point-set-topology

[visual rock]

you need something more general for hausdorff

[hollow geyser]

what is T1 again?

[visual rock]

given a pair of points, each has a nbhd not containing the other

[hollow geyser]

Oh I see. So like 0 would would have the interval (-1.5, 1.5) and 1 would have the interval (0.5, 2.5). They don't contain each other, but the intervals are not disjoint

[visual rock]

yes

[hollow geyser]

Okay thanks

So "my whole limit wording thing" if and only if Y is T1

[visual rock]

no

there just isn't really a converse to hausdorff implying unique limits from a particular space. this is highly dependent on X

[hollow geyser]

converse to hausdorff or converse to T1?

[visual rock]

like you want a statement such as: given X, Y, and every (cts) map from X -> Y has unique limits, then Y has property P. but there isn't really one

if you instead say: given Y, and every continous map from any top space X -> Y has unique limits, then you can show Y is hausdorff

[hollow geyser]

Oh I see. So it was more that I had a specific X, rather than allowing any possible X?

[woven trail]

why is this so similar to the 1 isomorphism theorem like is there a category theory thing that explains why it's the same thing

cus like even the proof is identical

[random sinew]

its factorisation of a map

basically taking any map, you kind of “force” it to be a bijection by identifying together the elements in X that map to the same element in Z via an equivalence relation, then restricting the codomain to just the image

[woven trail]

oh

[random sinew]

taken from Aluffi

[woven trail]

:0

I'll carry this theorem in my pocket so that I can just redo it everytime I see a "first isomorphism theorem" again

[random sinew]

yea and like u said the proof is pretty similar in any given structure

[hollow geyser]

Hmmm. I may have inadvertently come up with a separation axiom that is weaker than T0.

Someone please tell me if this is already a known thing, or if it is just completely pointless.

But let's say that a space is $T_{-1}$ if, for every pair $(x, y)$ of distinct points, there exists open subsets $U, V$ such that $x\in U$, $y\in V$, and $U\cap V\ne U$ (nor would it equal $V$)

[gentle osprey]

SWR

[hollow geyser]

Every $T_0$ space would be $T_{-1}$. Would the reverse end up being true as well?

[gentle osprey]

SWR

[hollow geyser]

Correction: $U\cup V\ne U$.

[gentle osprey]

SWR

[alpine nest]

Hm, but consider X = {0,1} with the open sets being {emptyset, {0}, {0,1}}. That space is T_0, but I don't see why it would be T_{-1}

It might be that I'm misunderstanding your definition

[hollow geyser]

Oh right. I was mistaking T0 as requring it in both ways.

[alpine nest]

That's T1

[hollow geyser]

I might be misunderstanding it also

[alpine nest]

Also I hate that I know that; non-Hausdorff spaces are the work of the Devil

And metric spaces are perfectly normal

[hollow geyser]

I think what I am trying to say is that, for any distinct points x and y, we can find a U containing x and and V containing y such that U is not a subset of V or vice-versa.

And as I say it that way, I realize that it must be T1

[plush folio]

btw, $U \cup V = U$ is equivalent to $V \subseteq U$, so the conditions $U \cup V \neq U$ and $U \cup V \neq V$ are the same as $U \not\subset V$ and $V \not\subset U$ (and it's the same with $\cup$ replaced by $\cap$)

[gentle osprey]

sheddow

[hollow geyser]

I normally spend a very long time thinking about something before asking about it here. And every time I ask about it first, I remember why I prefer to think it through first

Not that I feel any of you are being harsh or critical, I just dislike that my thoughts are so disorganized

[alpine nest]

Thinking about something for a while only to then find out the question wasn't really well-posed/the answer turns out to be trivial is a very common experience, don't feel badly about that.

[plush folio]

Sometimes the only way to think things through is to post it here

[hollow geyser]

Very true

I just so very much wish that I could one day have a truly original thought in mathematics.

Something no one has ever thought of before

But I guess I need a PhD to do that

[alpine nest]

I have a PhD and I've never had a truly original thought

It's not strictly necessary

[hollow geyser]

yeah, but wouldn't it be awesome

[alpine nest]

It probably would; but I've learned to be content in my adequacy

To be more serious, it probably will come easier with more time, experience and exposure

[hollow geyser]

I'm at least at the point now where I have profound ideas that have already been discovered 100 or so years ago.

SWR's Theorem will exist one day

[versed pivot]

It seems that that property is weaker than T1 but not comparable to T0

Is that right?

If so calling it T_{-1} would be appropriate

[hollow geyser]

That was my first thought, but on reflection, it would be exactly T1.

Let's relax my $T_{-1}$ to be that at least one of $U\cup V\ne V$ or $U\cup V\ne U$ must be true, and I think I can show that the space must be $T_0$.

[gentle osprey]

SWR

[plush folio]

What about the set X = {a, b, c, d} with the topology {Ø, X, {a, b}, {a, b, c}, {a, b, d}}? That's T_{-1} but not T_0 unless I'm mistaken

[versed pivot]

ya that's the one I had in mind too

and this one is T-1 but not T1

[hollow geyser]

ah because a and b can do one of {a, b, c} and {a, b, d}

I see I see

So it looks like I did have an original thought

(now, is it useful at all?)

Let's say $T_{-1}$ requires both $U\nsubseteq V$ and $V\nsubseteq U$, and let's define $T_{-2}$ as satisfying either of those. Easily, every $T_{-1}$ is $T_{-2}$. Is every $T_0$ space also $T_{-2}$ and is every $T_1$ space also $T_{-1}$?

[gentle osprey]

SWR

[hollow geyser]

I have some stuff to think about now

[warm kettle]

I very often feel the same way. It's very frustrating and I sometimes wonder if my head's pure chaos or what :')

[hollow geyser]

I suppose that is the source of good ideas though

Have one million crappy ideas and then one good idea may pop out too

[warm kettle]

Definitely. That's why I still believe I have a good chance of proving Riemann's hypothesis

[hollow geyser]

Prove that it is not provable

imo that's our only hope

[warm kettle]

dw, I'll develop a completely new field of maths where it'll be a simple corollary of a more general fact

But before I do that, I'll stop being overly off-topic here hah

[hollow geyser]

I had to dig for that

[cerulean oriole]

Is any finite T_0 topological space a quotient of S^n for some positive integer n?

[red yoke]

No

Quotient of connected spaces are connected

[wanton heart]

If we are considering a quotient topology, are the "singletons" every equivalence class or only equivalence classes with one element?

I am checking if a certain quotient space is T1 or not

it is R/~ where x~y if x=y or x=-y and |x|, |y|>1

so I want to check if each singleton in R/~ is closed or not

but I am not sure about fi the singletons are just each equivalence class, or the equivalence classes of x with x in [-1, 1] since these are the equivalence classes with only 1 element in them

like, [5]={5,-5}

but [1]={1}

[woven trail]

I suppose, extrapolating the data, that if you learn enough you'll start having profound ideas discovered yesterday and then the idea will have been discovered this day

holy shit just realized that since continuity is a topological property, than a function is continuous in S² iff it's continuous on the unit open ball in R² (with that quotient topology that collapses the boundary into a pole)

that's so cool omg

[gaunt crest]

guys

i need help

whete do i ho

go

[woven trail]

I mean, if you translate the continuous function anyways

[woven trail]

#❓how-to-get-help

[cerulean oriole]

OK, a connected T_0 finite topological space.

[cerulean oriole]

The former. So a quotient space is T_1 iff every equivalence class is a closed subset.

[hollow geyser]

For some point $x$ in a space $X$, an open subset $U$ of $X$ such that $x\in U$, we would say that "$U$ is a(n) (open) neighborhood of $x$". Is there a succint way to describe the set $U\setminus{x}$ other than "A(n) (open) neighborhood of $x$, but excluding the point $x$ itself"?

[gentle osprey]

SWR

[barren haven]

are you looking for intuition?

[hollow geyser]

intuition?

[barren haven]

I'm trying to understand what you're looking for

[hollow geyser]

I'm trying to find a less-wordy way to describe a set that is the difference of some open neighborhood of a point, but excluding that point.

Context:

[barren haven]

U \ {x}

like

that's what notation is for

[hollow geyser]

In first-year calculus limits, we have $0<|x-a|<\delta$, which is just a neighborhood of $a$, but excluding $a$.

[gentle osprey]

SWR

[hollow geyser]

Yes, but I like to say it in words also. And sometimes I'd like to say "some open neighborhood <blah blah>" in words, rather than putting it in pure math terms

[barren haven]

maybe you can make something up

[hollow geyser]

It's not always convenient to simply write U \ {x} without first saying in words why I am doing it

[barren haven]

hollow neighbourhood

[hollow geyser]

I was thinking of something like punctured neighborhood, or even your definition.

I just wanted to know if there was an established term already

[barren haven]

not that I know of, but I'm a topology rookie

[versed pivot]

punctured neighborhood is pretty common I think

[barren haven]

actually, I think Rudin describes something like that thinking about it

lemme check

[versed pivot]

https://en.wikipedia.org/wiki/Neighbourhood_(mathematics)#Deleted_neighbourhood

apparently also "deleted neighborhood"

[hollow geyser]

oh neat

I am also glad my intuition of "punctured neighborhood" was spot-on

[barren haven]

can't find it, maybe my lecturer mentioned it in passing

[hollow geyser]

"All of x's neighbors". Or "the local set of points who call noise complaints on x" if I can draw experience from my college years

[barren haven]

anyways, I myself have a (honestly really dumb question idk why I'm struggling so much). Let $A = {\frac{1}{n} | n \in \mathbb{N} } \subset \mathbb{R}$ prove that every point of $A$ is a boundary point (of $A$ in $\mathbb{R}$ with yhe metric topology). \
I know this is really simple but I'm struggling with the formalism. I'm struggling to formally prove that every open ball of x has a point not in r, does anyone have any clues to give me?

[gentle osprey]

ξor.

[hollow geyser]

I asked a lot of really dumb questions here

topology is funny like that

Take something very simple and intuitive and complicate it beyond measure. Literally "beyond measure"

[barren haven]

that's next year lol

[barren haven]

like say x = 1/n
clearly if 1/(n+1) isn't in the ball, then x + r/2 is in the ball and not in A, so all good
but if 1/(n+1) is in the ball, then I think what I should be showing is that the average between the two is also in the ball, but I can't manage to get the inequality to show the distance is smaller than r

"the ball" a ball of any radius around any point x in A as this is the metric topology

[versed pivot]

for 1/(n+1) to be in the ball means 1/n - 1/(n+1) < r

try writing out 1/n - (1/n + 1/(n+1))/2

[barren haven]

oh no

lmao

wait no

nvm

sorry, brain short

my tablet lost batter and I can't do this in my head one second

battery

ok I was right

[hollow geyser]

How I like to solve problems like this is by analyzing the definitions of what I need to find. So, for this problem, how are you defining boundary?

[barren haven]

1/2n(n+1)

[versed pivot]

I think you went too far

[versed pivot]

you want to be able to use this

[barren haven]

a boundary point of A is any point x that for every neighborhood N of x, N ∩ A ≠ ∅ ≠ N ∩ Aᶜ

[versed pivot]

get 2 as a common denominator but don't bother getting n(n+1) as a common denominator

[barren haven]

< r + 1/+2(n+1))
?

[hollow geyser]

So then you want to prove that every 1/n is a boundary point, and all other points are not, correct?

[barren haven]

yesyes
my problem is with the formalism that every open ball of x in A has a point not in A

it's an algebra skill issue not a understanding one I think

[versed pivot]

you should have 1/n - (1/n + 1/(n+1))/2 = (1/n - 1/(n+1))/2 < r/2

[hollow geyser]

So in your specific problem, you are struggling to say that every interval around any 1/n will contain some number r not expressable as some 1/k for a natural k?

[barren haven]

yes

[barren haven]

ok I really was going too far

huh

interesting

[barren haven]

wait where did the - before the 1/(n+1) come from

[versed pivot]

from the minus sign in the first expression

[hollow geyser]

I don't blame you for struggling on this tbh

I did this problem back in my topology days

And I literally wrote in the solution that this problem was a fair bit harder than the others

[barren haven]

oh shit, yeah

ok thank you!!!

I hate inequities

worst part of algebra

[hollow geyser]

Correction here: you do not need to prove that the non 1/k points are not boundary points

[hollow geyser]

Real analysis is calling

[barren haven]

huh
unfortunately I don't think this is the hardest question this week 😭

[barren haven]

this is intro to analysis

my professor is... unconventional

[hollow geyser]

Oh haha then have fun with inequalities

[barren haven]

yes I'm really sad we left topology behind and now all I see on the board is one inequality after the other

alright thanks ppl

[hollow geyser]

oh yeah you got it done?

[barren haven]

yup

I might be back with more though

[hollow geyser]

fun fact: there is one boundary point of A that is not in A

[barren haven]

0, presumably, as the next part of the question is to prove 0 is a cluster point

btw, is there any accepted shortening for neighborhood

it's a long word to write again and again lol

[versed pivot]

yes nbd

[fringe thorn]

punctured nbhd

oh wait, this was already answered

my mistake

[hollow geyser]

You're good

It's always a joy to hear from you

[versed pivot]

In the order topology...suprema of nonempty subsets (when they exist) are in the closure...is that true?
My proof feels a little more complicated than I'd expect is necessary so not sure I trust it and curious if there is a straightforward proof or counterexample

😊

Thank you!

It seems like an obvious theorem but the proof seems kind of indirect

Though it has the same flavor that Munkres uses to prove that dense complete orders are connected so maybe it's not so unnatural

[random sinew]

is the standard subspace topology the only topology that makes the inclusion map an embedding? (ie homeomorphic to its image)

actually is it even an embedding

oh nah cause u can have sets open in a subset but not the whole space

why do we choose the topology for a subspace S like that then, obvs we want inclusion to be continuous but that only tells us that we want S intersect U (U open) to be open in the subspace topology, why do we want only these to be open and nothing else

wait i think Lee has a bit about this in Topological Manifolds

[opaque scroll]

It is, but sort of by definition.

An embedding is a map that is a homeomorphism onto its image, and the image has the subspace topology so...

As to why the subspace topology is defined in that way.

For any continues map f:X -> Y you want it to factor as
X -> Image(f) -> Y
Then there's only one topology such that both of these maps are always continuous.

The easy way to see that is just to pick f the inclusion of a subspace (with subspace topology)

[iron bolt]

what you're thinking of is an open embedding. the inclusion of a set is that iff the set is open

[sonic crane]

You and boytjie both study rep theory?

[opaque scroll]

I think so. But I think we study different rep theory maybe...

[sonic crane]

Nice

[novel plank]

Let $X,Y$ be two hausdorff top. spaces, let $A\subseteq X$ and $f:A \to Y$ be a continuous function. Is it tue that if $A$ is locally closed, then the graph of $f$ is locally closed in $X\times Y$?

[gentle osprey]

Eduude

[cerulean oriole]

Yeah, the above proof seems convoluted to me. A nbhd basis of a := sup A is given by (b, c) for b < a < c if a is not the smallest element of X and by (-∞, c) for c > a if it is the smallest element of X. In the former case, b is not an upper bound for A so there exists a' in A with b < a' ≤ a, so a' in (b, c). In the latter case, a must be in A (or A would be empty), and a in (-∞, c).

[cerulean oriole]

If A = O ∩ C, then A ⨯ Y = (O ⨯ Y) ∩ (C ⨯ Y) is locally closed. And graph(f) = {(a, y) in A ⨯ Y : f(a) = y} is closed in A ⨯ Y since Y is Hausdorff (we don't use X is Hausdorff at all), i.e., graph(f) = (closed set) ∩ (A ⨯ Y) = (closed set) ∩ (open set) ∩ (closed set). So I think the answer is yes.

[random sinew]

what is I? and why should it be obvious that CS^n homeo to B^n?

[rain gyro]

Guys can you take a look of my solution at help channel 29 for a question that I attempted solving through discussion of topological structure? I only have a couple hours before deadline please

[warm kettle]

The quotient space has a geometric intuition that you take every equivalence class and reduce it to a point. In your case, the only nontrivial equivalence class is the base. You take the it and reduce it to one point. In a picture, the process could look like this:

[random sinew]

yeah i get that

[warm kettle]

Oh, and I usually denotes [0, 1]

[random sinew]

cool

why should it be obvious that CS^n homeo to B^n?

[warm kettle]

Oh, I misunderstood your question. Hm, I'm not sure about the intuition for the general case, but if we stick to the dimension from the picture, you could project everything onto the same surface and get a disc, which would be the 2-dimensional unit ball

[unreal stratus]

Pretty sure this is also wrong (it's off by a dimension) unless they have some weird convention with B^n

[unreal stratus]

Anyway, here's the intuition I have: take S^n and instead of imagining taking the cone "on the outside", imagine the point is the centre of S^n and you join each point to the centre. Then you get D^(n+1)

More rigorously, there's a natural map S^n x I -> D^{n+1} where you map (x,t) to tx (i.e. scale x by t)

This is clearly surjective, and you can check that it induces a homeomorphism CS^n -> D^{n+1}

:)

[random sinew]

ok cool im not going crazy, it should be B^n+1 right

[unreal stratus]

Yes

[random sinew]

yeah ok

[unreal stratus]

(The way to remember this is that the subscripts indicate dimension as a manifold w boundary, plus cone increases dimension by one)

[random sinew]

mhm

ty

[novel acorn]

S^{n-1} is the boundary of B^n and the cone is the interior basically

[random sinew]

yah

[unreal stratus]

Well, the cone is the whole thing

[woven trail]

is continuity in the product topology preserved by compositions of functions like
f: X x X -> X and g: X x X -> X x X
?

[plush folio]

The composition of continuous functions is always continuous, regardless of the topology

[odd gulch]

is it true that given a dense set D in a hausdorff space T we have |T| <= 2^|D|

or do you need the upper bound as 2^(2^|D|)

if so what's a counter to |T| <= 2^|D|?

[odd gulch]

my idea for a proof is you can define an injection T -> P(D) by
T(x) = {x} if x in D
T(x) = {d1,d2,d3,...} where (di) is a term-wise unique sequence in D converging to x

I think it's not too hard to show this is an injection

[unreal stratus]

Nvm I was thinking of smth else

[rocky olive]

Hello! I want to show that domain-restrictions on continuous functions are continuous (like functions between topological spaces). Would this channel be a good place to ask for someone to proof-read ?

I am following Lee, which I think is point-set topology strictly

[unreal stratus]

Sure or you can consider like the map {sequences in T converging in X} -> X which takes limits

This is surjective and the domain has cardinality <= D^N

Uhh is this enough

I guess this goes |X| <= |D|^N which is slightly different lol but stronger in general

[plush folio]

Yeah, just ask

[gentle osprey]

2.71828182845904523536028747135

[rocky olive]

My only concerns really are the assumption that g^{-1}(U) = X' intersect f^(-1)(U), which makes sense, but not for a super-logical reason...

at least in my head

Does this come somehow from the definition of X' topology?

I take it that T_{X'} = {X' intersect S : S in T_X} is by definition

[quick crane]

this is easy to show using the definition of inverse image

[rocky olive]

actually not by definition, it probably follows from the topology axioms

[rocky olive]

right, but is the assumption I made wrong, for what it means to be a restriction of f onto a subset of the domain

[rocky olive]

Let me think about that

Just from definition of pre image

[tender halo]

you do

stone cech compactification of N is normal and has cardinality 2^2^\omega

[tender halo]

the last part is incorrect, as it may happen that there are no sequences that converge to x

[tender halo]

alternatively, Cantor cube of weight 2^\omega is also normal, separable, but has cardinality 2^2^\omega

all the examples will necessarily be a little exotic, because you in the very least need a separable but not second countable/regular space and those are not particularly common

[odd gulch]

thanks!

[odd gulch]

mmm right so I cant use sequences without first countable?

I see the issue with my proof now

thanks

[sharp panther]

How do we show that the connected sum of two connected manifolds(let us assume it has no boundary) doesn't depend on the choice of removed open discs?

[iron bolt]

if you're curious, the property that says you can use sequences is more precisely the space being sequential

every first-countable space is sequential, but not the other way around

[tender halo]

well in this case they use Frechet-ness of the space, not sequentiality, although i am unsure if that matters

[white ginkgo]

I’m looking for a homeomorphism from RP^n to B^n/~, where x ~ -x for x ∈ ∂B^n. I just don’t see it — any hints? I know that B^n / ∂B^n ≈ S^n and that RP^n ≈ S^n / (x ~ -x), but composing these two homeomorphisms seems needlessly complicated.

Here B^n = {x ∈ R^n : |x| ≤ 1}.

[opaque scroll]

There's a continuous map B^n -> S^n given by
x |-> (x, sqrt(1 - |x|^2)) in R^n x R = R^n+1

Which picks out the upper hemisphere

[exotic hemlock]

Need help for 3.10 (3)

I can see why it is true if A is finite, since we can sort of pick small enough bubbles with tips on the x-axis that do not intersect

but im not sure how to prove it for the case where A is countable

[white ginkgo]

Oh, yes. This was just the hint I needed. Thanks!

[random sinew]

am i misunderstanding or is this actually a one line proof? U is a union of AxB, A open in X, B open in Y, choose any A that contains x? whats wrong with what i just did

i didnt even use compactness of Y?

nvm im thinking about this wrong

product space r confusing icl

in my head i imagine them as like 2 spaces side by side but thats probably more like a disjoint union space

[red yoke]

Counterexample: {|x| < 1/y} in R × R+

[random sinew]

how do u visualise an open subset of a product space, like in ur head

its easy for like RxR obviously cause its just 2 axes

[red yoke]

I assume it's R × R

Or some subset of R

[random sinew]

uhhhhhh oki

sounds kinda limiting but ill try that

[barren haven]

Just to make sure, this isn't more complicated than proving that a subspace of a Hausdorff space is also Hausdorff, right?

Seems odd they'd ask specifically about the Cantor set

[unreal stratus]

Yeah lol

It is just what you said

[barren haven]

Maybe it was going to be a 2 part question that relied on F being Hausdorff and then they scaraped part B

anyways thnaks!!

[unreal stratus]

Yeah idk lol but np!

[barren haven]

And I have another question... Can anyone give me a hint on this?

(note the set I is the index set, not a random partition lol, took me a few mins to see that)

Like if I have $A,B \subset \bigcup \mathcal{Y}$ then there's no reason to say they are a union of different sets $Y_i$, so I'm not sure how to use the property given

[gentle osprey]

ξor.

[opaque scroll]

Say the union isn't connected, and A and B are a disconnection.

Then think about AnYi and BnYi

(so A and B are open and disjoint and their union is everything)

[barren haven]

n = intersection?

I guess that would mean $(A \cap Y_i) \cup (B \cap Y_i) \neq Y_i$?

[gentle osprey]

ξor.

[barren haven]

Well assuming both intersections aren't empty

Im.. not sure how that helps me

Ok I think I got it!!

Thank you!!

thats really smart

[paper wedge]

since U is open, for each x there is a basis element (U_i x V_i) containing x contained in U. these V_is cover {x} x Y and hence there is a finite subcover. take the finite intersection of the corresponding U_is.

and this is your V

[random sinew]

yeah i got it

[alpine hound]

If $X$ has the cofinite topology, then is there an easy way to understand the product topology of $X\times X$?

[gentle osprey]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[warm kettle]

Every open set in that space is a sum of open sets of form $(X \setminus A) \times (X \setminus B)$ for finite $A, B$. So any open set $U$ satisfies
[
U = \bigcup_{i \in I} (X \setminus A_i) \times (X \setminus B_i) = \bigcup_{i \in I} X^2 \setminus (A_i \times X \cup X \times B_i) = X^2 \setminus \bigcap_{i \in I} (A_i \times X \cup X \times B_i)
]
In other words, you have a number of finite sets of lines parallel to either "axis". You intersect those sets and remove from the plane. That's your open set

[gentle osprey]

Thingoln

[warm kettle]

Note that intersecting a horizontal and a vertical line will give you a point. I'm pretty sure you could characterise open sets in that product topology as the whole plane minus a finite number of lines parallel to either axis and minus a finite number of points, but don't quote me on that

[granite slate]

Assume that a topological space G is both a 2-manifold and an n-manifold (i know it cant be). Would it be correct to say, if that were true, that for every x in G we have an open nbh $U_2$ of x homoemorphic to some open subset in $R^2$ and an open nbh $U_n$. Then $U_n \cap U_2$ is a non-empty neighborhood around x homeomorphic to both an open subset of $R^2$ and $R^n$?

[gentle osprey]

Dompa

[sonic crane]

If you have a homeomorphism f: X -> Y and a subspace A < X, is the restriction A -> f(A) a homeomorphism between A and f(A)?

[granite slate]

yes the restriction is homeomorphic to its restricted image

[alpine hound]

Can you tell me in words*and not mathematical notations) how to understand the CLOSED sets in that topology?

[warm kettle]

It would be sets containing a finite number of lines (parallel to either axis) and points in the plane X^2, I think

[alpine hound]

So if I want to prove that a set is not closed in that topology

I dont think Ill be able to prove that there exists a point that does not have a contained neighberhod in the complement

So I think what can I do.. Hmm

I thought underatand the closed sets will help me somehow, but it doesn't

Thank you anyways

[warm kettle]

No worries

[inland thistle]

hello just have one quick question

for a closed rectangle, [a1, b1] x [a2, b2] x ......, can we trivially assume that a point itself is also a closed rectangle?
For example, can we let a_i = b_i for all i, and say this point itself is a closed rectangle in R^n ?

[unreal stratus]

Sure, unless your textbook makes some specific assumption against that

[barren haven]

Ello! We proved in class that $K \subset \mathbb{R}$ is connected by assuming otherwise thus $\exists A,B \subset K$ a disconnection of $K$, taking $a \in A$, $b \in B$ and showing that $A',B'$ as defined in the image added are a disconnection of $[0,1]$ which we proved is connected. The problem is we didn't prove $A' \cup B' = [0,1]$ and I'm not sure how to do that myself, any hints?

[gentle osprey]

ξor.

[barren haven]

or maybe that's not needed and I'm missing something

we showed $A',B' \neq \emptyset$, disjoint and open, is that enough?

[gentle osprey]

ξor.

[barren haven]

ohh nvm we assumed K is convex

[granite slate]

Problem. Let $\phi, \psi$ be two continuous maps $\mathbb{S^1} \rightarrow \mathbb{S^1}$ with different degree. Prove that there exists a $z \in \mathbb{S^1}$ such that $\phi(z) = -\psi(z)$

Would it be feasible to solve it like this $g(z) := \frac{\phi(z) + \psi(z)}{\mid \phi(z) + \psi(z) \mid}$ and then showing that $\phi \cong g \cong \psi \Rightarrow deg(\phi) = deg(\psi)$ i've typed out the proof with an explicit homotopy but i worry that the homotopy isnt necessarily continuous for all $t \in [0, 1]$ so i wonder if i should pursue this proof further or scrap it

[gentle osprey]

Dompa

[granite slate]

demonic sphere

it should say 1 but i guess the bold font makes it go crazy

actually i think this proof works, only way we get zero division in my homotopy is if $\phi(z) = -t\psi(z)$ for some $z \in \mathbb{S}^1$ but since $\mid \phi(z) \mid = 1$ and $\mid -t\psi(z) \mid < 1 \forall t \in [0, 1)$

[gentle osprey]

Dompa

[odd gulch]

what does Lee mean by apply 2-17 to Z, considering Z isn't a manifold

ah I guess it is a 0-manifold?

doesn't the construction just become Z - {0}, or am I missing something?

bit confused here

[fathom flax]

Topological spaces need not be manifolds. Lee is asking for you to define a topology on \Z as given in 2-17. If you sent an image of 2-17, that would be helpful

It seems that you sent 4-17 instead of 2-17

[odd gulch]

well, that would be my mistake lmao

thx

[fathom flax]

of course

[prime elbow]

Any hint how I can find an example for g?

[fathom flax]

$(x,y) \mapsto \min { d(x,y), e(x,y) }$ satisfies the first 3 properties of being a metric, i.e.

• a distance from a point to itself is 0
• positivity
• symmetry
  so the only possible condition that can be broken is the triangle inequality

[prime elbow]

Yes

[fathom flax]

what examples of metrics have you tried?

[prime elbow]

I tried the usual metric on R and discrete metric

And I found that if d(x,y) ≤ e(x,y) for all x,y in X then triangular inequality holds therefore we have to take metric such that which doesn't follow that condition

[fathom flax]

Are you taking d or e to be the discrete metric?

Either way, neither bounds the other from below

[prime elbow]

Yes but this doesn't not work

[fathom flax]

I would suggest working with a finite set, just try it on a set with a few elements and give it your own metric

It shouldn't be too long before you find a counterexample

[prime elbow]

I don't know how to assign the metric on the finite set, it makes discrete topology

[fathom flax]

a metric is just a function, so you can just define some function on elements of some finite set

Let $X = {a,b,c}$, and let $d : X\times X \to R$ be defined by d(a,b) = ...

[prime elbow]

But we have to verify all those three conditions to make it metric

[gentle osprey]

Darrion

[fathom flax]

Luckily that's not too bad to do on a very small set

[prime elbow]

So you mean I have to define d and e on a finite set?

[fathom flax]

Not that you have to but that it might help in defining a counterexample

[prime elbow]

Okay

Can I get an example of such a function which is not isometry? I know that function will not be linear transformation

I proved this one but I am not sure when both don't exist then how inequality works here ?

[mighty hull]

diam is usually defined to take values in the extended reals, so diam(A) always exists. It's either a finite real number or infinity.

[prime elbow]

okay thank you i got it

[keen seal]

Hey, I'm not sure if this is a result of general topology or not or if someone is familiar with a similar type of result

Suppose $\tau$ is the smallest topology which contains the topologies ${\tau_i}{i \in I}$ on some set $X$.
Is it true that $Cl\tau(A) = \bigcap_{i \in I} Cl_{\tau_i}(A)$

[gentle osprey]

Trivial Lemma

[keen seal]

It is clear that the set on the right is closed and contains A

[iron bolt]

smallest as in finest?

[keen seal]

set theoretic, so coarsest

[iron bolt]

ah, ok

[keen seal]

it is clear that the finest topology which contains all of those topologies is the power set of X, as it's always the finest topology you can give a set

[iron bolt]

right. I confused myself there for a second

[prime elbow]

If X has more than 2 points, say x,y,z.

Assume d(x,z) is the diam( {x,y,z} ) then take B = {x,y,z} and A = {x,z}, right?

Or is there any other condition?

[keen seal]

That is a very strong condition (in the sense that the conditions are strong and thus barely applicable)

It is enough to restrict X to such a B, as nothing depends inherently on X

only on B

so you want to find sufficient conditions on (X,d) such that there exists a subset of X with the same diameter

An example of such a condition is denseness

if A is dense in X, then diam(A) = diam(X)

if X is unbounded, then any other unbounded subset (which must exist) also has infinite diameter

you may also be able to do funny things with borel measures I think

[iron bolt]

consider X = {0,1,2} and the two topologies in which {0} resp. {1} are the only closed sets. then the closure of {0,1} under each of those topologies is the whole space {0,1,2}, but the closure of {0,1} under their supremum is {0,1} itself

[keen seal]

non-trivial closed sets you mean?

[iron bolt]

yes

[keen seal]

{{0},emptyset,X}

the non-trivial opens on the first is just {1,2} and the other {0,2}
it generates the topology with non-trivial opens {1,2},{0,2},{2}.

[iron bolt]

also {2} because finite intersections

[keen seal]

so the closure is {0,1}

oh right

any Hausdorff examples?

[iron bolt]

same thing if you argue via closed sets directly - the topology generated by {0} and {1} must contain {0,1} because of finite unions

[keen seal]

yeah

makes sense

what about if all the topologies are Hausdorff?

A hausdorff topology on a finite set is discrete so there aren't finite examples

[iron bolt]

yup

I don't know if there's larger ones

I only considered non-Hausdorff examples because I found it easier to think about this for spaces with few open sets - but it could well be that there's counterexamples for Hausdorff spaces too

[keen seal]

nvm the spaces I'm considered are non-Hausdorff

so it doesn't matter

thanks

[iron bolt]

maybe if you take something like the two topologies on R that are generated by intervals [a,b) resp. (a,b]

nevermind, I don't think that would work

I'm curious, what is it that you're trying to prove? maybe you can avoid all of that

[keen seal]

it's not a general topology question

and would take a while to give the prereqs

clearly denseness in the sup implies denseness in each topology

I wanted to see if this was just a P implies Q or if the other direction was also true

[nocturne plume]

Can someone check my proof please.

The question wants me to give an example of a pseudo-metric on a set

The properties in red define a pseduo-metric

I showed that on RxR, d(x,y) = | |x| - |y| | defines a pseudo-metric.

[alpine hound]

I am sorry for asking such "dumb" question, but if I have a topological space $X$ and $Y,{X_i}{i\in I}$ are topological subspaces of $X$, such that $X_i\subseteq Y$ for all $i\in I$. Does it mean that $\cup{i\in I}X_i\subseteq Y$?
I am unsure if it is correct for all $Y$'s or just for closed $Y$'s. Thanks.

[gentle osprey]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[unreal stratus]

Yes, this is just a matter of set theory

But you should check this just from the definitions

[alpine hound]

Yes, like... if $x\in\cup_{i\in I}$ then there exists $X_i$ such that $x\in X_i$ but $X_i\subseteq Y$ so $x\in Y$

[gentle osprey]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[unreal stratus]

yes exactly

[alpine hound]

The problem is, because $I$ is not finite, then I wasn't sure if "then there exists $X_i$ such that $x\in X_i$" is correct

[unreal stratus]

This still just follows from the definitions

[gentle osprey]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[alpine hound]

Yeah, I just think that everything I do is wrong lol.
Thanks

[unreal stratus]

Np!

I should say like i'm saying "this just follows from the definitions" to sort of reassure you nothing magic is happening / that what you did was fine

[alpine hound]

To make things funnier, I was trying to prove that if X_i is a chain of irreducible spaces, then their union is irreducible. And this is where I got stuck at. LOL

[unreal stratus]

Lol fair enough

[prime elbow]

No idea about borel

[prime elbow]

I am not sure about when a is not an isolated point of X and we have to show a is in the boundary of S.

We know a is an isolated point so dist(a, S{a} ) ≠0 and a is isolated point implies dist(a , X\{a} ) = 0.

And X\{a} = S^c union S\{a}.

We need to show dist(a, S^c) = 0.

[prime elbow]

Boundary points of {1/n | n in N } union {0} is same as { 1/n | n in N } union {0}

Right?

[pallid comet]

sorry for interrupting - ive recently encountered something im not sure i understand in my topology textbook:

(7.58, 7.59)

so the thing is

i may be missing the point of 7.58, because from what I'm thinking, this is quite trivial and unhelpful to proving 7.59: f^-1[0,r) is automatically open in A, and so must be the intersection of A and a set U open in X. Thus we could select this U for our U_r, since (the closure of U) intersect A is contained in f^-1[0, r].

and so the specified open sets exist.

however, i can't seem to use this to prove tietze extension theorem, and i've been questioning whether I understood 7.58 correctly at all since i never used the normality lemma.

can anyone help me with this?

(for good measure, here's a pic of 7.56:)

(I know there's a "standard" proof for the Tietze extension theorem - it's directly below the statement in the book - but this one seems just a little too intriguing to miss so.... please help if possible 💜)

[pallid comet]

(nevermind, i got it! thanks anyways!(

[alpine nest]

Glad we could help!

[pallid comet]

(this is not quite math but everytime i post a question on here my mental strain decreases by like 50%, which actually helps a lot with problemsoving, so sometimes i'd just as a question that seemed impossible for me to figure out, only to solve it like 1 hour later)

[odd gulch]

doing this Q rn, what're some motivations for studying this thing

why do we care about compactification

[alpine nest]

Because compact spaces have a lot of very useful properties that other spaces don't, so if we have a non-compact space we sometimes benefit from being able to "make it compact"

(I think the most notable example is the Riemann sphere, which is the one-point compactification of the complex plane)

[odd gulch]

I see thanks

[prime elbow]

In the second part, I think if S is a closed dense set then S must be equal to X because X\S is open and if it is non-empty then S must have non-empty intersection with it, but it is not so X\S is empty.
Right?

[alpine nest]

Yeah, a closed dense set is the whole space, so I find the second part weird.

Maybe they meant open dense?

That would make sense and is a useful observation despite its simplicity.

@prime elbow ping because I forgot to use the reply feature

[prime elbow]

Yes

[prime elbow]

There are countable disjoint open balls in R, right? Because we can make injective mapping (a,b) -> r, where r is the rational number between a<r<b

[alpine nest]

Your statement is ambiguous, but if you mean "a collection of pairwise disjoint open balls in R is at most countable", then yes

And a countably infinite collection of disjoint open balls in R can indeed be constructed

(analogous statements apply for R^n)

[prime elbow]

Yes

[granite slate]

isnt there some nbh of 1 such that the preimage is, say, (0.5, 1.5) and therefore its evenly covered?

[unreal stratus]

Consider how many preimages there are of points near 1

[prime elbow]

I am thinking about something related to the separable part, if metric has uncountable collection of disjoint open sets then it cannot have a countable dense subset, right?

[thorny agate]

Why are $\mathrm{GL}_n(\mathbb{C})$ and $\mathrm{SL}_n(\mathbb{C})$ connected in the Euclidean topology?

[gentle osprey]

Spamakin🎷

[unreal stratus]

There are funny arguments

[plush folio]

It boils down to the fact that C \ {0}, unlike R \ {0}, is connected, right?

[unreal stratus]

For GL_n(C) a fun argument is that the matrix exponential is a surjection exp: M_n(C) -> GL_n(C) ("every matrix has a logarithm")

[prime elbow]

I thought about the continuous image of the connected set

[prime elbow]

What is a matrix exponential?

[unreal stratus]

Exactly the argument here yeah

[unreal stratus]

The map $A \mapsto \sum_{n \ge 0} \frac{A^n}{n!}$

[gentle osprey]

Prismatic Potato

[unreal stratus]

But anyway, a simpler argument I would give is that you can apply elementary row operations "continuously"

As in, you can get a path between any two elementary row operations within GL_n(C)

[prime elbow]

Is it guaranteed that the image in GL_n(C) ?

[unreal stratus]

Yes, by the same arguments that exp(x) is nonzero

i.e. basically you check that exp(A) has inverse exp(-A)

[prime elbow]

Oh it is something about Taylor's expansion of exp(A), right?

[unreal stratus]

ye sure

usually this is how i would define exp anyway

[prime elbow]

But their expansion is an infinite series, right?

[unreal stratus]

ye

[prime elbow]

How does it work in a finite case?

[unreal stratus]

Idk what you mean

[prime elbow]

This summation is infinite?

[unreal stratus]

Yes

[prime elbow]

Oh

[unreal stratus]

But that's fine cause we're in GL_n(C)

which has a topology

(open subset of M_n(C), which we give a topology via an identification with C^(n^2))

[prime elbow]

I don't getting much but okay

Prismatic Potato, I am doing Searcoid right now, what do you think which book should I have to do for metric space? I done Carothers metric part

[unreal stratus]

To be honest I've never heard of either of those books, sorry. I learnt a bit about metric spaces through a uni course, but iirc Munkres has a nice section on them too

I guess it depends on what you want to learn about metric spaces for

[prime elbow]

Okay I will do Munkres

[thorny agate]

oh that is fun (but the elementary row ops is much more intuitive for GL_n(C) lol)

thanks

what about SL_n(C)?

Because images of row ops don't necessarily live in SL_n(C)

[unreal stratus]

I guess there's a continuous surjection from GL_n(C) -> SL_n(C) where you divide the top row by the determinant, say

[thorny agate]

ah

thanks

that's clever

[gentle osprey]

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

[wise mist]

i have proof but are the statements correct at all ? it was fill in the blank!

[iron bolt]

statements 1 and 3 seem good, 2 and 4 less so

[stray stratus]

does anyone understand what this is saying here? what does "so that the embedding restricted to each L_i is an epsilon-imbedding, then there is ..." does that mean each L_i on its own is an epsilon-imendding? or is there some more complicated relationship between them

like is it a disjoint union of links L (links L_i) where each L_i is on its own an epsilon-embedding, then there is a constant a > 0 s.t. N < ...

[granite crane]

its not very common for me to see \sqcup to see as disjoint union of sets. But yeah its a nice notation, can i ask what book are you following

[stray stratus]

It’s a short 3 page paper someone sent me

“How many links can you fit in a box”

[granite crane]

Sounds interesting

[stray stratus]

fr

idk where it goes tho. pst channel felt right but

[granite crane]

whenever it comes to shapes like this i also think of algebraic top

but i am not sure if your question fits there

[chrome plank]

Don't know if it's too basic for this channel, but well, I'll post it regardless

REALLY struggling with this, I have no idea how to proceed

I started by considering a z in the preimage(which will be a point (z1,z2) in R^2) and then, considering its function, m(z), since (1,2) is open, we know there exists some δ>0 such that (m(z)-δ,m(z)+δ)⊆(1,2)

Which implies that the preimage of the aforementioned open ball is also in the preimage of (1,2)

However, I don't know how to proceed any further, no idea on how to show there exists a disk around every point on the preimage

[worldly narwhal]

have you covered how continuous functions and open sets interact?

[chrome plank]

I know continuous functions map open sets to open sets, but I can't assume that the multiplication function is continuous here

I mean

I know it is

But fairly sure the book wants me to show the existence of a disk on the preimage of (1,2)

[worldly narwhal]

is it that you are unsure how to show it is continuous ?

or you want to avoid using that information?

[chrome plank]

I don't know how to show that mapping is continuous, indeed

[chrome plank]

I mean, I know it is, but since I don't know how to prove it(that's the whole thing I'm struggling with), fairly sure I cannot use the fact

[worldly narwhal]

you may have proven some results at some point you could use here

but otherwise epsilon delta for this isnt too bad

also, have you actually found what the pre image is, i forgot to ask tht above?

[chrome plank]

Most of them assume an already continous function

[hoary breach]

use the fact that sequential continuity is equivalent to continuity in metric spaces

[chrome plank]

Since the function takes two variables, most I could do was determine the interval ((1/y,y),(2/y,y)

Which, well, now that I think about it, may be wrong

[chrome plank]

I assume you mean the definition of every sequence of points converging to x converging to f(x)

Don't know how I could use that here, well, first I haven't used that definition at any point so far due to how restrictive the condition of “every sequence“ is

And I have to admit I'm not all that familiar with sequences either

[worldly narwhal]

i mean more that functions in multiple variable constructed in this way, the proofs usually use projections

if you havent then try an epsilon delta argument

[chrome plank]

As in, fixating the y and considering the interval ( 1/y, 2/y)?

[chrome plank]

Well, tried already, but haven't gotten anything

[alpine nest]

Did you sketch that preimage, as the problem suggests?

[worldly narwhal]

what have you tried?

[worldly narwhal]

also this^ ive always found skteching the regions to be helpful

[chrome plank]

If I'm being honest, I thought that by sketching it just meant “finding“ or “bounding“

[alpine nest]

No, it meant literally make a drawing

[chrome plank]

So no, haven't tried graphing it

[alpine nest]

A drawing is not a proof, of course, but it can inspire you as to how the rigorous argument might work

[chrome plank]

First of all what I said below the initial image

[alpine nest]

Also, it sounds like you're doing topology without having done a real analysis course before? How did that happen?

[worldly narwhal]

do you know how epsilon delta proofs go?

[chrome plank]

Self studying, also education works differently on each country

[chrome plank]

Done some, yeah, but not really related to R^2

[alpine nest]

If you're self-studying, then I definitely recommend putting topology on hold and going through a real analysis book (such as Abbott) before

[chrome plank]

So regardless of my self studying, I'd have to go through a course on topology before a course on real analysis

[worldly narwhal]

it works in the exact same

[worldly narwhal]

🤨

[alpine nest]

Sounds like the exact opposite of how it normally goes

[chrome plank]

In about 2 month's time

[chrome plank]

That's how it is around here

[worldly narwhal]

can i ask where?

[chrome plank]

Granted, the subject is called “elementary topology“, but doubt that helps all that much

[chrome plank]

Spain

[worldly narwhal]

weird, im also spanish

ive not heard of it going around that way before

[chrome plank]

Depends of the university, but I assure you it goes that way

Well, I mean

They usually have one subject related to “real analysis“ but it's mostly a slightly more rigorous calculus

[alpine nest]

Yes, that's exactly what you should do before topology

[chrome plank]

As one would expect of a first year undergrad course

[worldly narwhal]

perhaps, ive lived mainly in the UK, so only familiar with some specific courses

[chrome plank]

That, I'm taking, though maybe not as rigorous as I'd like it to

[alpine nest]

I'm not talking measure theory, I'm talking "real analysis" as in "rigorous approach to limit and continuity of functions of one real variable"

[chrome plank]

Yeah, I've taken that

I know the definitions and I've used em

[alpine nest]

Hm, then you should be familiar with epsilon-delta, and with sequences

Maybe a refresher is in order

[chrome plank]

The book also mentioned them and used ε-δ proofs

Several times

[alpine nest]

Either way, start by making that sketch of the preimage.

And show that it's open directly from definition

[worldly narwhal]

these are from my notes, the definitions are basically the same as in the case of R

[chrome plank]

I mean, I know what a sequence is, just an indexed array of points defined by a function f:N->A

[hoary breach]

this is not exactly what the book wants, recall the definition of an open set

[chrome plank]

The thing is that I haven't used the sequential definition (every converging sequence A to xn converges to f(xn)) due to how hard it is to find all the converging sequences

[alpine nest]

The idea is never to "find all the converging sequences", the idea is to make an argument general enough to apply to every possible converging sequence

There's typically uncountably many converging sequences, so going through all of them would indeed be impossible

[chrome plank]

Alright, for the usual topology:
Given a set A⊆R, A is open iff for all z∈A there exists some δ>0 such that B(δ) centered in z is contained in the set A

[alpine nest]

Indeed

So you need to show that the set {(x,y) : 1< xy <2} satisfies that property

[chrome plank]

Hm, well, then I guess it's just my limited knowledge of sequences, that's the only part of basic real analysis I didn't look into myself, and we haven't gotten there yet so, yeah

[chrome plank]

Well, I'll give it another shot for now, I'll ask for help again if I get stuck

[stray stratus]

On frog

Wait

This is geo top

Bruhh it literally says knot theory in the description

[granite crane]

[prime elbow]

How can I prove that total boundness implies every sequence has a Cauchy subsequence?

How can I go for ii) to i), if for each eps>0 S can be covered by the finite collection of open Balls of X of radius eps, then how do I find a finite collection of open balls in S of radius eps which covered S ?

[tender halo]

take balls of radius eps/2

[prime elbow]

I got it

[prime elbow]

Is there a subset S of R such that S and R\S both are discrete space as subspace of usual topology on R?

[tender halo]

no

out of S and R \ S take the one which is uncountable

it has too many points to be discrete because its a metric space that is second countable

therefore any discrete subset will be countable

[tender halo]

that is in fact equivalent to second coutability in metric spaces

[prime elbow]

Subspace of second countable set is the second countable, right?

[tender halo]

yep

[prime elbow]

I got it thank you

[sonic crane]

what is the wedge product? Just the subspace of X x Y of X x {y0} U {x0} x Y?

but you can also realize this by an attaching map?

[sonic crane]

whats the difference between the wedge and smash product

[willow sandal]

google it

[sonic crane]

Obviously I did

I dont really get the difference which is why i'm asking here

[willow sandal]

what dont you get

[opaque scroll]

They're the same thing. Though wedge product is also used in different contexts (for different things)

[sonic crane]

I think this is why im confused

they are actually the same thing ??

you just identify some point of X,Y (x0,y0) and then the wedge /smash product is the quotient space X [_] Y / ~ with ~ being defined as (x0, 0) ~ (y0, 1), and every other point is its own equivalence class?

So you are basically just gluing X and Y together at the point (x0,y0)?

idk I tried to write the symbol [_] for disjoint union lol

[opaque scroll]

That would be the wedge sum

[sonic crane]

Yeah

But smash product is the same thing? I saw smash product is the quotient X x Y by the wedge sum X V Y but i dont know how to think about this

[novel acorn]

It's kind of tricky to think about
But try for example to visualize what the smash product of two S^1's looks like

[opaque scroll]

The smash product and wedge sum are different.

One is like a product the other is like a sum.

For suitably nice spaces, the smash product is adjoint to the mapping space, so that could be a good way to think about it.

Otherwise you can also just draw the XxY and imagine contracting XvY to a point

[unreal stratus]

I agree in some way but i think it's probably best to think of smash product as like a tensor product (in particular because of that adjunction you mentioned, and because it isn't a categorical product)

which is probably what you mean but i thought i would mention this for others' sake lol

[unreal stratus]

Wait wdym

I have never heard somebody say wedge product for smash product

wedge product to me is just a thing for like the exterior algebra etc

maybe some people do say this though but it must be relatively rare

[opaque scroll]

Well, Wikipedia has a redirect for it.

And it's a product denoted with the wedge symbol

[unreal stratus]

sure

[opaque scroll]

And it's like the product version of wedge sum. I think wedge product is a reasonable name

[unreal stratus]

Hm fair i just have never heard anyone say that lol

[chrome plank]

I'm finding this problem quite strangely worded

I mean, I'm not sure it's always true, for example:

Consider the set: {1,2}, let us consider B the collection of subsets only containing itself, so B={{1,2}}, it is satisfied that 1): all elements of the set are within at least one of the subsets of B(the only one) and 2): it is closed under finite intersections, since any set intersected with itself is itself, which is within B

[alpine nest]

The empty set is the union of an empty collection of sets from B, and together with {1,2} it does form a topology.

[chrome plank]

Then, the problem says that the set of all unions, which I'll denote U, is a topology, however, for this case:
U={{1,2}}, once again, and since ∅ is not in U, it's not a topology

[alpine nest]

I agree they might have mentioned the empty set more explicitly because it's a common tripping point in these definitions

[chrome plank]

Isn't the problem saying we can choose whichever collection of subsets we want?

Cuz the empty set is precisely what was bugging me here

[alpine nest]

We take all possible subsets of B (including the emptyset), and take unions of every such subset.

The collection of such unions forms a topology.

In symbols, $\mathcal{T} = \left{\bigcup_{A\in \mathcal{A}} A : \mathcal{A} \subset B\right}$

[chrome plank]

Doesn't it say B is A collection of subsets?

[gentle osprey]

Outsider

[alpine nest]

It is

And you produce a new collection based on B

And that new collection will be a topology

[chrome plank]

I mean, if B is the power set of T, then yeah

[alpine nest]

No, the whole point is that B can be much smaller than that

[chrome plank]

The union of two nonempty sets can't be the empty set, though

Just choosing a B that doesn't include the empty set breaks all of it, that's what I'm having trouble getting

[alpine nest]

Yeah, which is why I agree the phrasing of 3.11 isn't great; the point is that "unions of sets in B" means "anything you can get if you take any amount of sets that are an in B and form their union", and that includes the case of "take no sets at all" (in which case the union will be an empty set as well)

[chrome plank]

Ohhh, it meant union of subsets, geez, since B was already a collection of sets, I assumed it meant union of elements

[alpine nest]

If you prefer you might also claim that 3.11 is wrong and should say "the collection of all unions of sets in B, together with the empty set, forms a topology on T", or you can strengthen the assumption by requesting that the emptyset be an element of B

But "the union of an empty collection is the empty set" is a fairly standard convention

[chrome plank]

That's what I was thinking of doing, really

[alpine nest]

(and the intersection of an empty collection is, again mostly by convention and for convenience, the whole space)

[chrome plank]

Was about to ask why that was

But guess that makes sense, due to the empty set being quite literally an assumption for convenience's sake

[alpine nest]

Well, technically if $\mathcal{A}$ is a collection of subsets of a space $X$, you can define their intersection as $\bigcap_{\mathcal{A}} = {x\in X: \forall_{A\in\mathcal{A}} x\in A}$. If $\mathcal{A} = \emptyset$, then the condition is vacuously true for every $x\in X$, and thus the intersection will be equal to $X$.

[gentle osprey]

Outsider

[alpine nest]

And it so happens that this interpretation makes sense, in that various identities for unions/intersection will remain true.

[chrome plank]

Hm, alright

[alpine nest]

You can also get this conclusion using De Morgan's laws

[chrome plank]

Well, I think I'll first provide the counterexample, then add the restriction of the empty set being in B and then prove from there

[chrome plank]

Alright

Thanks btw, I found this quite confusing

[alpine nest]

Hm, or maybe not, but either way, the union of an empty family is the emptyset, and the intersection of an empty family is the whole space.

[alpine hound]

If X is a set with the cofinite topology, and n is a natural number, does that mean every CLOSED set in the product topology of X^n has to be finite?

[red yoke]

No

X × finite set is closed

[alpine hound]

Ohh ok

So every finite set is closed?

[red yoke]

Yea

[alpine hound]

And how can I prove that every finite set can be written as the finite union of closed sets?

[red yoke]

?

[alpine hound]

You said that every finite set is closed.
But for a set to be closed it has to be written as a finite union of some closed sets

[red yoke]

Well yea it is the union of itself

You mean closed sets of the form [closed set] × [closed set]?

[alpine hound]

Yes

Or X x [closed set]

[alpine hound]

That is better

[red yoke]

Every singleton is of this form

[alpine hound]

And there are finite

[red yoke]

So finite sets are finite unions of this

[alpine hound]

I am sorry fr

Thanks

[red yoke]

Math be like dat sometimes

[hardy marsh]

https://media.discordapp.net/attachments/867389710132707378/1318169630015225917/image0.jpg?ex=676158a2&is=67600722&hm=e5d7cffb53045b4c60a1261fb2602b42e2f749399b28eaa672bbefc04f4b412a&

I just feel really dumb but i don't see this

I understand that f is a identification map from the conditions

But by separated they mean it's not connected right

[mighty hull]

Separated means Hausdorff aka T2

Separated as in "separation axiom"

[hardy marsh]

Oh

Yeah I mean the connected thing wouldn't even be true just take a hausdorff connected space and identity map

[mighty hull]

It's strange not to say "Hausdorff" or T2 though. I only know what it's asking offhand because a space X is Hausdorff iff X^2 is closed.

[hardy marsh]

Yeah I am aware of that fact

[mighty hull]

This is basically the same thing, yeah

[tender halo]

do you even need openness though, isnt being quotient already enough

[mighty hull]

They're probably making it easier for the purpose of the exercise

I actually don't know that it makes it easier, looking at it. Maybe the textbook just hasn't defined quotient map generically yet.

[hardy marsh]

It's dieck

This is the intro chapter

[mighty hull]

Ah then probably not

[hardy marsh]

He has defined quotient map

[mighty hull]

This will essentially be a proof that the quotient of a Hausdorff space is Hausdorff.

[hardy marsh]

Yeah ig

But I was really for like 4 fcking hours thinking why would Y be disconnected lol

Cause most books i have read before mean not connected by separated

Thanks a lot

[mighty hull]

It's possible this is true, but in my experience decent textbooks are careful to use terminology consistently, even if their choice is misleading or distracting for extra-textual reasons.

I think the instinct to develop is to always keep "I'm misinterpreting the statement" as a low-ranking hypothesis for why you're not making progress. You want to double check terminology often.

[mighty hull]

To wit, Proposition 1.3.3 contains the third instance of the word "separated" in the text. The first two are here:

An expensive way to learn what kind of textbook this is, but hopefully a price you only have to pay once!

(Everyone has done some version of this more times than they can count, FWIW.)

[hardy marsh]

Oh i usually forget terminologies sadly

I should write them down somewhere og

Ig

But like i have done point set before and didn't bother

But ig that's the way to go

[mighty hull]

All the more reason to double check every time, just to be sure. Especially for core definitional proofs, I make a kind of mathematical mise en place every time.

Whatever terms are mentioned, I get those definitions out and put them literally in front of me.

[hardy marsh]

Yeah I can agree it's good practice

[mighty hull]

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

[hardy marsh]

"let him cook"

[mighty hull]

Haha yeahhh

[prime elbow]

Find a surjective open mapping that is not a closed mapping. I need in metric space, any hint? I know X and Y cannot be discrete metric space

[hardy marsh]

How about (x,1/x) x in R+ to R

Via projection

Both of them are metric spaces too

[alpine nest]

That one is not surjective, but it's a good starting point

[hardy marsh]

Oh right they need surjective

So yeah look at projections ig

[prime elbow]

but then how to show it is not closed, R^2 -> R such that (x,y) -> x, it is open mapping but i saw on mse they say about image of hyperbola xy = 1 is not closed

[opaque scroll]

Yeah, so there you go. It's open but not closed

The hyperbola is closed, but it's image is not

[hardy marsh]

Can you find other exotic shapes whose image is not closed

[prime elbow]

here i am not sure image of hyperbola

[hardy marsh]

Mse doesn't have to be all end all

[prime elbow]

how to visulize this?

hyperbola is closed

[hardy marsh]

When I thought of this

[prime elbow]

do i need to sequence in image of hyperbola which limit is not in image

[hardy marsh]

One can see that you can get around with some sort of asymptotic kind of thing

[prime elbow]

1/n?

[opaque scroll]

Which x are such that xy=1 has a solution?

[prime elbow]

yes i think this one work

[prime elbow]

which is non-zero

[opaque scroll]

Right, so the image will be R\{0}

[hardy marsh]

You can also look at 1/x 's graph too

[prime elbow]

i think sequence 1/n in image gives contradiction

[prime elbow]

yes got it

[hardy marsh]

Nah

[prime elbow]

why?

[opaque scroll]

That's literally the same thing

[hardy marsh]

Yeah

Just saying

Or any other weird curve you can draw that has an asymptote

[prime elbow]

i don't know the theory of asymptote

[hardy marsh]

My god I just mean it should go to a line but never touch it

Nothing to do with theory of asymptotes

[prime elbow]

okay

[opaque scroll]

How about, pick an enumeration of the rational numbers r1, r2, ... and consider the set
{(rn, n) : n in N}

[hardy marsh]

Yeah this is cool too

[prime elbow]

you mean this set is closed but its image is not, yes image is Q in R not closed

oh there is no convergent sequence in this set, right?

i mean non-ultimate constant sequence

[alpine nest]

Hm, will the projection be an open map in this case? Singletons are open in this space, but the image of a singleton (open in this space) will also be a singleton (not open in R)

[opaque scroll]

I'm just talking about closed sets in R^2. The mapping is still R^2 -> R

[alpine nest]

Ah, yeah, I get you

[sharp aurora]

are discrete topologies sigma algebras?

nvm

apparently they are

[tender halo]

that is not true

take a Hausdorff space that is not normal and two closed sets that are cant be separated with open sets, take the quotient that sends each of those sets to a point, then the quotient won't be Hausdorff

[unreal stratus]

Note discrete topology is just the whole power set which will be closed under everything you need for a sigma algebra by definition. so really this has little to do w topology

[unreal stratus]

Yeah in fact pretty sure if you change "continuous open surjection" with "quotient map" then this is still an iff with Y being separated

[proper salmon]

hello . i want to understand the mesh topology that is on 3d modeling but with the mathmatic understanding so if u can tell what i need to understand first like precalculus

[opaque zodiac]

Mesh topology isn't terribly directly related to point-set topology

There are things you can say but mostly it's just a matter of them both being about 'shape' in some abstract sense

[ebon flower]

you'd want graph theory instead, right?

[warm kettle]

Pin

[prime elbow]

[wise mist]

not compact

unboundedness

[opaque zodiac]

Yeah basically

[wise mist]

i just found this a few days ago

[wise mist]

in anyways, to appreciate Furstenberg topology, they wants me to look at the infinitudes of primes and profinite integers

can i get good source(s) to go through all that individually or in a package

[prime elbow]

There are many sources on the internet

Maybe this one is good

[wise mist]

if not for subasis i would have never known that trivial intersection is a thing

[wise mist]

well what is the key distinction bw subasis and a cover on a set ?

after the first lemma what else is there to be said

[opaque scroll]

Just contextual. Like a subbasis is something you intend to use to create a topology

[wise mist]

right!

from what i have got when we are talking about subbasis, topology can or cannot be mentioned right

and the wording gets a lil different in two cases, idk if im being dumb here

[wise mist]

$S = {[a, \infty) : a \in \mathbb{R}} \cup {(-\infty, b] : b \in \mathbb{R}}$

[gentle osprey]

yeshua

[wise mist]

so this is coarser than usual right ?

if yes then i think i am kinda getting the feel of subbasis now

any cover would atleast suffice to generate the coarsest topology on the given set, is that a valid corollary ?

[opaque scroll]

Notice that by intersecting you can obtain the set {a}, so this will in fact give the discrete topology which is the finest topology there is.

You could drop the middle union though, just do
S = {[a, infinity) : a in R}

[wise mist]

right, i didnt not notice we can get to the singleton, but wait its finite intersection tho

is it not generating [a,b]

[opaque scroll]

Will [a, infinity) intersect (-infinity, a] is {a}

Same as [a, b] for a=b I suppose

[wise mist]

right!

grrrrrr, all my construction gone in vain

lets say we have {T1, T2 .... Tk} on X and they are arranged to be from finest to coarsest without degeneracy,

is there a valid statement we can say that if my subbasis can generate Ti th topology then it will generate Ti to Tk

[opaque scroll]

So a subbasis only generates one topology.

But if T2 is coarser than T1 and you have a subbasis for T1, then every open set in T2 can be written as a union of intersections of things from your subbasis. But this is just because T2 is a subset of T1. Any open set in T2 is also an open set in T1. That's just what coarser/finer means

[wise mist]

oh right, my dumb brain clicked finally then

[exotic hemlock]

Could someone help me with this, I feel like I have some ideas but im not quite there.

[gentle osprey]

somethingwrong

[alpine crown]

Could someone explain to me the definition of continuity in topology

Idk if it's uniform continuity or continuity at a interval whichever has that v set

[iron bolt]

"v set" isn't really standard terminology, just a variable name, so it's not quite clear to me what you mean

continuous just means that the preimage of every open set is open

[marsh goblet]

we intuitively have a very good understanding of what continuity "should" mean for functions R -> R

now the question is how do we generalise that defn?

are you aware of like metric spaces? well if we have a metric space then an obvious generalisation would be the same one as for continuity R -> R but we replace | . | with "distance"

but what happens if we deal with weirder spaces where we don't have this notion of distances?

[marsh goblet]

you can look at your previous defn and see that actually, the pre-image of open sets is open is equivalent to being continuous

[marsh goblet]

so the idea is that for generic topologies, we use this as the definition of what it means to be continuous

[alpine crown]

This is what I'm talking about

[marsh goblet]

this is just normal continuity (not uniform)

[iron bolt]

yup, it's one of many equivalent phrasings

[alpine crown]

My prof labeled it as the topological definition

[marsh goblet]

it's the topological defn because it doesn't rely on the idea of what 0 or 1 or | | or < or etc.

[iron bolt]

ah, you're probably doing something related to metric spaces or real analysis right now, right? and just briefly touching on topology with this?

[alpine crown]

It's so confusing to me 😔

[marsh goblet]

for a generic topological space, we don't have ideas of "size" or "distance" etc. but open sets and closed sets are defined

[alpine crown]

Yes

[marsh goblet]

ah right

[alpine crown]

Real analysis 1

[iron bolt]

the idea is basically that many properties related to metric spaces can be characterised just in terms of open sets. for example, continuity at a point is equivalent to what you just posted

[marsh goblet]

i hoped i provided some light motivation why we should care about this definition, if ur just starting out analysis you probably wanna just stick with the normal "if x gets close to y, then f(x) gets close to f(y)" but keep in the back of your mind that we have an equivalent definition given in terms of open and closed intervals

when you start doing metric spaces, you should be thinking about open/closed sets more to then generalise to topology

[alpine crown]

No I understand why it's useful, I'm at the end of my course right now and reviewing

[iron bolt]

so you could do a lot just by knowing what the open sets are - topology does that, by forgetting about metrics and instead defining a topological space to be a set together with a well-behaved set of sets that will be called open

[alpine crown]

You provided good insight

I understand pre images

Q

I don't understand what v

[iron bolt]

it says right in your screenshot: a function is continuous at 0 iff for every open set U containing f(0), there is an open set V such that...

so for U = (-1,1), there should be an open set V containing 0 such that f(V) ⊆ (-1,1)

for U = (-1/2,1/2) there should be such a V with f(V) ⊆ (-1/2,1/2)

it asks that for every U, there exists a V

[heady skiff]

sanity check: S^0 is not connected right? (since it can be expressed as {1} U {-1}, which are both open in the subspace topology on S^0)

[wise mist]

i was thinking of nZ but the 3rd condition fails

any hint ?

nevermind it looks like nZ satisfies the conditions

[quick crane]

correct

[wise mist]

if we inherit some nice subspace topology, a cont non constant surjective map can be constructed on it to {-1,1}

i hope !

[opaque scroll]

So if you split Z into A and Z\A, then what you have is two countably infinite sets.

Then I guess it's just a question of which countable Hausdorff spaces you know

[novel plank]

Let X,Y be compact T2 spaces, and f:X -> Y be continuous. I want to show that f is surjective. Is there any result that you know of about continuous functions on compact spaces that may help with this?

[novel acorn]

this doesn't sound true

are there no other conditions?

[novel plank]

I mean, In what I'm doing the sets X and Y are not any two arbitrary compact spaces, but it's a bit hard to define them here 😅 but do you think that compactness is not enough to prove this then?

[novel acorn]

nope

take X = Y = S^1 and f maps all of X to a single point

[opaque scroll]

The image is closed, so maybe you can show it's dense...

[novel plank]

Do you know of any properties that would imply this (maybe my spaces have them)

[red yoke]

Epimorphisms of CH spaces are surjective

[novel acorn]

I feel like you need more conditions on the map instead of on the spaces

[red yoke]

Surely state the whole question

[novel plank]

yeah, that's probably true

well, thank you for you help anyway 🙏

[silver ridge]

I'm struggling in particular on (a), (b) and (f). I'm not even sure where to begin with to be honest. I'd appreciate any type of hint because I am very stuck.

[opaque scroll]

I have an example for (f), but I'm not sure how to make it into a good hint.

Something to think about might be how you can make somewhat simple discontinuous functions. The easiest might be to take some connected space and "cut" it into two pieces.

And yeah I guess once you have (f) you can just compose with some map that is open not closed or closed not open to get (a) and (b)

[silver ridge]

Thanks jagr, I'll digest this later because my brain has just crashed and lost all will to think.

[prime elbow]

I am not sure here, if X is R and d is usual metric on R and y let any symbol say ∆, then how can I assign the number d(1, ∆ ) ?

[opaque scroll]

So you can just assign d(1, Delta) to be anything positive completely arbitrary. After that you just have to make sure d satisfies the triangle inequality.

So for any x you should have d(1, x) <= d(1, Delta) + d(x, Delta)

I guess the easiest would just be to define d(x, Delta) = d(1, x) + d(1, Delta)

[warm kettle]

@opaque scroll are you a scholar

[opaque scroll]

I guess so

Or what do you mean?

[warm kettle]

Oh, I was just curious if you worked at a uni. You seem to be very knowledgeable in algebra

[opaque scroll]

I'm a grad student yeah

[sonic crane]

You are working on your Phd?

[opaque scroll]

Corectumondo

[sonic crane]

Nice

[opaque scroll]

Will graduate in the summer. Then we'll see where the wind takes me

[sonic crane]

Oh exciting! Are you considering leaving academia?

[opaque scroll]

Yeah, I mean I'll take the job I can get I guess

But if you know of any postdocs in rep theory adjacent fields in Europe then I'm all ears

[sonic crane]

Hahaha

i dont unfortunately

[wise mist]

https://cdn.discordapp.com/attachments/1054780485408133170/1319061517420728402/attachment.gif

[warm kettle]

There's not that many profs that work in that field in Europe? Or do they just not want to recruit postdoc students?

[opaque scroll]

Nah, it's fine. Applying for jobs is just hard in general

[sonic crane]

Frrrr

[warm kettle]

I see, fair enough

[prime elbow]

Okay thank you

[prime elbow]

If it is possible, take x ≠ y, then since this is discrete metric therefore |x-y| = 1 but then use the property of norm, for r > 0 which is not equal to 1, |rx - ry | = r ≠ 1 contradicts that it is discrete space.

Is it correct?

[slender glen]

yeah that looks correct

[prime elbow]

Any hint?

[slender glen]

Other than the ordinary metric space properties, what properties does a metric induced by a norm have

In particular, stuff like if d is a metric induced by a norm what can we say about d(cx,cy) for example

Or d(x+b,y+b)

[prime elbow]

I see

But it is the necessary condition right, not sure about sufficient condition

[slender glen]

The opposite way; take a metric d and define ||x|| = d(x,0).

Lmk what properties we need d to obey for ||x|| to be a norm

[prime elbow]

d(x+y, 0) ≤ d(x, 0) + d(y,0) and d(cx, 0) = |c| d(x, 0)

So this is equivalent to d(cx, cy) = |c|d(x,y) and d(x+b, y+b) = d(x,y)

[slender glen]

Yeah, that shows it's necessary and sufficient

[prime elbow]

Yes but I am confused here you mean we take the norm |x -y | = d(x,y), right?

Because if I have | x | = d(x,0) then I am trying to show | x- y |= d(x,y)

[slender glen]

Mmhmm

[prime elbow]

If I define |x| = d(x,0) and with d(cx,0) = |c|d(x,0) and d(x+y,0) ≤ d(x,0) + d(y,0), then can I prove | x-y | = d(x,y)

[prime elbow]

No because if there exists a metric space X such that there are exactly three clopen sets then there exists non-trivial proper subset U such that it is clopen but it implies X\U is non-trivial proper subset which is clopen, contradicts that there are exactly three clopen sets, right?

[wise mist]

{\emptyset, {x}, X} ?

[prime elbow]

It is not metric space, unless X is singleton set

X is finite then it is not

I am not sure about infinite but yes it has no Hausdorff property

So it cannot be possible

[tender halo]

i mean clopen sets exist in pairs

so you cant have an odd number of them

[rancid umbra]

(0,1) U (1,2) gives one with three non-empty clopens fwiw

[prime elbow]

Yes

[prime elbow]

Sorry ?

[rancid umbra]

(0,1), (1,2), and (0,1) U (1,2) are the only non-empty clopens in this space

[prime elbow]

Empty set?

Oh I got it

You mean non-empty

[rancid umbra]

is that not what i said?

[prime elbow]

Yes but in question they didn't mention non-empty

But yes thanks

[prime elbow]

Is that argument correct?

[rancid umbra]

X\U could be one of the given clopens. and you have to rule this out.

[prime elbow]

It cannot because U is non-trivial proper subset

Non-trivial means non-empty

[rancid umbra]

yea, then it works. i would be more explicit about the fact that the other two clopens must be X and the empty set in this case

[prime elbow]

Yes

[wise mist]

it sort of implies the nontrivial spaces being not connected ig

is clopen and regular open/close equivalent ?

yes, i hope so

[prime elbow]

No

Take discrete metric space

[prime elbow]

What is regular open/close ?

[tender halo]

no?

a ball in Rn is regular closed but hopefully not clopen

[prime elbow]

Hi, bussy Beaver have you recently changed your profile picture?

[wise mist]

i meant to say regular open and closed together

[tender halo]

I have yes

[tender halo]

oh then sure

[wise mist]

\biglist\3\15

[wise mist]

no and the containment is easy to see

but when does the equality hold

for regular open/close this is just A

but that is very strong, is there anything weaker that still holds this

[warm kettle]

The only non empty set that’s simultaneously open and closed is the whole space

[alpine nest]

I mean, depends

[warm kettle]

Oh, discrete topology…

[alpine nest]

Any topology that isn't connected, really

Let X be the union of (-2,-1) and [2,3] (with the standard topology), then (-2,-1) is a clopen subset, and so it [2,3]