quick crane]

You are proving existence of the completion?

[gritty widget]

Yes

[mighty hull]

I'd call this an outline of a proof, but the main steps are all correct.

[quick crane]

Why is $f_x$ bounded?

[gentle osprey]

L

[gritty widget]

Because | d(x,t) -d(a,t) | < d(x,a) which is independent of t

[gritty widget]

Okay

[quick crane]

This works

[mighty hull]

Really what this proves is that you can isometrically embed any metric space M into a complete metric space M'.

In order to warrant talking about the completion of M, you'd also want to prove that any two such M' are isometric.

[gritty widget]

You mean I have to show uniqueness?

[mighty hull]

Up to isometry, yes.

[gritty widget]

Okay but in the textbook only two conditions are given for completion, first M' needs to be complete and second M is isometry to the dense subset of M'.

[mighty hull]

I'm just saying what this proves. You want to show that any two metric spaces into which M embeds is isometrically as a dense subset are themselves isometric, otherwise this could just be one "completion" among many

So it'd be "a" completion and we wouldn't be warranted in talking about "the" completion.

All this proves is that there's "a" completion, which is maybe all you care about.

[cedar jungle]

$\partial A \cap \partial B \subseteq \partial \left( A \cap B \right)$

[gentle osprey]

Gapi

[cedar jungle]

is this formula true?

[cedar jungle]

can you give me a counter example? I was trying intervals on the real number line but it doesn't work

ahh, thank you

[alpine nest]

even more fun, let A be the set of rational numbers, and B the set of irrational numbers

[woven trail]

I think I just understood why a topology that contains another is said to be finer

like, the nomenclature

it's cus the set is broken up in more pieces

[plush folio]

Yeah, I like to think of sand vs gravel. Gravel is coarser, the open sets are bigger and there are fewer of them

[heady skiff]

can somebody explain to me why we need the U_i in the first place?

oh i guess it's to show every point has a locally euclidean neighborhood

ah i think i get it now

Define $f: U_i^{\pm} \to \mathbb{R}$ by $f(x) = \pm \sqrt{1 - (x_1)^2 - \dots - (x_{i - 1})^2 - (x_{i + 1})^2 - \dots - (x_{n + 1})^2}$. Then $f$ is continuous. Let
[\Gamma(f) = {(x_1, x_2, \dots, x_{i - 1}, f(x), x_{i + 1}, \dots, x_n) \mid x \in U_i}] We claim that $\mathbb{S}^n \cap U_i^{\pm} = \Gamma(f)$.

I'm trying to verify this, can somebody check if this is the right idea?

[gentle osprey]

okeyokay

[wise mist]

hint on part d,e

also what is the motivation for such extension

[pallid delta]

d) d(x, y) = d(phi(x), phi(y)): any bijection induces a homeomorphism by defining the appropriate topology
e) show that ||the open balls are open intervals||

Motivation: it's a metric on a 2-point compactification of R which induces the same topology on R
that's pretty neat, as then going to infinity is just converging to a different point

[alpine nest]

Yeah, from my point of view the main motivation of the extended real line is to enable unified treatment of finite and infinite limits.

Since as Bezier points out, divergence to infinity can be restated as convergence to the "infinity" point in the extended reals.

So in settings where "infinity" shows up as a possible "value" (such as measure theory, where measures of sets or integrals of functions often have the potential to be infinite), it's helpful to be able to treat it almost the same as "ordinary" real values.

[fiery gazelle]

hi! should i use contradiction to prove that Q (set of rational numbers) is not open in (R,d)?

[wise mist]

thanks for the input but i think i dont have enough tools to get the motivation for now

[prime elbow]

Yes

[wise mist]

in each interval in reals
u are going to get atleast one element from Q and R\Q

$\mathbf{x}_k = \left(r \cos\left(k\theta\right), r \sin\left(k\theta\right)\right)$

[gentle osprey]

yeshua

[wise mist]

is this a valid counter, can i get more counter pls ?

[prime elbow]

| |x_k| - |x | | < | x_k - x | < e

[wise mist]

use bot 😭

nvrmind its fine now

[prime elbow]

oh sorry i proved the question not converse part

[wise mist]

thats ok

[prime elbow]

you can take sequence , -1,1,-1,... in R

[wise mist]

alt harmonic would work too right ?

$\mathbf{x}_k = \frac{(-1)^k}{k}$

[gentle osprey]

yeshua

[prime elbow]

Can you describe your sequence?

[wise mist]

$\mathbf{x}_k = \frac{(-1)^k}{k}$

[gentle osprey]

yeshua

[prime elbow]

But here x in R^n right?

So x_n are in R^n

[fierce lily]

I can not understand why this proof can makes me know that W(p(K)) is open set here, can anyone help me see it?

The proof is about projection map here is closed

[mighty hull]

This is the tube lemma: https://planetmath.org/tubelemma

[mighty hull]

The way it's written here is confusing to me. For example, they say K is a subset of X×W, but have x ∈ X and write (y,x) ∉ K, suggesting that x is in W?

What is y? Seems like maybe they mean to write w instead of y? And switched halfway through to talking "as if" the lemma was stated using W×X instead of X×W? Even then, what is p(y) if p is a function from X×W? Did they mean the preimage p^-1(y)? Is it some kind of shorthand?

What book is this?

The tube lemma is straightforward enough, I'd read one of the many proofs out there and apply it to prove this lemma. It's like 2 lines w/ the tube lemma.

[wise mist]

i mean this in just R but u can extend if u like

[fierce lily]

I believe he has many typos here

this is a book written by our instructor, and I always think what he writes is truly not clear enough to read

[fierce lily]

I have a doubt here, it seems tube lemma is obvious to show that for every x in X and {y_0} in W such that (x, y_0) not in K, we have open set= intersection of all U_i containing y_0 in W. But is it adequate to say y_0 in W\ p(K) always has open set containing y_0 as well

[mighty hull]

If K is closed in X⨯W and w in W\p(K) then (X⨯W)\K is an open set containing X⨯{w}. Apply the tube lemma to get the desired neighborhood of w.

[fierce lily]

I know this, since we can get the open set containing {w}. what i am not sure is will the projection map really project the same element {w} in X x W to w in W?

Namely, {w} in image of W \ p(k) is really the same as {w} not in K or not?

[mighty hull]

How is the projection map defined?

Because normally it'd just be defined as p(x,w) = w for all (x,w) ∈ X×W.

Like...it's a particular map.

That one

[fierce lily]

for part 2: proving (b)->(a), I have questions of universally closed implies f is proper map. f:X->Y is universally closed implies for any W->Y, X x_Y W->W is closed map. let K be compact in Y such that X x_Y W is homeomorphic to f^(K) and f^-1(K)->K is closed. Since K compact, then we have finite subcover for K, then I am confused about how I can proceed or if there is alternative method here.

[heady skiff]

how do we know x_{n + 1} = 0?

isn't x on the sphere so its third coordinate doesn't need to be zero right

nvm

why couldn't we just say the subspace where the last coordinate is zero tho

[mighty hull]

Subspace of what?

[heady skiff]

of R^{n + 1}

[wise mist]

for the 16. how does it want me to generalise ?

for every n \in naturals

[rancid umbra]

i think so?

i don't recognize any other obvious generalization

[wise mist]

is it super obvious or am i missing something

[prime elbow]

Let M be compact and let f: M -> M satisfy d(f(x), f(y) ) ≥ d(x,y) for all x,y in M. Prove that f is isometry of M onto itself.

I proved that if we have the same condition and d(x,y) = d(f(x), f(y) ) for all x,y in M then f is onto.

So we have to only show that d(f(x), f(y) ) = d(x,y).

I am working on a given hint: Consider x_n = f^n(x). Here f^n denotes the composition of f.

Now since M is compact so there exists convergent subsequence, we can assume (x_n) converges. Since (x_n) is convergent so it is also cauchy.

Thus for eps > 0 there exists N such that for n>m>N we have d(f^n(x), f^m(x) ) < eps.

Now we get d(f^(n-m), x) < eps implies that (x_n) converges to x, right?

Now take any x,y in M. Let x_n = f^n(x) and y_n = f^n(y) using the previous result, we get d(f(x),f(y))≤ d(f^n(x), f^n(y)) ≤ d(f^n,x) + d(x,y) + d(f^n(y), y) it follows that d(f(x), f(y) ) ≤ d(x,y).
Hence d(f(x), f(y) ) ≤ d(x,y)

Did I get it correctly?

I think there is a mistake to show x_n converges to x

I mean subsequence

[pallid delta]

yeah because then it's only small for some n as you have a subsequence
So it might be d(f^n(x), x) and d(f^n(y), y) are never both small at the same time

[prime elbow]

How can I correct that?

[sonic crane]

Covered top groups, g spaces, g orbits and all this mumbo jumbo

I uhhh definitely need to review these notes

[wise mist]

im i(2.) wrong?
or
the statement in the book(1.) is wrong ?

[alpine nest]

Why do you think one of them is wrong?

[wise mist]

its seems to me that i have proved quite the opposite

or i am being super dumb the whole time !

[alpine nest]

I mean, (2) is a bit sloppy because the radius of the ball is not considered.

[wise mist]

my concern is just the statement

[alpine nest]

But the logical thrust of both is the same

[wise mist]

im arriving at quite to the opposite no ?

[alpine nest]

You are arguing a ball B_1 can never be a subset of a ball B_infinity, which is what 1.2.58 is asking you to establish.

On the other hand, a sufficiently small ball B_infinity will be a subset of a ball B_1

[wise mist]

i cant prove the initial statement

[opaque scroll]

A set is open if every point has an open ball around it.

So the L1 ball being open in infinity norm amounts to showing that it contains certain Linfinity balls, like you did.

[wise mist]

i think im super dumb to still not see the most obvious thing

[alpine nest]

Write down, rigorously, the definition of an open set. Then let A bet the open unit ball in the || ||_1 norm, and show that A satisfies the definition of open set in the || ||_\infty norm.

[wise mist]

i got it dear sirs,
maybe i was tripping so badly to not see that

[sonic crane]

does every top space have a basis?

[red yoke]

The topology is a basis of the topology

[paper wedge]

yes

[paper wedge]

🦊

[sonic crane]

o

ngl im tired

just a lot of math everyday and its all really hard now

[cedar jungle]

Let $f : X \rightarrow Y$ be a continous function and let $x$ be a limit of some sequence of elements of set $A \subseteq X$. Is then $f \left( x \right)$ a limit of some sequence in $f \left( A \right)$ ?

[gentle osprey]

Gapi

[paper wedge]

yeah

what is ur prime candidate

[cedar jungle]

How can I prove it?

[paper wedge]

@cedar jungle

[paper wedge]

.

what do u suggest the sequence

that would have f(x) as its limit

given that say x_n converges to x

[cedar jungle]

x_n = x?

[paper wedge]

let me rephrase

suppose we have x_n converges to x

[cedar jungle]

yeah

[paper wedge]

we want to show that there exists a sequence that converges to f(x)

what sequence comes to mind

[cedar jungle]

f(x_n)

[paper wedge]

yes exactly

now i will assume ur working over metric spaces

what does f being continuous tell you in regards to distances

[cedar jungle]

if x is close to y, then f(x) is close to f(y)

[paper wedge]

kinda yeah

then ig can you write the proof?

details

unfold the definitions

[cedar jungle]

i think so

[civic verge]

TRUE

[paper wedge]

good luck

[cedar jungle]

thanks

[paper wedge]

yw

[tribal palm]

girls am i losing it or does every point p in a space X locally hom to R^d, have a nbhd hom to all of R^d

[paper wedge]

no

ur not losing it

[tribal palm]

why then does every def i've seen of "locally hom" insist on the every nbhd being hom to a open set in R^d

[paper wedge]

check this out

https://en.wikipedia.org/wiki/Topological_manifold#:~:text=A topological space X is,additional requirements on topological manifolds.

[tribal palm]

oh there it is

wonderful, thanku

[opaque scroll]

I guess it gives you slightly more freedom in choosing your neighborhoods

[paper wedge]

yes

[opaque scroll]

Why make something hard when you can make it easy

[paper wedge]

kinda used that in computing local homology of a manifold

geting Z

getting Z (IIRC?)

[tribal palm]

to me having all of R^d seems easier :S but i suppose i have no idea what i'm talking about

[tribal palm]

it was fun drawing some figures to try convince myself of this

[opaque scroll]

I mean R^d is open in R^d, so you're not robbed of that option. You just have more choices

[real star]

deranged

[paper wedge]

cool

[tribal palm]

i could've reused the picture i already had of R^2, but i feel that would've been properly deranged

[real star]

lol please do

[cedar jungle]

what about if I'm doing it on a topological space

[paper wedge]

it's still true

the converse isn't tho

[cedar jungle]

What do you mean?

[paper wedge]

if f sends convergent sequences to convergent ones it need not be continuous

[plush folio]

One example where this definition is easier is when you want to show that if U is an open subset of a manifold M then U is also a manifold: if phi is a chart on M, then the restriction phi_U is a chart on U (to an open subset of R^n). If you need phi_U to be a homeomorphism to all of R^n then you may need to restrict the domain further (as far as I can understand, I'm still very new to manifolds)

[cedar jungle]

but f is continuous

[paper wedge]

yes im talking abuot the ocnverse

[tribal palm]

right that makes perfect sense thanku

[cedar jungle]

so does it converge in topological spaces too?

[opaque scroll]

Try writing down the definition of a sequence converging to f(x), and seeing if it holds

[cedar jungle]

what is the topological definition?

[opaque scroll]

A sequence xn converges to x if any neighborhood of x contains a tail of the sequence

[sonic crane]

Thx for the cat love ❤️

[woven trail]

just to be sure, it's also a homeomorphism in the box topology right?

cus like, h((y1,z1)x(y2,z2)x...) = (a1y1 + b1, a1z1 + b1)x(a2y2 + b2, a2z2 + b2)x... since the extremes are obvious and the in betweens are also easy and also the function is monotone for each component

like, idk this question is worded in a way that makes me believe that the answer should be that in the box topology it would be different but it pretty cleraly works out to not be

[paper wedge]

yes

[robust drum]

In practice when doing stuff with manifolds the freedom to switch between all 3 definitions is very useful

[sonic crane]

Typo in 2) a)? Why is U0 never mentioned again

[opaque scroll]

Think maybe U1 is supposed to be contained in U0 and that U0=U

[sonic crane]

Ok thanks!

[opaque scroll]

What's your definition of topological group here?

[sonic crane]

A topological space that satisfies T1 axiom(?) (says finite sets are closed) and the map a,b to ab^-1 continuous

[opaque scroll]

Okay, so you have T1 baked in

[sonic crane]

Yeah

[opaque scroll]

Was just curious because of b)

[sonic crane]

Is it not always like tha?

[opaque scroll]

Idk, I would allow non-Hausdorff groups

[sonic crane]

Oki doki

Did yall know i completely missed a question on my topology midterm i thought we only had to do 3 out of the 4

I WOULDVE GOTTEN 18/20 otherwise if i did the question czuz i knew how to do it 😭

So i got 13/20 in reality. So dumb man

Thats never happened to me but ofc it has to happen now that im in grad school -_-

Rant over!

[paper wedge]

what was the problem

[pallid delta]

I once left the midterm with my paper in my bag
I just talked to a classmate, packed up and left. Didn't actually hand the paper over to the prof.
Noticed 2.5 hours later and didn't get punished for it, fortunately

[sonic crane]

Show that the topologies generated by metric d and metric d’ = d/(1+d) is the same

Idk if the formula for d’ i did was correct but i think u know the question

[sonic crane]

Dang

[fierce lily]

is there any elements in this set that is not open?

[opaque scroll]

There are lots of sets that are not open yes, if that's what you're asking

[fierce lily]

which kind of sets here are not open?

because i find that inf can not appear in R

[opaque scroll]

Singletons for example

[fierce lily]

you mean stuffs like {0} * {2} etc?

[opaque scroll]

Yeah, or {(0, 2)} maybe I would write

[fierce lily]

besides singletons, do you think if either a or b is singleton, then the set will not be open here?

[opaque scroll]

Depends.

Like {a} x [0, 1] isn't open, but {a} x (0, 1) is

[fierce lily]

what kind of other sets are not open here?

[opaque scroll]

Most really. It's easier to describe the sets that are open

The open sets are ||unions of {a} x U for U open||

[fierce lily]

so is it correct to say if no closed or half open interval appears and guarantee at least one open interval, then it is open sets here?

[opaque scroll]

Well, I guess there can be more complicated sets than just half open or closed intervals as well. But morally I guess that's the answer.

Not sure what you mean by "guarantee at least one" necessarily

[heady magnet]

Suppose $X \subset Y \subset Z$ are subspaces and $X$ is dense in $Y$ and $Y$ is dense in $Z$, does that mean $X$ is dense in $Z$?

[gentle osprey]

stable compass needle

[quartz horizon]

Yes, I think this is true

[heady magnet]

nice

[paper wedge]

yeah

[heady magnet]

sweet

[paper wedge]

ig just

using the A is dense in X iff every point in X has a sequence of elments in A converging to to this point

is the way to prove this

right?

[heady magnet]

I don't know I havent done anything with sequences in a very long time

this is true for all topological spaces ?

[paper wedge]

metric spaces

[heady magnet]

hmmm

[paper wedge]

or no ig that would work for any top spaces too

[heady magnet]

ok sweet

[paper wedge]

yeah if the closure of A is X then every point in X is either a limit point or a point in A what am i thinnking haha

😄

sorry

[heady magnet]

nice thanks

[paper wedge]

yw

[quartz horizon]

I don’t think this works in general

Sequences aren’t enough to probe any topological space

[paper wedge]

ur right

this needs first countability

or

use nets

[quartz horizon]

sure, if you know how to use them

(i do not)

[paper wedge]

yeah theyre pretty cool ig they are just sequences from arbitrary sets instead of N so u do not need first countability for this

also who deals with spaces that aren't even second countable..

😦

[quartz horizon]

Right, but aren’t there like a proper class of “arbitrary sets”

How the heck do you even show something works for all nets

[paper wedge]

directed setrs

ur using a "directed set" instead of N

[quartz horizon]

Yes but there are a huge number of directed sets

A proper class probably

[quartz horizon]

So my Q remains

[mighty hull]

What's the definition of density you're working with?

[quartz horizon]

I assume it’s to do with topological closure

[fierce lily]

is it correct to argue that because cantor set is subspace of [0,1], [0,1] is hausdorff, then cantor set is hausdorff. and because cantor set is compact, and every compact hausdorff space is normal, so cantor set is normal?

[rancid umbra]

yes, that is fine

[unreal stratus]

Yeah

[heady magnet]

The closure is the whole set

[mighty hull]

Well, then:

cl(X) = cl(cl(X)) = cl(Y) = Z

But that's not usually what "X is dense in Y" means.

Instead it means that Y ⊆ cl(X)

e.g., "the rationals are dense in (0,1)"

[heady magnet]

I mean that if i take the closure of X in Y i get Y and if i take the closure of Y in Z i get Z

[mighty hull]

The closure of ℚ∩(0,1) is not (0,1) — (0,1) isn't even open.

But it is [0,1]

(Assuming ℝ is the ambient topological space.)

[heady magnet]

the closure of Q intersected with (0,1) in (0,1) is (0,1)

right

[heady magnet]

im not working withing R

[mighty hull]

I'm talking about my example.

Let X and Y be subset of a topological space S.

"X is dense in Y"

Does that mean cl(X) = Y or cl(X) ∩ Y = Y?

In general, A∩B = B is equivalent to saying B ⊆ A

[heady magnet]

sure

[mighty hull]

If you mean cl(X) ∩ Y = Y then that's what I said. "X is dense in Y" means Y ⊆ cl(X), which is equivalent to cl(X) ∩ Y = Y.

[heady magnet]

im working with the subspace topology though

[inland thistle]

That’s cool thx

[gritty widget]

is there a dense, open set in R with uncountable complement?
can someone help me to find a way to do this?

say U is dense open subset of R, then R\U is closed set which is nowhere dense set

and R\U is uncountable

[dire dove]

||complement of cantor||

[gritty widget]

ah thank you

[gritty widget]

Let M be complete, and let E be an F-sigma set in M. Prove that E is a first category set in M if and only if M\E is dense in M.

I proved second direction but i have a problem in first direction.
E is first category then M\E is dense, how can i show that?

[ocean canyon]

Is it obvious that $[f] = [f_1] \bullet \cdots \bullet [f_n]$ in the last paragraph here? It seems "intuitively right" but I find it hard to think of an explicit path-homotopy. Does one need to construct one to see this?

[gentle osprey]

Benjamin

[alpine nest]

M\E is the intersection of countably many open dense set, so by Baire's theorem it has to be dense.

[gritty widget]

so E = cup F_i i over countable index set I, where F_i is nowhere dense set, so we take E \ subset cup cl(F_i) over I?

no

[alpine nest]

My mistake, I should have said M\E contains the intersection of countably many open dense sets.

The complement of the closure of F_i is an open dense set

The union of F_i's is contained in the union of their closures.

[gritty widget]

yes

[gritty widget]

yes

[alpine nest]

Oh wait, I was right, since you're talking of an F-sigma set

Closures aren't needed, the F_i's are already closed

So their complements are indeed open dense

[gritty widget]

but forget it, because now i want to show that in complete set a meager set has empty interior

[alpine nest]

It's the same thing, its complement is dense.

A set has empty interior if and only if its complement is dense

[gritty widget]

but it is F-sigma set how can we gurantee F_i is nowhere dense

[alpine nest]

If any of the F_is had nonempty interior, then so would their union.

[gritty widget]

yes therefore now i want to prove general statement

[gritty widget]

yes

[alpine nest]

What's the wording of Baire's theorem that you use? "in a complete metric space the intersection of countably many open dense set is dense"?

[gritty widget]

yes that one

[alpine nest]

Right, so equivalently the union of countably many nowhere dense sets will also have empty interior.

[gritty widget]

so a now i want to show it is equivalent to a meager set has empty interior

[alpine nest]

It's just De Morgan, basically

[gritty widget]

yes i proved that

wait

[alpine nest]

Yeah, I mistyped

It doesn't have to be nowhere dense (e.g. the rationals in R aren't nowhere dense)

[gritty widget]

i proved that countable union of meager sets

countable union of meager sets is meager set

[alpine nest]

Yes, and a meager set has empty interior.

[gritty widget]

but how can i show that?

[alpine nest]

By considering complements

if F is a closed set with empty interior, then its complement is an open dense set.

[gritty widget]

yes

[alpine nest]

If F is a subset of a closed set with empty interior, then its complement contains an open dense set.

[gritty widget]

so you mean if E is meager set then i have to take cl(E) such that here F = E is subset of cl(E)

[alpine nest]

The closure of a meager set can be the whole space, so that won't work.

But a meager set is contained in a union of countably many closed nowhere dense sets.

So all you need to show that such a union always has to have empty interior.

[gritty widget]

how can i show that?

[alpine nest]

Isn't that the definition?

[gritty widget]

meager set is countable union of nowhere dense set, right?

[alpine nest]

Yes

And a nowhere dense set is one whose closure has empty interior.

So the closure of a nowhere dense set is also nowhere dense.

[gritty widget]

you said countable union of closed nowhere dense set

[gritty widget]

yes

[alpine nest]

Yes, and I said contained in, not equal to

[gritty widget]

oh i see

[alpine nest]

If we show that our set is a subset of something with empty interior, it will also have empty interior.

[gritty widget]

yes

[alpine nest]

So, now consider how the complement of "union of countably many closed sets with empty interior" can be described.

[gritty widget]

intersection of open dense set, which is dense by Baire Category theorem so we get our result

thank you @alpine nest

[gritty widget]

empty set is first category, right?

so doen't necessary that complement of first category is second category unless underlying metric is complete

similarly complement of second category set not necessarily a first category set, because we can take discrete metric space every non-empty set is second category set , say N with discrete metric space then {2} is second category and its complement is also

[round oyster]

Let $(X,\tau )$ be a connected topological space,
There are no open and closed subsets, apart from $X$ and $\emptyset$
Proof??

[gentle osprey]

MadAbdo

[round oyster]

\begin{proof}
Assume that $X$ is connected, and
assume for contradiction that there is $A \subset X$,
where $A \neq \phi,X$
,such that $A$ is open and closed, so we get,
$X \backslash A$ is open and closed, and :
[
(X \backslash A) \cap A = \phi \quad
(X \backslash A) \cup A = X \quad
X \backslash A \neq \phi, A \neq \phi
]
therefore we have a contradiction. (X not connected)
\end{proof}

[gentle osprey]

MadAbdo

[mighty hull]

There's no need to talk about contradiction, just remove that language and prove the contrapositive.

\\

Also, I'd use something like $\emptyset$ or $\varnothing$ for the empty set.

[gentle osprey]

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

[woven trail]

when studying subsets of R^n, is the standard just using the subspace topology for them?

like, when studying a sphere

[alpine nest]

Yep, unless specified otherwice, subsets of R^n are by default considered with the Euclidean topology (i.e. induced by the Euclidean metric inherited from R^n, which will be the same as the subspace topology).

[woven trail]

cool

one of the things that was most terrifying me is how we define a topology on curved surfaces without doing some calculus shit but after seeing a random example on the topology textbook I was like oh sure ig

[alpine nest]

Yeah, topology is fine, it's geometry that's the issue

[woven trail]

who likes geometry smh

so cringe

who cares

[alpine nest]

I mean, if you're doing calculus on surfaces, you pretty much need to.

[woven trail]

*dies

[fringe thorn]

[red yoke]

Real geometry is algebra

[woven trail]

who likes anything that's not pure topological properties smh

[red yoke]

Real topology is algebra

[woven trail]

oh

[forest pasture]

Hello

Reading rudin's book, I saw this theorem

He proof is quite simple, but it is explained why finiteness is essential in c and d

My question is, what is the proof no such counterexample on infinite sets exists for a and b?

How do we know they'll work until infinity?

Asking this question I think I know the answer now, because in the proof for c we take the minimum of positive values, which at infinity can be 0. We don't do so for the proof in a

Is that right?

[paper wedge]

for c, take (-n,n)

[forest pasture]

I figured

The example was (-1/n, 1/n)

For natural n

[paper wedge]

yeah

for d) u can think of it more that in T_1 spaces the singletons are closed

and so any set really is a union of singletons which is a union of closed sets

[forest pasture]

My question was about the non-existence of a counterexample for a and b

I gave a hypothesis about that

[paper wedge]

anyy union of open sets is open

[forest pasture]

Is that right @paper wedge

[paper wedge]

this really depends on what topology ur on, this is probably on metric spaces

so ig yeah

[forest pasture]

It is on metric spaces

And its archimedean and everything

[paper wedge]

yeah

whats ur question

[forest pasture]

That

I think I answered it

That's why I hate asking freaking questions

Coz I just get the answer

And then it's awkward

Thanks though

[forest pasture]

I'll now proceed to combust

[paper wedge]

good luck

[civic verge]

Has anyone read the book by the author Stefan Friedl ?

[woven trail]

tbh I love when this happens

just the act of formulating a question is an amazing understanding tool

since you have to dig a little into your own mind to explain to others what you don't understand, but when you expose it you kinda broke the confusion up into a more manageable piece so sometimes you kinda just get it

[forest pasture]

Exactly

I always reword the exercises I get to solve them

[woven trail]

trying to explain why what you already tried didn't work is also great

[forest pasture]

If they are like, hard

[woven trail]

ym, when you reword them you solve them?

[forest pasture]

Well it's an exaggeration