point-set-topology

[uneven bronze]

why is $d(\lim_{n\to\infty}s_n,\lim_{n\to\infty} t_n)=\lim_{n\to\infty} d(s_n,t_n)$ equivalent to $d(f(s),f(t)) = \rho(s,t)$?

[gentle osprey]

psie

[tiny obsidian]

the RHS is just the definition of rho, the LHS is just rewriting f(s) and f(t) as what you've defined it to be

[uneven bronze]

but the LHS is the metric on Y, whereas the right hand side the metric on X, or?

[tiny obsidian]

oh

yeah good point

[uneven bronze]

I'm not sure I'm right though, I might be wrong 🙂

I guess it is the metric on Y, but when we restrict to X, it becomes the metric on X

[tiny obsidian]

yeah

By d I mean the relevant metric

I was typing out that $d_Y$ restricted to $X$ is $d$, so we can just use the same symbol for both metrics and might as well take the easier $d$

[gentle osprey]

Edward II

[uneven bronze]

indeed 👍

[tiny obsidian]

anyway you know something about certain distances from the limits of s and t (I'm going to call them S and T from now)

because they're limits

[uneven bronze]

ok, showing that f is an isometry looks like to be equivalent to the continuity of the metric, or?

[tiny obsidian]

yes

[grave solstice]

I don't know a lot about simple connectedness. Is it true that an open subset U of R^2 is not simply connected if and only if you can find a point a not in U and r>0 such that the punctured disk centered at a of radius r is contained in U?

[unreal stratus]

No, for example consider either 1) a disjoint union of two discs - this is pedantic but simple connectedness has path connectedness as part of the definition and 2) an annulus e.g. { z : 1 < |z| < 2}

The converse is true though

Beyond that you can also think of drawing a random squiggle in R^2 (and then ensuring it is open by thickening it up); for example an open figure of 8

[grave solstice]

Right. I'm just trying to see if "simply connected" is "equivalent" to "it doesn't have holes". So I guess I can rephrase the question as: Let U subset R^2 be open and connected, is it true that U is not simply connected if and only if you can find a point a not in U such that every ray emanating from a intersects U?

[dire dove]

This is false

The point has the property you want but the set is simply connected

In R^2 iirc a nice characterization for compact simply connected sets is the complement being connected

[grave solstice]

oh yeah, that was dumb. I'm mainly wondering about the other direction tho

[unreal stratus]

Yeah

Nice

So suppose that U is simply connected, then for every point not in U there's a ray not intersecting U, you mean?

Hm

[dire dove]

No, the other one

Which has (part of) jordan's curve theorem as corollary

Ok no not really

[dire dove]

I thought it would imply the "interior" of a jordan curve being nonempty but the boundary of this is a counterexample

[dire dove]

This should be "bounded open" instead of "compact"

[grave solstice]

nice

[woeful coral]

Are there fixed point theorems in S^n with points removed

[umbral panther]

It’s tough to have fixed point theorems for non compact spaces because you can push all the interesting behavior to infinity. For example, if you subtract a point from S^1, you get R, which has the translation x |-> x+1, which has no fixed points

[red yoke]

S^n with points removed is just R^n with points removed

[uneven bronze]

I am working the following problem: \

Problem: Prove that the set of isolated points of a countable complete metric space $X$ forms a dense subset of $X$. \

To solve this, I put $U_x=X\setminus {x}$ for $x$ a non-isolated point, which is open and dense in $X$. By Baire category theorem, the intersection of the $U_x$ are dense in $X$, and this intersection are the isolated points of $X$. \

My question; where in my solution did I use the fact that $X$ is countable? I don't see where.

[gentle osprey]

psie

[tender halo]

the amount of Ux's is countable

[uneven bronze]

you're right 🙂 thanks!

[prime elbow]

If E is bounded and closed in R^k then E is compact, right? Because if E is bounded then we can use the general version of Bolzano Weierstrass theorem on E, so for any sequence in E there exists subsequence which is convergent, now E is closed so limit is in E.

[gritty widget]

What is the version of Bolzano Weierstrass you know of?

[prime elbow]

If sequence in A\subset R^k is bounded then there exists subsequence such that it is convergent in A\subset R^K

[gritty widget]

That's not true: let A= (0,1) and a_n = 1/n

Then a_n is bounded but no subsequence is convergent in (0,1)

[prime elbow]

I meant in Bolzano Weierstrass theorem replace R by R^k

[gritty widget]

R is R^1

[prime elbow]

Limit not necessary in A it can be in R^k

[gritty widget]

I think this works then

Erm

[prime elbow]

Okay thank

[gritty widget]

Technically that only says that E is sequentially compact

[prime elbow]

Yes

I don't know why Rudin not introduce sequentially compact

[gritty widget]

Probably because he only works with metric spaces

And it's equivalent for those

If you mean Principles of Mathematical Analysis, that is

[prime elbow]

Yes

If every infinite subset of E has a limit point in E then E is sequentially compact, right?

Because if I take any sequence in E then if the set { x_n | n in N } is finite we can make a subsequence which is convergent.

And if the set is infinite it has a limit point then we have a subsequence which convergent in E

[gritty widget]

Yes, I believe Rudin has a theorem precisely about that?

[prime elbow]

It is the theorem itself

[uneven bronze]

I am trying to prove that the Cantor set is uncountable, using a) that the endpoints make up a dense subset of the Cantor set, b) that it has no isolated points and c) the following exercise: \

Exercise: Prove that the set of isolated points of a countable complete metric space $X$ forms a dense subset of $X$. \

I have solved this exercise, but I don't see the connection between proving the uncountability of the Cantor set using this exercise. Is this clear to someone?

[gentle osprey]

psie

[uneven bronze]

This procedure, to use the exercise, is given as a hint in my book. I thought perhaps they meant some contrapositive of the statement in the exercise, but I can not work out the logic; it seems to be saying if X is a countable complete metric space, then the set of isolated points form a dense subset of X. I am not sure what the contrapositive of this is.

Maybe one could argue by contradiction somehow; suppose the Cantor set C is countable. Since it is complete, the set of isolated points should form a dense subset of C. But C has no isolated points. So the closure of the empty set is the whole space, which is empty. A contradiction I guess?

[gritty widget]

I believe this works

[potent prism]

Hi guys, I have recently been reading about separation axioms. I understand there are different conditions of increasing strength, leading to the T0, T1, T2, etc, classes. From what I gather, for finite set topologies, all of them will be either non-T0, T0, or T1 (the discrete topology), but basically, separation axioms dont make much sense in this context. However, the different topologies of finite set points do not all seem to be "structurally" equivalent, e.g., for N=3, one could argue that the topology {(), (0), (0,1), (0,1,2)} is more complex than the topology {(), (0), (0,1,2)} . Is there any described measurement of this different complexity of finite set topologies?

[gentle girder]

by “more complex” do you just mean “superset of”? people sometimes use the words “coarser” and “finer”

which refer to the subset partial order of topologies on a set

[potent prism]

Hmmm, well, im not sure about that. For example, see the topology {(), (0), (0,1), (0,1,2)} . In order to create this topology, you must take 2 decisions: first which length-1 subset to included, and then which length-2 subset including the previous length-1 subset to include. A topology of same size, {(), (0), (1,2), (0,1,2)} , only requires 1 decision, since the topologies of this homeomorphism class all include 1 subset of length 1, and then the subset of length 2 that does not include it

It becomes more complex with increasing N

By that concept of complexity, also, the discrete topology would be of the same (and lowest) complexity as the indiscrete topology, as it requires 0 decisions to generate it

So that seems to be different to the partial order thing

In fact, so far the best definition I have been able to think of for the concept of complexity I have in mind is how many decisions you must take to make that topology, or equivalently, how many member topologies its homeomorphism class has

Im looking for some better definition / measurement that can be generalized to different values of N

[gentle girder]

it’s 6 am so i’m just gonna give my shot in the dark and hope it sticks, but maybe someone else has an idea, but you can take, for each T a topology on X, the size of Aut(X, T), since a member of Aut(X,T) represents a different arrangement of elements where the same topology fits, and work with that somehow.

[potent prism]

Sorry for my ignorance, but what is Aut(X,T)

[gentle girder]

it just consists of maps f:X->X where f is a homeomorphism

[potent prism]

So you mean, to group together the topologies by the ones that have equivalent structure but putting the different elements of the original set in different places?

[quartz horizon]

Yep

[potent prism]

if so indeed, that is the lines Im trying to go along! by doing that I could get a metric of how complex a homeomorphism class of topologies is, which is through the number of topologies that belong to it

But this is not ideal, because that metric is dependent on the value of N

E.g, for N=4, the largest classes have 24 members, but for N=3, 6, in general N!. So it is not generalizable across values of N, unlike the separation axioms

however it partially works, as it gives a value of 1 to both the discrete and indiscrete topologies

also doing that led me to getting this graph for the homeomorphism classes of topologies, for N=4:

it seems to have interesting structure, so im trying to explore it in more detail

im also surprised by the very little amount of literature (compared to other topics) that there is about finite set topology :S

[gritty widget]

What do the edges and their colours represent?

[potent prism]

a metric of the distance between 2 homeomorphism classes of topologies

we can see multiple interesting properties but one of them is that the connection between the discrete topology (the top node) and the closest one to it, is always the longest edge in the graph, and there is no other of this length. I verified this for N=3 also

unluckily cannot verify for N=5 since generating all the topologies by brute force takes unreasonable time

also I generated that graph by taking all the minimum spanning trees from a complete graph that contained an edge between every possible pairs of homeomorphism class of topologies. so originally there was an edge between any 2 topologies. But that was not very informative. so this one shows in a way the shortest way to be able to go to all topologies

I also further pruned it to leave only the shortest paths between indiscrete and discrete topologies:

which revealed interesting properties. but i have no maths background so i struggle to read the literature some times. because of notation

[prime elbow]

If I take (0, (n-1)/n), n in N and n≥2 then set of all such interval are open covering of (0,1) but no finite subcover covers (0,1), right?

[plush folio]

yep

[prime elbow]

Okay thank you

[prime elbow]

They take E \subset R^k, but in general is it true?

[gritty widget]

What do you mean by in general?

[alpine nest]

Either way, in a general metric space a closed bounded set is not guaranteed to be compact.

However, in a complete metric space, a closed and totally bounded set is compact.

(and the proof goes much like the Heine-Borel/Bolzano-Weierstrass ones, because in R^k boundedness and total boundedness are equivalent)

[prime elbow]

Yes

[prime elbow]

I mean if every infinite subset of E has a limit point then E is compact in Metric space X

[alpine nest]

Try proving it.

[prime elbow]

If every infinite subset of E has a limit point in E then E is sequentially compact, right?

Because if I take any sequence in E then if the set { x_n | n in N } is finite we can make a subsequence which is convergent.

And if the set is infinite it has a limit point then we have a subsequence which convergent in E

[alpine nest]

In metric spaces sequential compactness is equivalent to compactness.

[prime elbow]

And to prove last statement what if I take x_n such that d(x_n,x) < 1/n

[prime elbow]

Yes

[alpine nest]

And yes, that's the proof, the set of terms of your sequence has a limit point, and you can show that there's a subsequence converging to that limit point

[prime elbow]

Okay thank you ❤️

And let X be metric space in which every infinite subset has a limit point then X is compact because we can take E = X

[alpine nest]

Yep! a category theorist would probably call it an application of Yoneda's Lemma

[prime elbow]

So compact also implies that every infinite subset of E has limit point in E

Both are equivalent?

[tender halo]

you can say that after every statement

[uneven bronze]

I am reading about product metric spaces. There are various metrics we can put on a product, but my book singles out two defining properties a metric on a product space should enjoy. 1) That a sequence converges in the product space to a limit iff the component sequences converge to the components of the limit and 2) that each component metric on the component space is less than or equal to the metric on the product.

[gentle osprey]

psie

[uneven bronze]

I am basically trying to verify property 1 for the max-metric.

[uneven bronze]

I think I have figured out a way to motivate this. Thanks!

I would be interested in knowing though if there is an upper bound to d(x,y) in terms of d_k for some k. It seems like the maximum of the d_k times any number greater than 1 would be an upper bound, is that right?

[prime elbow]

In normed vector space, if x_n converges to x and y_n converges to y then x_n+y_n converges to x+y.

But let there be metric space X such that addition is defined in X then is that result true ?

[tender halo]

addition is defined in X

what does that mean

addition of what

[prime elbow]

Addition of elements

[tender halo]

just like

a random abelian group structure on X?

generally we define a topological group so that addition is continuous, but that is not particular to any concrete metric on the space

[alpine nest]

Not necessarily.

[prime elbow]

Example?

[alpine nest]

Look up the jungle-river metric on R^2

And try to show an example that addition isn't continuous in that metric

Hm, you can probably even do it on R with the metric d(x,y) = |x-y| + |sgn(x)-sgn(y)|

[prime elbow]

So I need to find two sequences such that x_n converges to x and y_n converges to y but x_n + y_n is not convergent to x+y

[winter falcon]

Is there someone here working on pointless topology (frames & locales)? Frames generalize topology and now there is this new thing called partial frames, where only certain subsets in the meet-semilattices are allowed to have joins. So partial frames kinda generalise frames. How can one also generalise separation axioms like Hausdorfness in this setting? Some authors have done the relatively easy to translate separation axioms from frames like regularity. Also another question would be what would be the benefit of partial frames?

[uneven bronze]

I am reading a proof of the fact that a sequentially compact metric space $X$ is complete and totally bounded. I have already shown it is complete. But I am stuck at the totally bounded claim. \

Proof: We wish to cover $X$ by a finite number of open balls of radius $\epsilon$. Let $y_1\in X$ be arbitrary. If $B(y_1,\epsilon)\neq X$, let $y_2$ be a point in $X\setminus B(y_1,\epsilon)$. Having chosen $y_1,\ldots,y_n$, we let $y_{n+1}$ be any point in $X\setminus \bigcup_{j=1}^n B(y_j,\epsilon)$, provided the union does not coincide with $X$, otherwise we stop. The points $y_1,y_2,\ldots$ satisfy $d(y_k,y_j)\geq\epsilon$ for $1\leq j<k$. If the procedure of selecting the $y_j$'s does not terminate, then the sequence $(y_j)$ has no convergent subsequence. ... \

I fail to see why $(y_j)$ has no convergent subsequence. Why is this?

[gentle osprey]

psie

[gritty widget]

I believe it is because of the condition on the distances making it so no subsequence can be Cauchy

[uneven bronze]

hmm ok, do you know what it mean for no subsequence to be Cauchy (symbolically)?

[gritty widget]

Any subsequence of y_i is of the form y_n_i where n_i is a strictly increasing function from N to N.

[uneven bronze]

I am reading a proof of the fact that if $E\subset\mathbb R^n$ is bounded, then it is totally bounded. It suffices here to prove that $T=[-b,b]^n\supset E$ is totally bounded, as any subspace of a totally bounded space is totally bounded. For this, let $\epsilon>0$. Then the balls $B(\epsilon j,\epsilon)$ cover $T$, where $j=(j_1,\ldots,j_n)$ ranges over all integral lattice points of $\mathbb R^n$ which satisfy $\epsilon |j_i|\leq 2b$, $1\leq i\leq n$. (Recall, an integral lattice point is a point whose coordinates are integers.) \

How can I verify that the balls $B(\epsilon j,\epsilon)$ cover $T$? In particular, why the condition $\epsilon |j_i|\leq 2b$, $1\leq i\leq n$?

[gentle osprey]

psie

[gentle osprey]

psie

[keen seal]

Seems true

Every infinite subset of E has a countable infinite set

which has a well order isomorphic to that of the naturals

a limit point of this countable infinite set will be a limit point of the sequence defined by the ordering

[tender halo]

that is definition of countable compactness

in metric spaces (or, in general, first-countable spaces (or, even more in general, sequential spaces)) sequential and countable compactness are the same

[alpine nest]

oh cool, I didn't realize that was another variant of compactness, although of the course it is. what's the example of a space where sequential and countable compactness aren't equivalent?

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

never mind, looked it up

[alpine nest]

For example, the product of continuum-many closed intervals [ 0 , 1 ] with the product topology is compact and hence countably compact; but it is not sequentially compact.

So there exists a sequence in this space which has an accumulation point but no subsequence convergent to that point?

[opaque scroll]

Yes that's right. The problem is that points in this space doesn't have a countable basis of neighbourhoods. So you can't do the naive thing of filtering out elements "too far" away from the acumulation point.

[alpine nest]

I'm still trying to come up with an explicit example

Not great with intuitions in nonmetrizable spaces 😦

Although, take [0,1]^[0,1], enumerate the rationals in [0,1] as (q_n) and take a_n(x) = 1 if x = q_n and 0 otherwise

And then 0 is an accumulation point of this

But there's no subsequence convergent to 0

Am I thinking right?

A typical neighborhood of 0 consists of functions that are 0 for all but finitely may arguments, right?

No, wait; "close to 0 on finitely many specified coordinates, and arbitrary otherwise"

But then every cylinder around 0 would contain all but finitely many points of my sequence, so that would mean the sequence does converge to 0

[opaque scroll]

Take X = {0,1}^N and let your space be [0, 1]^X. Take fn to be the nth coordinate function.

[alpine nest]

Yeah, I see that

[opaque scroll]

Take any function for which the value at c is 0 or 1 and is not one that only oucors finitely many times.

Then specifying the value at finitely many points you should always be able to find coordinate functions that give the same value

Hmmm, wait. I need to think about this a little more

[alpine nest]

I mean, I can see how this sequence has no convergent subsequence, and that the space in question is compact.

I'm just struggling with seeing the accumulation point(s).

[prime elbow]

Countable compactness?

[opaque zodiac]

every open cover has a countable subcover

also called being Lindelöf

[dire dove]

This is lindelof but not countable compactness

Countable compactness should be open countable cover has finite subcover

[tender halo]

lindelofness is sigma-complete filters converge

countable compactness is countable filter-bases converge

[opaque scroll]

So at least it's easy to argue that any infinite set has an accumulation point.

If it didn't, then because of Hausdorffness the complement would be open, so the set would be closed hence compact. But a compact set can't have infinitely many isolated points.

But it does seem tricky to find an explicit example

[tender halo]

not only it has that, the space in question is separable

so we have a countable dense subset with no (non-trivially) convergent subsequences

[opaque scroll]

What set is that?

[tender halo]

[0; 1]^c

oh

what dense subset

uhhh

dont think i can actually describe it easily

[opaque scroll]

Well, can you argue that it exists?

[tender halo]

not really, the only proof i know is an explicit construction

[opaque scroll]

I see, so you can describe it, just not easily

[tender halo]

its the Hewitt-Marczewski-Pondiczery theorem

actually i think you can do it pretty easily in this case

maybe

ah, no, unfortunately not

yeah you have to go through the full construction

[tender halo]

you can also think of Stone-Cech compactification of N

the same thing is true there

[alpine nest]

Well, I'm glad it's so complicated because this means I don't need to feel bad about not coming up with an example quickly

[plush folio]

Let $f_x : X\times Y \to X$ and $f_y : X \times Y \to Y$, and let $f : X \times Y \to X \times Y,; f(x, y) = (f_x(x, y), f_y(x, y))$. Then $f$ is continuous if and only if both $f_x$ and $f_y$ are continuous, and this follows from the universal property of the product topology, correct?

[gentle osprey]

sheddow

[fringe thorn]

I'm fairly confident, yes

[plush folio]

Thanks! Follow-up question: is there a way to write $f$ in terms of $f_x$ and $f_y$ in a more point-free style? Like, does $f = f_x \times f_y$ make sense?

[gentle osprey]

sheddow

[fringe thorn]

what does $f_x \times f_y$ denoting here?

[gentle osprey]

higher!

[fringe thorn]

I'm used to that notation denoting that product map $f_x \times f_y$, which is given by $f_x \times f_y(x_1, x_2) = (f_1(x_1), f_2(x_2))$

[gentle osprey]

higher!

[fringe thorn]

where each f_i is a map from an X_i to a Y_i, that is

[fringe thorn]

I will admit though, I'm also probably not the best person to ask for this question

[alpine nest]

I've also seen this kind of product map denoted as $f_x \otimes f_y$

[gentle osprey]

Outsider

[alpine nest]

To go with how the product sigma-algebra tends to be denoted

[fringe thorn]

is that so?

interesting

[paper wedge]

yes

with measures too

[fringe thorn]

[paper wedge]

the measure defined mu (+) mu_2 for example

[alpine nest]

Interestingly the product measure I've usually seen denoted as $\mu_1 \times \mu_2$, with the traditional product symbol.

[gentle osprey]

Outsider

[paper wedge]

iirc i saw it in folland

or maybe a problem on the internet ig

whatevers ig

[plush folio]

So two ways to combine functions would be $\otimes : ((X \to X) \times (Y \to Y)) \to (X \times Y) \to (X \times Y)$ and $\times : ((X \times Y \to X) \times (X \times Y \to Y)) \to (X \times Y) \to (X \times Y)$

[gentle osprey]

sheddow

[plush folio]

if you think of $\times$ and $\otimes$ as higher-order functions

[gentle osprey]

sheddow

[plush folio]

This is slightly incomprehensible tbh. But for clarity it is perhaps best to just write f(x, y) = (f_x(x, y), f_y(x, y))

[tender halo]

its called the diagonal of the two maps

that is, the combination of maps X -> Y and X -> Z into a map X -> Y x Z

one of the standard notations is big delta

i.e. an up-facing triangle

the one you dubbed with the tensor product symbol is the cartesian product of maps

usually dubbed by a cross

[plush folio]

I see, so $f = f_x \Delta f_y$? I am familiar with the diagonal of X being the set of points (x, x) in $X^2$

[gentle osprey]

sheddow

[tender halo]

exactly

and actually the diagonal of two maps is the composition of natural inclusion of X into its diagonal and cartesian product

[quartz horizon]

Oh nice universal property of the product

I’ve also seen $f = (f_x, f_y)$ for this

[gentle osprey]

Pseudonium

[plush folio]

Talking about universal properties and Pseudo shows up, like a bee to honey

[plush folio]

I like this notation, this might be my favourite

[quartz horizon]

Hehe, well y’know I love them

[opaque scroll]

What is a matrix if not a map from a coproduct to a product.

[unreal stratus]

A womb

[prime elbow]

Is there infinite topological space such that there exists finite dense subset in that space

[quartz horizon]

You can take the indiscrete space

[prime elbow]

But infinite topological means infinite open sets right?

[alpine nest]

Depends whether by "infinite topological" you mean "the space is infinite" or "the topology is infinite"

Every topology on an finite space will be finite, but on an infinite space you could have finite topologies

[prime elbow]

Yes

So if I have infinite topological space means the topology is infinite then is there any finite dense subset exists?

[quick delta]

There can be

Just adjoin a single point to any topological space, and add it to every open set

Then the closure of that point is the whole space

[quick delta]

Every non-empty

(And make that point be open)

[prime elbow]

So in R if I add 0 to every non-empty open set then it works?

{0} is dense in R?

[alpine nest]

More directly, let your space be R and let the topology consist of all sets that include 0 (and of the emptyset)

[prime elbow]

I see

[alpine nest]

You can't just add 0 to the opensets in the Euclidean topology (although you could do this is you used a point that isn't already in R)

[prime elbow]

i, complex number

[prime elbow]

But why can't I add 0?

[alpine nest]

Because in the Euclidean topology {0} is closed

[prime elbow]

But now it is a new topology

[alpine nest]

Oh wait, you meant replacing all the existing open sets with versions that also include 0

Then yes, that would probably work

[prime elbow]

And also containing {0} because we can add 0 to empty set

[prime elbow]

Thank you ❤️

[alpine nest]

Basically the key thing is that you want to have some element x that's in every nonempty open set

And then the closure of {x} is the whole space, because that's the only closed set that contains x

[prime elbow]

I see

Thank you

[uneven bronze]

I have been thinking a lot about a specific problem, and this question arose and I am somewhat tired of thinking about this problem, so this question might be silly. Consider the open ball $B(0,\epsilon)$ of radius $\epsilon>b$ in $\mathbb R^n$ with the maximum metric. Does it contain the cube $T=[-b,b]^n$?

[gentle osprey]

psie

[uneven bronze]

Here b>0 is some real number.

[gritty widget]

It almost does but no

It is the open cube ]-b,b[^n

er

I misread sorry

It does

It is the open cube ]-epsilon, epsilon[^n

If x is in [-b,b]^n then it is of the form (x1, ..., xn) where each xi is in [-b, b]

[uneven bronze]

right

[gritty widget]

And since -b > -eps and eps>b, xi is in ]-eps, eps[

[uneven bronze]

ok, thank you 👍 I have to read up on why B(0,\epsilon) is an open cube ]-\epsilon,\epsilon[^n

[gritty widget]

For brevity I will show the 2 variable case

[uneven bronze]

that would be great 🙂

[gritty widget]

|(x,y) - (0,0)| < eps iff max(|x|, |y|) < eps. That is equivalent to |x| < eps and |y| < eps. And this is equivalent to -eps < x < eps and -eps < y < eps, which is the same as (x,y) is in ]-eps, eps[^2

[tender halo]

how do I prove R is not a product space of a collection of nontrivial topologies?

is it true even

[opaque zodiac]

I think you can show this using some reasoning about path-connectedness

Basically if R can be written as a product, then it can be written as a binary product AxB

This gives you (open) projections of R onto A and B

I think you should be able to roughly speaking show that [0,1]^2 has to embed into A x B

which is impossible

[tender halo]

i got that far

hmmm

[quartz horizon]

Hmmmm how about this

[tender halo]

oh yeah true

[quartz horizon]

You can use the path connectedness

A product of path connected spaces is path connected

And also a continuous image of a path connected space is path connected

So A and B are path connected

Then I think if A and B are nontrivial, removing a single point from A x B still leaves it path connected

But removing a single point from R doesn’t leave it path connected

Maybe that works?

[opaque zodiac]

This seems right but how do you actually show that this is true?

[quartz horizon]

I think maybe you can use a similar argument as you’d do for R^2

[opaque zodiac]

I guess you literally just go around the point

[quartz horizon]

Yeah

Change coordinates one at a time

[opaque zodiac]

right but we don't know what A and B look like that well

*would have to look like

[quartz horizon]

But we know they’re individually path connected

I’d have to write this out though

Or perhaps maybe you’d want connectedness rather than path connectedness

[tender halo]

changing coordinates one at a time works though

[quartz horizon]

Neat

[tender halo]

one caveat though you need to wiggle your points around a bit

which is only possible if both of your spaces are not made out of one point

[quartz horizon]

Right, which I think is fine because we said they were nontrivial?

[swift sierra]

damn i need to learn topology for no particular reason i just feel like it

[quartz horizon]

After all R is a product of R with the singleton, topologically

[tender halo]

yeah

[prime elbow]

How to show sequential criteria is equivalent to continuity?

I know we can show that inverse image of closed set maps to closed set.

But I want to find U such that if x in X and then around any open ball of f(x) in Y then there exists U \subset X such that x \in U and f(U) \subset of that open ball.

X and Y are metric such that f: X->Y

We have if x_n converges to x then f(x_n) converges to f(y)

I don't want to use stronger results here

If A is compact in metric space then A × A is compact or not ?

[gritty widget]

Yes, Tychonoff's theorem says the product of an arbitrary collection of compact spaces is compact

[prime elbow]

We can use sequentially compactness, also right? But how

I am not sure

Because let (x_n, y_n) sequence in X × Y, where X and Y are compact spaces in metric space

So x_n have subsequence x_n_k converges to x in X and y_n has subsequence y_n_t converges to y in Y

Now how can I choose the subsequence of (x_n, y_n) ?

[dire dove]

Choose a convergent subsequence xnk and then choose a subsequence of the ynk

[prime elbow]

But subscripts are different, right?

[dire dove]

No

[prime elbow]

So when I choose (x_n_1, y_n_1) but what if y_n_1 not in y_n_t

[dire dove]

You want both the x's and the y's to converge at the same time

[prime elbow]

Yes

[dire dove]

So you make the x's converge

[tender halo]

you choose subsequence of a subsequence

[prime elbow]

Yes

[dire dove]

And then choose a subsequence of those indexes such that also the y's converge

Subsequence of a convergent sequence is convergent so with the x's you're fine

[prime elbow]

I see

So first apply to x

Get x_n_k

Then apply on y_n_k

[dire dove]

Yes

[prime elbow]

Thank you ❤️

[dire dove]

By the way, this argument extends to all finite products by induction, and for countable products you can diagonalize (iirc uncountable product of sequentially compact spaces is not necessarily sequentially compact so you can't go further)

[prime elbow]

If I want to show that if f: E -> E then F= { (x, f(x) | x in E }.

If E is compact and f is continuous on E then F is compact.

Because we can map g:E-> E × E then F is g(E) and also continuous so the image is compact.

Right?

[dire dove]

Yes

[prime elbow]

g is continuous because if x_n converges to x then (x_n, f(x_n) ) converges to (x,f(x) ) because f is continuous

[dire dove]

More generally it's true that a function into a product is continuous if and only if each component is continuous (even for general topological spaces where you can't use sequential continuity)

[quartz horizon]

Mhm, that’s the universal property

[prime elbow]

Yes by using projection maps, right?

[dire dove]

Yes

[prime elbow]

Is it good exercise if E \subset R is closed and f is continuous on E then there exists continuous g: R -> R such that g(x) = f(x) for all x in E?

[prime elbow]

Now I want to prove the converse part.

Let F be compact and E be compact then f is continuous

So let x_n converge to x

I need to show that f(x_n) converges to f(x)

Now let (x_n, f(x_n) ) in F

Since F is compact

So there exists subsequence (x_n_k, f(x_n_k) ) converges to (x,f(x)) because it converges in F and x_n converges to x so x_n_k converges to x

It shows that f(x_n_k) converges to f(x)

But how to show f(x_n) converges to f(x)

[livid narwhal]

$i:(Y,\beta)\to (X,\alpha)$ is a mapping between the topological spaces and $Y\subseteq X$ and $i$ is inclusion. Now my question is if $i$ is continous then is $\alpha\subseteq\beta$? Remember we know the converse that if $Y$ is equipped with subsoace topology then i is continuous.

[gentle osprey]

Vishnu das

[prime elbow]

Inclusion mapping is always continuous

But not necessarily $\alpha\subseteq\beta$

[gentle osprey]

Notknow🙇

[livid narwhal]

No i have not said that the topology $\beta$ is a subset of $\alpha$?

[gentle osprey]

Vishnu das

[prime elbow]

What is your question?

[livid narwhal]

Give me a example?

[prime elbow]

Take [0,1] to R inclusion mapping

[livid narwhal]

Topology?

[prime elbow]

What is alpha and beta, topology right?

[prime elbow]

[0,1] with subspace topology of standard topology on R and R with standard topology

[livid narwhal]

Sorry my question is alpha a subspace topo of beta or not if i is continuouS.

[prime elbow]

Y is subset of X, if Y is proper subset of X then how alpha is subspace topology of beta? I don't understand your question

[plush folio]

Like Notknow says, take [0, 1] with the subspace topology in R. The inclusion map is continuous, but the topologies on [0, 1] and R are incomparable; [0, 0.5) is open in [0, 1] but not in R; conversely, (1, 2) is open in R, but not in [0, 1]

[rancid umbra]

yes, this one is good. you need to figure out a way to extend E to all of R

[prime elbow]

Let f be a real uniformly continuous function on the bounded set E \subset R. Prove that f is bounded on E.

Since E is bounded so d(x,y) < M, for some M>0.
By uniformly continuously, there exists e>0 such that d(f(x), f(y) ) < e for d(x,y) < M.

So f is bounded on E.

Is it correct?
I think no how can I choose e>0 for M?

[rancid umbra]

take a limit point e of E. there is a sequence of points e_n in E converging to e. f(e_n) is a cauchy sequence due to uniform continuity, so it converges to something. use this idea to extend f to a bigger function F on cl(E).

[livid narwhal]

I am putting the question let $Y$ is a subset of $X$. Now the inclusion map $i:(Y,\beta)\to (X,\alpha)$ between these topological spaces. Now my question is if $i$ is continuous then the topology $(Y,\beta)$ is the subpace topology of $(X,\alpha)$?

[gentle osprey]

Vishnu das

[livid narwhal]

i.e. that mean can we write every member of beta as a intersection of $X$ and some member of alpha?

[gentle osprey]

Vishnu das

[rancid umbra]

1. The inclusion map is always continuous when beta is the subspace topology coming from X

2. alpha is generally not a subset of beta, since, for example, if Y is a proper subset of X, then X is in alpha but not in beta

3. beta is not necessarily a subset of alpha. for example, take Y = [0,1] and X = R. then [0,1/2) is open in Y but not open in R

what is the confusion?

[livid narwhal]

No, no. I never said Y is a subspace topo. This beta is some arbitrary topo on the subset Y

[rancid umbra]

what do you mean by inclusion map then?

[livid narwhal]

Set theoretic map

[rancid umbra]

do you mean an injection?

[livid narwhal]

Yes since Y is a subset of X

It is well defined

And we know the converse of my question is standard .

[rancid umbra]

ah

okay

misread your question and got confused with some of the comments

[rancid umbra]

you are asking if beta makes the inclusion map continuous, then must beta be the subspace topology, correct?

[livid narwhal]

Correct exactly this is the question. @prime elbow

[prime elbow]

Not exactly I think. Because then there can be more open sets which are not of the form X intersection open set of alpha

{1,2} maps to {1,2} and take first topology with discrete topology and second one with indiscrete topology

[prime elbow]

So then since cl E is closed and E is bounded its image also bounded

Because we can use compact

[rancid umbra]

correct. im trying to think of a good example tho.

consider the topology generated from the basis of all open intervals and all open intervals minus {1/n : n in N}

this is called the K-topology

it is strictly finer than the usual topology

[plush folio]

The subspace topology is the coarsest topology that makes the inclusion map continuous. You could assign the discrete topology to Y for example, that would also make the inclusion map continuous

[rancid umbra]

so endowing, say Y = [0,1] with the K-topology and X = R with the usual topology makes the inclusion map continuous

but the K-topology on Y is not the subspace topology inherited from the usual one on X

[gritty widget]

I wanted to prove bijective local homeomorphism implies homeomorphism

So let f: X->Y be bijective local homeo

Let U an open of X

[fringe thorn]

[gritty widget]

For each x in U, there is an open V_x such that f is homeomorphism from V_x to f(V_x)

Denote by U_x the intersection of V_x with U

Since f homeomorphism on V_x and U_x subset V_x, f is also homeomorphism from U_x to f(U_x)

so f(U_x) is open

And U is the union of the U_x's for all x in U

[paper wedge]

yeah

[gritty widget]

So f(U) = union of f(U_x)'s is open

So f is open so f is homeomorphism

This works right?

And also this same proof shows that local homeomorphisms are open maps right?

[paper wedge]

yeah

this works

[quartz horizon]

Yeah local homeo being open map feels cleaner

And then bijective implies homeo follows

[fringe thorn]

you can quickly show that a local homeo is continuous by invoking the local criterion for continuity too

[quartz horizon]

Mhm sure

[gritty widget]

I didn't even notice you don't require continuity in the definition of local homeomorphism

[fringe thorn]

are you learning topology at the moment too, Bequi?

[gritty widget]

I'm doing the appendix chapters of Lee's smooth manifolds

That review stuff needed for the book

[fringe thorn]

ah, I see

[prime elbow]

In metric space, if l is subsequential limit of sequence x_n then l is limit point of { x_n | n in N }

[limpid fern]

x_n may be a constant sequence

[prime elbow]

I see

[prime elbow]

How it contradicts that f(x_n) not converges to f(x), if I have subsequence f(x_n_k) converges to f(x)?

[prime elbow]

I don't understand how uniformly continuous helps me to extend this function?

[rancid umbra]

it tells you that f(e_n) is Cauchy

@prime elbow

are you asking for help showing how uniform continuity implies f(e_n) is Cauchy?

[prime elbow]

No

[prime elbow]

So if x in cl E then there exists x_n in E which converges to x.

Now since f is uniformly continuous so it maps Cauchy sequence to Cauchy sequence

And Cauchy sequence is convergent in R

[rancid umbra]

yes

[prime elbow]

So lim f(x) exists?

Now g(x) = lim f(x) for all x in cl E ?

Now f is continuous so for all y in E lim f(y) = f(y)

[rancid umbra]

you define f(e) to be the limit of the cauchy sequence f(e_n) for any choice of convergent sequence e_n --> e

yes

[prime elbow]

Yes

[rancid umbra]

yep

[prime elbow]

And to check continuity of g

[rancid umbra]

on E, g is continuous

[prime elbow]

At E' points ?

[rancid umbra]

yes

[prime elbow]

Yes

But on E'?

Means limit point which is not in E

By definition of g

Because we define g(x) as lim f(x)

So any x_n converges to x, so lim f(x_n) also converges to lim f(x)

Because lim f(x) exists

[rancid umbra]

this feels imprecise

[prime elbow]

Yes

[rancid umbra]

right

so you can actually show that g is uniformly continuous

and this is a bit easier

[prime elbow]

I sew

See

[rancid umbra]

let ep > 0. from uniform continuity of f on E, there is a d > 0 such that for any x,y in E with |x - y| < d, we have |f(x) - f(y)| < ep/3

set d' = d/3. let e,e' in cl(E)
there are four cases:
(1) e,e' are both not in E
(2) e is in E, e' is not in E
(3) e is not in E and e' is in E
(4) e and e' are both in E

in case (1), suppose |e - e'| < d' = d/3. let e_n --> e, e'_n --> e'. We can choose n large enough so that |e_n - e'_n| <= d by using the triangle inequality

try to show now that |g(e) - g(e')| <= ep by using the triangle inequality and the fact that f(e_n) converges to g(e), f(e'_n) converges to g(e')

cases (2) and (3) are symmetric, and similar to (1), just less complicated

case (4) is trivial

[prime elbow]

Got it, thank you ❤️

I think we can do it in just one case, right? But yes it helps to understand if we do work on cases

So by this exercise we can say that if E is dense set in metric space X, and let f be a uniformly continuous real function defined on E, then f has unique continuous extension from E to X

Unique because if there exists g then f=g on E, so f = g on X

What if I replace R with another metric space Y ?

[rancid umbra]

well, we are using completeness here

[prime elbow]

Oh yes

So we can replace it with any complete metric space?

[rancid umbra]

only the codomain has to be complete

but yes

[prime elbow]

And what about X ? What is the weaker condition we need

Need just topological space

But in topological space I don't think so any concept of Cauchy sequence

[rancid umbra]

metric space

[prime elbow]

@rancid umbra thank you ❤️

[prime elbow]

Yes

[rancid umbra]

there are other generalizations, but they are a bit out of scope

this is a good generalization

[prime elbow]

Ah

What is that?

[rancid umbra]

there is this concept called a uniform space, where the same statements hold, the codomain has to be a complete uniform space

[prime elbow]

Okay

[rancid umbra]

they generalize top spaces and metric spaces in some sense

but like, the idea is the same

[rancid umbra]

@quartz horizon this the Yoneda lemma?

[quartz horizon]

Wait what is

[rancid umbra]

this guy

uniformly continuous map from a metric space to a complete metric space extends uniquely to a map on the completion

uniformly continuous map from a uniform space to a complete uniform space extends uniquely to a map on the completion

[quartz horizon]

Oh that sounds like a universal property?

So if you call X the uniform space, and X bar the completion

Then uniformly continuous maps X -> Y for Y complete naturally correspond to uniformly continuous maps X bar -> Y

Setting Y = X bar and following the identity X bar -> X bar then gets you the inclusion X -> X bar

[rancid umbra]

where is the natural transformation hiding here?

[quartz horizon]

Ah so

In the category of complete uniform spaces, with uniformly continuous maps between them

[rancid umbra]

ah

[quartz horizon]

You have the hom-functor Hom(X bar, -)

This is naturally isomorphic to the functor sending Y to the set of uniformly continuous maps X -> Y

[rancid umbra]

okay

neat

that was cute

[quartz horizon]

Yoneda in this case (as in every other case) tells you how one direction of the natural iso works

Precomposition with the inclusion

[red yoke]

I think you might be looking for reflective subcategory

[quartz horizon]

Yes this an example of one

[gritty widget]

This looks a lot like an universal property but I don't see how to phrase it as one

Can it be phrased as an universal property?

[quartz horizon]

Yeah this is the universal property of subset

Continuous functions Y -> S naturally correspond to continuous functions Y -> X whose image is contained in S

[gritty widget]

This isn't just talking about continuous functions but about functions in general though

[quartz horizon]

It’s equivalent

This also works in the category of sets

And lots of other categories

[red yoke]

For any morphism f: Y → X with image in S, there exists a unique morphism g: Y → S such that ig = f

Or

The functor sending Y to the set of morphisms Y → X with image S is represented by S via composing with S → X

[quartz horizon]

Yeah you can get those descriptions using yoneda

The continuous functions S -> S transforms to the inclusion map S -> X, whose image is contained in S

[prime elbow]

Let f_E: X -> R such that f_E(x) = inf{ d(x,e) | e in E }.

X and E are metric spaces, E \subset X.

Now if E and F are disjoint such that F is closed then f_F: E -> R is a positive real valued function, right?

[quartz horizon]

Think so yeah

[prime elbow]

So if E is compact also so there exists M>0 such that M < d(p,q) if p in E and q in F.
Because f_F is continuous mapping so it maps to compact set in R and it will be [a,b], where a>0, right?

Not [a,b]

[candid bronze]

a union of closed bounded intervals yes

cuz otherwise u can construct a sequence using definition of infimum

[prime elbow]

So can I say that there exists M >0 such that M < all elements of that compact set

What is the characterization of compact sets in R? Closed and bounded

[candid bronze]

yea closed and bounded

[prime elbow]

But generalize?

[gritty widget]

I didn't read the whole conversation but there are compact subsets of R that aren't unions of closed bounded intervals like the cantor set

[candid bronze]

true

and also singletons

[prime elbow]

I see

But let 0 not be in my compact set, then ?

[candid bronze]

but yea i'm pretty sure u can say M < image(f)

[gritty widget]

Singletons are intervals of form [a,a]. I guess technically every set is an union of closed bounded intervals if you include those

[candid bronze]

honestly the convention around intervals is kinda weird

[low flame]

whats this mean by open sets?

when it says U is said to be open

does it mean like (x,y)?

[tender halo]

open sets are elements of tau

its what they are, definitionaly

[rancid umbra]

a naming convention

[tender halo]

also i find that paragraph to be a piss poor definition

[rancid umbra]

of what?

[fringe thorn]

why is it a bad definition?

[low flame]

okay so its just a naming convention, its just the sets that form the topology?

[rancid umbra]

yup

[low flame]

okay im getting hung up on like why couldnt i use closed sets and form a basis that way.

so i think its squatched now thank you

[rancid umbra]

you can do all this from the perspective of “closed” sets as well

[quartz horizon]

You can phrase topology in terms of closed sets too if you want, though the definition of basis would have to be a little different

[plush folio]

I found that definition needlessly confusing when I first read it, I wish they would just write that a family of sets is a basis if every open set can be written as a union of sets in the basis

[rancid umbra]

you just invert the conditions on union and intersection

[austere dirge]

doesnt really motivate it

[rancid umbra]

what would motivate it better?

[quartz horizon]

Yeah i think there’s often a difference between “easy to check definition, hard to see consequences” and “hard to check definition, easy to see consequences”

[austere dirge]

dunno i dont have strong opinions on trivial things

[rancid umbra]

lol

[quartz horizon]

I think in general mathematicians tend to prefer the former

[fringe thorn]

what makes it different from most definitions then?

[quartz horizon]

yeah…

[fringe thorn]

I usually see most definitions dropped with little to no motivation anyways

[austere dirge]

most definitions explain themselves

[quartz horizon]

hm, this has definitely not been my experience

[austere dirge]

but the motivation is always only found in the exercises section

[plush folio]

To be fair, two pages later he writes in a lemma that if B is a basis for a topology T, then T equals the collection of all unions of elements of B. But when I'm confused, I'm not just gonna skip ahead two pages. I prefer if the easy to understand definition came first, then the more practical one later

[austere dirge]

idk i just woke up

[rancid umbra]

i think it’s a skill to try and reinterpret abstract definitions

[quartz horizon]

yeah, i think i tend to have this preference too

it’s part of why i prefer physics to math

[austere dirge]

yes this is frustrating

[rancid umbra]

i found it to be the opposite in my physics classes

like

some stuff was all heuristics

maybe i’m getting at a different issue

[paper wedge]

who cares about definitions

[rancid umbra]

but i hated it it bugged me

[paper wedge]

intuition wins

[fringe thorn]

which book is this from, by the way?

[low flame]

munkres

[fringe thorn]

[quartz horizon]

yes i much prefer this

[plush folio]

Like, when you know the basis has to generate the topology under unions, both points of the definition make perfect sense. You need point (1) because the X itself is open, and you need (2) because the intersection of two sets are open. But if you don't know that, the definition seems kind of arbitrary

[rancid umbra]

why?

[quartz horizon]

yeah, lots and lots of arbitrary definitions pushed me away from maths

[quartz horizon]

i tend to prefer focusing on what something “does” than what something “is”

physicists don’t tend to make precise definitions of things because this is generally not very helpful for doing physics

[rancid umbra]

that’s what made it feel inaccessible to me

[quartz horizon]

hm, whereas maths felt quite inaccessible to me

[rancid umbra]

i found myself asking my prof a lot: “literally what r u talking about?” when they would gloss over some stuff. ig i just couldn’t let it go by without understanding some stuff first

weird

now look at us

both in a math server

[paper wedge]

u kinda need both guys

imo

im new to math so idk really but like

u get new ideas and like solve problems

not with ur definitions tbh 😄

with ur like

feel

or intuition

when u write it out then it's l;ike definitions and shit

[quartz horizon]

I mean, im still very much a physicist

I am not a mathematician

[quartz horizon]

yeah, I think different people will have different preferences here, because this is how I felt in my math classes

[rancid umbra]

i think we’re dual-ing

[quartz horizon]

dual-ing?

[rancid umbra]

play on words

didn’t land ig

[quartz horizon]

hm, but im not trying to fight you or anything

[rancid umbra]

no ik ik

lol enjoying the convo

[tender halo]

cuz thats the wrong way to think about bases for a person whos first learning the subject

[quartz horizon]

there tends to be this tension between what’s more efficient, and what’s better pedagogically

[tender halo]

also there wont be a clear parallel if you will be introducing bases at a point

also if you are going to do the generator definition you kinda need to prove that it generates the least topology that contains that basis

which is implicit and obvious if you are mature enough of course, but if you arent its not very good

[unreal stratus]

I should say lol

I have found the "generating under unions" to be the standard one and I see no reason to prefer the other

It is more intuitive and catchy and usable lol

[gritty widget]

Perhaps the other definition should be a proposition instead, a corollary from the "generating under unions" definition

[quartz horizon]

this is usually the style I’d prefer - present the more usable definition first, and then supply the more efficient definition as a proposition

[fringe thorn]

that's precisely what Lee does in ITM

haha

I think Munkres does it the other way around then?

[austere dirge]

yes

it was confusing until i realized that the union defn was way more fundamental

our course followed munkres so close except for the ultrafilter stuff

u should read on ultrafilters sometime higher its pretty cool

[fringe thorn]

I see

[quartz horizon]

oh yeah ultrafilters can be used to do compact Hausdorff spaces or something

[austere dirge]

ultrafilters are super cool they convinced my friend and i to read kunen

[quartz horizon]

what’s kunen

[austere dirge]

set theory

[quartz horizon]

ah, right

going to a forcing course is probably the most set theory I’ll ever do, I imagine

[austere dirge]

read "notes on forcing axioms" by todorcevic i hear its great

and by great i mean horrible

[quartz horizon]

i think I’m good…

[prime elbow]

Any hint? Let D be dense subset of a metric space, and suppose that every Cauchy sequence from D converges to some point of M. Prove that M is complete

[rancid umbra]

each point in the sequence is approximated by a point in D

you can make the approximation arbitrarily precise as n grows

[prime elbow]

I see

[uneven bronze]

How can one show that the open balls with rational radii form a base for some metric space X? An argument I've seen is that every open ball with irrational radius p is the union of open balls with rational radius r<p, and so every open set is a union of open balls with rational radius (here we have accepted that the open balls with real radius form a base). Does this argument work? Is every open ball with irrational radius a union of open balls with rational radius because the rationals are dense in the reals?

[unreal stratus]

Yes

You can do this more explicitly though

B(a,r) is the union of B(a,s) over all rational s < r

And then yes this follows by density

[uneven bronze]

ok, I see 👍

[unreal stratus]

You can also show that e.g. balls of radius 1/n form a basis though

Where you can't use the same trick but yeah

[uneven bronze]

ah interesting. I believe this is the next theorem in the book

[granite crane]

Ah I remember this hint in Rudin chapter 2

[gritty widget]

It is impossible to. If that was true then in particular (1+1)x1 = (1+1)x(1+1) which is false

[warped bloom]

let m,m,p be a cardinality then (m+n)p=mp+np.
It this a false proposition?
To proof this, I think one-to-one corresponding f: (A∪B)×C → (A∪B)×(A∪C) (card A=m, card B=n, card C=p)

[gritty widget]

Maybe you meant to say (AxC) ∪ (BxC) instead of (A∪B)×(A∪C)?

[uneven bronze]

Consider the following definitions in my book: \

A family $(U_{\alpha}){\alpha\in A}$ of sets is said to cover a set $S$ if $S$ is contained in the union of the $U\alpha$'s. An open cover of a metric space $X$ is a family of open subsets of $X$ that covers $X$. And $X$ is compact if every open cover has a finite subcover.\

So is it true that the finite subcover, or the open cover for that matter, will always equal $X$? Seems strange somehow, but this is what I infer from these definitions.

[warped bloom]

Sorry. You are right.

[gentle osprey]

psie

[gritty widget]

The union of the subcover will cover X

Else it wouldn't be a (sub)cover

[uneven bronze]

ok, so we have that X is contained in the open cover by definition. But since every set in the open cover is a subset of X, the open cover is also contained in X. So hence equality.

[gritty widget]

No

The (open) cover consists of (open) subsets of X

not of elements of X

For example {X} is always an open cover of X

[uneven bronze]

hmm, but a union preserves inclusion, or? If U_i are a collection of subsets of X, then their union is still a subset of X I think.

[prime elbow]

If I take my set X = N union {0}, and d_1 is discrete metric on X and d_2 is standard metric induced on X by R. Then d_1 and d_2 induce same topology but d_1 and d_2 are not bi-lipschitz, right?

[gritty widget]

Yes

[gritty widget]

There's not even a lipschitz surjection from (X,d_1) to (X,d_2) since d_2 can get arbitrarily big while d_1 is always at most 1, right?

[prime elbow]

Yes so it is not bi-lipschitz

What is lipschitz surjection ?

[gritty widget]

A surjective function that is lipschitz

[prime elbow]

Okay

So that example works?

[gritty widget]

I believe so

[prime elbow]

Thank you ❤️

[prime elbow]

In b part, I think we need to define y_n as recursively, right?

Since e_n is positive so there exists k_n such that d(x_i, x_j) < e_n for all i,j ≥k_n.

So let y_n = e_k_n. Now for e_(n+1) there exists k_(n+1) such that d(x_i, x_j) < e_(n+1) for all i,j ≥ k_(n+1).

So if k_n ≥ k_(n+1) then we can choose y_(n+1) = x_(k_n +1) and if k_n < k_(n+1) then choose y_(n+1) = k_(n+1).

Is it correct?

[prime elbow]

Anyone?

[rancid umbra]

surely this is a typo, right? xn is in a metric space. what does it mean to multiply?

[prime elbow]

No it means | x_ny_n | = d(x_n, y_n)

Author used this notation

[rancid umbra]

that is really horrific

[rancid umbra]

y_n is a subsequence of x_n. did you mean to set it to a real number?

[prime elbow]

no

Wait

[prime elbow]

I meant y_n with the corresponding x_k_(n+1) or x_(k_n +1)

[keen seal]

consider a Cauchy sequence in M, call it a = (a_n). You can modify it to be in D by considering b = (b_n) contained in D such that d(a_n,b_n) < 1/2^n (by density). This will still be a Cauchy sequence ( d(b_n, b_m) < 1/2^(min(n,m)-1))

[keen seal]

You can construct such a sequence:
Let (i_n) be the sequence of the y_n it is much easier to construct it in terms of the indices.
Take i_n = max{min{k: d(x_r,x_s) < eps_n forall r,s >= k}, i_(n-1) +1}

then take y_n = x_(i_n)

such a sequence constructed this way will convergence incredibly fast btw

these sets {k: d(x_r,x_s) < eps_n forall r,s >= k} are always of the from {k0,k0+1,...}

Which are especially nice to work with, cause there's no gaps

[prime elbow]

Yes then let b_n converge to x so d(a_n,x) ≤ d(a_n,b_n) + d(b_n,x) < e, right?

[prime elbow]

Is my proof correct?

[keen seal]

Yeah

[red folio]

Does anybody know how to prove this? I've been stuck on it for a good amount of time.
Let $(X,\tau)$, $(A, \mathcal{U})$ be topological spaces, $F \subseteq A \subseteq X$, and A be a subspace of X. Then F is closed in A iff there exists a closed set F' in X such that $F=F' \cap A$.

[opaque zodiac]

Do you just want the solution?

Also I'm guessing the topology on A is supposed to be the subspace topology?

[red folio]

Yes, A is a subspace topology. Forgot to add that

[gentle osprey]

snus

[red folio]

I just want hints

[opaque zodiac]

Can you do either of the directions?

[red folio]

wdym?

[opaque zodiac]

The problem is asking for an if and only if

In other words you're supposed to prove:
\begin{enumerate}
\item If $F'$ is a closed subset of $X$, then $A \cap F'$ is closed in $A$.
\item If $F$ is closed in $A$, then there exists a closed $F' \subseteq X$ such that $F = F' \cap A$.
\end{enumerate}
Can you do either of these?

[gentle osprey]

Exomnium

[red folio]

I've only been able to prove this: let $\tau_A$ be the subspace topology for A, then $A \backslash F \in \tau_A$ iff there exists a closed set F' in X such that $F = F' \cap A$

[gentle osprey]

snus

[opaque zodiac]

But that's equivalent to what you're supposed to show

A set is closed iff its complement is open

[granite crane]

Can I prove the converse like this: Suppose $U\subseteq Y$ is open, we want to show $f^{-1}(U)$ is open. Let $x\in f^{-1}(U)$ then there is a ngbh $V_x$ such that $f \restriction_{V_x}$ is continuous. Thus $(f\restriction_{V_x})(U)=f^{-1}(U)\cap V_x$ is open. Notice $$f^{-1}(U)=\bigcup_{x\in f^{-1}(U)} (f\restriction_{V_x})^{-1}(U)$$ is open.

(I am talking about proposition 2.19-- local criterion for cont)

wait I wrote line 1 incorrectly

I mean " let U sub Y now we want to show f^-1(U) is open"

my bad

[gentle osprey]

Afzal

[rancid umbra]

yes you can do it like this

take a look at the proof of 2.8 g tho

its the same argument

[red folio]

are you saying I already proved it?

[granite crane]

oh yes yes. I see, that bullit point of prop 2.8 is same since $A=\bigcup_{x\in A} U_x$ where $U_x$ is ngbh of $x$

[gentle osprey]

Afzal

[granite crane]

thank you c squared

[opaque zodiac]

With like one more line, yes

[red folio]

does this mean showing F is closed in A iff $A \backslash F \in \tau_A$?

[gentle osprey]

snus

[opaque zodiac]

Yes

[red folio]

may I have a hint? I feel stuck

[opaque zodiac]

What is a closed set?

[red folio]

A subset A of B is closed iff $B \backslash A$ is open in B

[gentle osprey]

snus

[opaque zodiac]

And what does being in tau_A mean?

[red folio]

$\tau_A = {U \cap A | U \text{ is open in } B}$

[gentle osprey]

snus

[prime elbow]

If f is continuous mapping from Compact metric set X to metric set Y. Then f is uniformly continuous on X.

Now since f is continuous on X.

f is continuous at x_i, for all e>0, there exists \delta_x_i >0 such that d(x_i,x) < delta_x_i implies d(f(x), f(x_i ) < e.

Now let open covering B(x, delta_x ) over x in X then it covers X. So finite open covering covers X.
So let B(x_i, delta_x_i/ 2 ) , i = 1,2,...,n. It covers X.

So let y in X, so y in some B(x_i, delta_x_i/2 ) so d(x_i, y) < delta_x_i/2 and let x in B(x_i, delta_x_i/2) so d( x,y ) < delta_x_i.

Also d(x_i, x) < delta_x_i implies d( f(x_i), f(x) ) < e
And d(x_i, y) < delta_x_i implies d( f(x_i), f(y) ) < e
So d(f(x), f(y) ) < 2e.

Thus for every e>0, there exists delta>0 such that if d(x,y) < delta then d( f(x), f(y) ) < e.

Is it correct?

[rancid umbra]

just a note: you can just write d_i, no need to double subscript

when you say, “so let y in X, so…”, just say, “let x, y in X with d(x,y) < min(d_i/2)”. it’s more clear that way

otherwise it looks fine

[prime elbow]

Okay thank you

[forest pasture]

Hello

Question if you may

[chrome dew]

the picture you've drawn is not that helpful for thinking about ultrametric spaces

when you say "the midpoint" it's not really a thing

for instance, every point of a ball is its center in an ultrametric space

can you expand a bit on that part below the line |AC|_p <= max{|AB|_p, |BA|_p} did you mean |BC|_p for the second term instead?

I gotta go, feel free to ping me, but you might also like to try proving that if |x| != |y| in an ultrametric space then |x+y| = max(|x|,|y|) (the strong triangle inequality becomes an equality) and also try proving that fact about the center of balls earlier

[forest pasture]

No, the terms are what I meant

I think I made two mistakes

I knew I did, I'm just checking what the correct thing is

[forest pasture]

I have no clue how to prove them, if you may link me to good resources please

Also is there anything specific I don't know about plotting ultrametric spaces or projecting them onto orthonormal ones?

If I'm not wrong, by ostrowskis lemma, all ultrametric spaces are equivalent to p-adic ones sorta. Can't we plot p-adic points on the real number line by reading them right to left? (Basically treating them for example as numbers from 1 to 0)

Though I think then p-adic "straight lines" would look like fractals

Is that what the cantor set is about?

I'm so sorry for asking so much I'm really lost and not nearly equipped with the necessary prerequisites

[chrome dew]

not quite what ostroski's lemma says, and also sort of unrelated to that way of plotting p-adic points. You can do that but it's not really particularly useful to think of p-adic numbers with their digits reversed thrown on the real number line that way

I'd say get an intro book to p-adics like Gouvea's Intro to p-adic numbers or Katok's p-adic numbers compared with reals, or whatever they're called

[forest pasture]

Thanks so much

[chrome dew]

Some other good books are Koblitz's p-adic and zeta function book, Alain M. Robert's book (forgot the title), Schikhof's ultrametric calculus but may be a bit tougher

[forest pasture]

Thanks a lot

You are the best

[chrome dew]

you're welcome lol, there are quite a few people better than me here I assure you 🤣

[forest pasture]

You have no clue how horizon expanding this is

Imagine that 2 days ago the most mathematically knowledgeable person I knew was my highschool teacher

[uneven bronze]

I am trying to wrap my head around how $\mathbb R^2$ and $\mathbb C$ share the same open sets (I am studying Folland's real analysis text, and out of the blue, he just states that the Borel $\sigma$-algebra on $\mathbb R^2$ is the same as that on $\mathbb C$). If we equip both spaces with the usual metrics, the Euclidean one, then they have the same open balls, hence the same open sets. But can we equip the spaces with metrics such that they have different open sets?

[gentle osprey]

psie

[craggy cedar]

Their standard topologies are the same (the one induced by the euclidean metric) but you can give them different metrics that result in different topologies sure

[merry geode]

R^2 is in bijection with R, so you can certainly equip topology from that as well

[craggy cedar]

But this is not something we usually care about in the context of real analysis

In that context we're basically only worried about the standard topology

[uneven bronze]

Ok 👍 so I assume Folland has simply assumed R^2 and C are equipped with the standard topology

[craggy cedar]

If someone talks about their open sets or metrics without additional information then that's pretty much always what they're assuming yeah

[uneven bronze]

When someone says that arctan(x) is a homeomorphism between the extended reals and [-pi/2,pi/2], how can that fact help me determine/characterize the open sets of the extended reals?

I don't know a lot about homeomorphisms, except the definition. I know what the open sets in [-pi/2,pi/2] look like. Since it is subset of the real line, an open set U in [-pi/2,pi/2] is just U= [-pi/2,pi/2] cap V, where V is open in the real line. However, with the extended reals, I feel like I can't make the same argument, since, at least to my knowledge, it is not a subset of some metric space I know the open sets of.

[gritty widget]

If f: A->B is a homeomorphism then a subset U of A is open in A iff f(U) is open in B

[uneven bronze]

Ok, are all homeomorphisms open maps, i.e. mapping open sets to open sets?

[opaque zodiac]

no

wait

yes

[alpine nest]

By definition, yes

Although there's a subtlety to keep in mind when one space is homeomorphic with a subset of another.

Since then the homeomorphism will map open sets in one space to ones that are open in the subspace topology (but not necessarily in the whole-space topology)

[uneven bronze]

Ok, so in my case arctan(x) (or really, the continuous extension of arctan(x)) will map open sets in the extended real line to open sets in the subspace topology of [-pi/2,pi/2], i.e. sets of the form U=[-pi/2,pi/2]\cap V, where V is open in R?

[alpine nest]

Yes.

A homeomorphism is an open map because its inverse is continuous

[uneven bronze]

Ok 👍

[uneven bronze]

sorry if this is obvious, but why will this occur? Why will the homeomorphism map open sets in one space to ones that are open in the subspace topology (as opposed to sets that are open in the whole-space topology)?

[alpine nest]

The real line is homeomorphic to the set {(x,y): y = 0} as a subspace of R^2

The image of (0,1) is the set {(x,y): 0 < x < 1, y =0} in R^2

Which is open as a subset of {(x,y) : y = 0} in the subspace topology inherited from R^2, but is not open in R^2 as a whole

The homeomorphism is between space X and some space Y; and if space Y is a subspace of some bigger space Z, the homeomorphism still only cares about the subspace topology.

[uneven bronze]

ok

[cyan iris]

I've been trying to think of a counterexample: For X and Y topological spaces, if Y is regular then it is well-known that the space of continuous functions C(X,Y) with compact-open topology is regular. However if we consider the set of all functions F(X,Y), then books (e.g. kelley, willard) claim this is not regular with compact-open topology even if Y is regular

best I could find regarding this in MSE was this unanswered question: https://math.stackexchange.com/questions/4186696/how-to-prove-that-cx-y-is-completely-regular-but-not-normal

item 2 is my question (kinda), and a coment just say to "use a regular space that is not locally compact like the rationals"

I dunno, all functions is too large of a space to think of something useful, does anyone know a counterexample?

[merry geode]

Did you try the rationals, as in the comment?

[cyan iris]

Well I suppose he meant Y = Q? or both spaces being Q, anyway I'm not sure where this would lead, perhaps contradict the fact that Q is not locally compact somehow?

[rapid osprey]

Probably dumb question, but in proving 2.12, how can i be sure that a neighborhood of a contains infinitely many points of X-A?

[opaque zodiac]

If there is a neighborhood of a that doesn't contain infinitely many points of X-A, then a is in the interior of A

[rapid osprey]

Oh

That's very simple indeed

And, to be clear, it's because you can take the open ball of radius smaller than distances of a to a finite set and that would contain no elements of the exterior

[opaque zodiac]

yes

well no elements of the complement, technically the word 'exterior' is sometimes used to mean the interior of the complement

[rapid osprey]

I guess so, i used ext cuz complement is more letters

[rapid osprey]

Having difficulty with 2.13 as well

[craggy cedar]

You mean with figuring out why the construction satisfies it?

[rapid osprey]

Yeah

I feel like i have a mental picture but i can’t solidify it

nor describe it

[craggy cedar]

Looking back at this, 2.13 seems... not true?
Define R to be the elements of {U} contained in W' and S the elements disjoint from it. This should be a partition of {U}. We should also have \bigcup R and \bigcup S as a partition of A. Then \bigcup R and \bigcup S are open, which should imply that A is disconnected (which is not true for arbitrary X,A)?

[granite crane]

Does it look fine?

Later part of ex. 2.22

[rapid osprey]

I think you might have misread it

[rapid osprey]

It says that the intersetion of U and W' is nonempty implies that U is contained in W

[rapid osprey]

It is definitely not false

[craggy cedar]

Hi sorry apparently I'm illiterate. Just in case you haven't figured it out yet, you need to construct a small enough ball around a and then apply the triangle inequality a couple times.

[rapid osprey]

Not sure what you mean

Could you spell it out a bit more

[craggy cedar]

Let's say we have a ball B of radius r around a, a U in {U} such that B\cap U is nonempty, and a point x in U. What's an upper bound on d(x,a)?

[rapid osprey]

oh my goodness

Much better way to think about this

I think i’ve all but lost my mind overthinking this

[rapid osprey]

Let B(a,r) in W, then W’ = B(a,r/3) will suffice

[rapid osprey]

How do i verify this

I think i basically want to show that if x_U is in W', then x_U is not in any other U

[rapid osprey]

Damn

I think it's just incorrect

$F(x_U)$ need not equal $f(a_U)$, since $x_U$ can be in multiple $U_i$'s. Either way, it doesn't matter.

[gentle osprey]

aNDY

[rapid osprey]

I can show $f(a_{U})\in V$, and that's okay

[gentle osprey]

aNDY

[heady skiff]

in the axioms of a topology when we say any arbitrary union of open sets is open that union can be countable or uncountable right

[sonic crane]

Yeah

[heady skiff]

cool thx

[rapid osprey]

Turns out it was very fucking hard to prove

[topaz pebble]

Hello, I have been working my way through John M. Lee's Introduction to Topological Manifolds and I'm stuck on Problem 4-6 part b).

I've been trying to prove that the long line L is locally Euclidean to R, however there is a problem then I try to get a neighborhood of a point of the form (y, 0) be homeomorphic to an open interval in R, for an arbitrary y in the set Y.

Since for the point to be locally Euclidean, I need to find a y' such that it is the biggest element of Y that is smaller than y.

But such an element does not necessarily exist. Since we can order, for example, the natural numbers such that all the even numbers are smaller than any odd number, and 1 won't have such a smaller number, despite having countably many elements smaller than it.

And without that best I can do a homeomorphism to the closed interval $[0, \infty )$

[gentle osprey]

Zugr

[tiny obsidian]

Why? Why specifically do you need the biggest one?

[alpine nest]

Yeah, you can't just take the "previous" segment, because as you note, one might not exist

https://math.stackexchange.com/questions/700178/long-ray-proof-of-being-locally-euclidean

There's a discussion and explanation here, on how to handle the limit ordinals

[topaz pebble]

So that I can build an open neighborhood around that point. If I pick an arbitrary y' < y to glue onto {y} x [0, 1), then it won't be an open neighborhood.

[tiny obsidian]

Why not?

Also details on what you mean by "gluing y' " would be helpful

[topaz pebble]

Because of the order topology, it's basis is made up of the sets of elements that are strictly smaller or bigger than any element. There are no finite intersection of such sets that gives us an open set {y} x [0, 1).

And by gluing I mean just taking the union of {y} x [0, 1) with a {y', (a, 1)} for some a and calling it a an open neighborhood of (y, 0)

[alpine nest]

Yep, and that works great if y is the successor of y'

For limit ordinals you need to be more tricksy, as described in the MSE post I quoted

[topaz pebble]

Thank you linking that. I'm currently trying to wrap my head around it. From what I understand, it takes a countably infinite number the the segments that are below (y, 0) as the neighborhood and contruct a homeomorphism from that to (1, 3). Though I'm still trying to wrap my head around the given function.

It should also work for my specific set L, I'll just need to do a cut-off at (y_0, 0)

[alpine nest]

Yeah, thinking in ordinals takes some getting used to, I often struggle with intuitions as well. But for an intuitive example of how these intervals work: think of the union $\left(\bigcup_{n=1}^\infty \left[1-\frac{1}{n},1-\frac{1}{n+1}\right)\right)\cup[1,2)$

[gentle osprey]

Outsider

[alpine nest]

You've got countably many intervals which add up to [0,1), and then you glue [1,2) to it at the end

So 1 has a perfectly nice neighborhood in this set, even though [1,2) doesn't have an immediately preceding interval

And the whole thing is locally euclidean (0 aside)

[topaz pebble]

That makes perfect sense, thank you.

[alpine nest]

What's described on MSE is basically this, except you need to prove that you can do this kind of interval packing for an arbitrary countable ordinal.

[topaz pebble]

I'll get right to it then.

[alpine nest]

And yeah, it gets very hard to visualize very quickly

[rigid spoke]

Hi, idk if this is the correct chat, but does anyone have a demonstration of why The Sorgenfrey Plane is not T4 ? Ty.

[craggy cedar]

Consider the sets {y=-x:y,x\in Q} and {y=-x:y,x\not\in Q}

[tender halo]

its a separable space with a closed discrete subspace of uncountable cardinality

that doesnt happen in normal spaces

[sonic crane]

"And the sphere (surface of a ball) can be constructed by taking a disc and collapsing
its entire boundary to a single point; see Figure 22.2"

wut

[alpine nest]

it's true

[sonic crane]

i dont understand it really

what does collapsing boundary to a single point mean

instead of the boundary being distinct points, we now identify the whole boundary as one single point?

[robust drum]

Yes

Precisely, it means the quotient space of the disc by the smallest equivalence relation such that all points on the boundary of the disc are equivalent

[alpine nest]

https://en.m.wikipedia.org/wiki/Quotient_space_(topology)

serendipitously it's the first example in the quotient space article on Wikipedia

[sonic crane]

ok so there is some sort of stretching going on too?

so that quotient set under quotient topology is homeomorphic to sphere?

I guess it would prob be nice to see it all worked out and to see a homeomorphism

[civic verge]

hahaha I hardly understand the quotient in a group.

[opaque scroll]

I mean, the animation in the picture is the homeomorphism.

If it's easier one dimension down, it's exactly the same idea as of you take a line segment and identify the endpoints you get a circle.

[quartz horizon]

Using spherical coordinates will help write out an explicit homeomorphism

Together with the universal property of the quotient

[alpine nest]

How does that work here?

Ah, every continuous map from the disk into some space Z which is constant on the boundary, corresponds to an unique continuous map from the sphere into Z

[quartz horizon]

That’s roughly the idea

You do still want a continuous map from disk to sphere

I was going to suggest following geodesics

[alpine nest]

Yeah, but then use the universal property to avoid working with the quotient topology directly.

I like this.

[quartz horizon]

Mhm

Then you can use topological inverse function theorem

Since you’re mapping from compact to Hausdorff

So you just need a continuous bijection

[alpine nest]

I think this is the first time when I genuinely find this kind of reasoning advantageous.

[quartz horizon]

Nice lol

[alpine nest]

So far every time I saw universal properties mentioned, I didn't find that approach much more useful than arguing directly.

But using quotient topology directly is annoying.

[quartz horizon]

Mhm, it’s often easier to look at what the quotient topology “does” than what it ”is”

[alpine nest]

True, since the most intuitive way of looking at quotient topology is that it glues together certain areas of your space.

[alpine nest]

I also agree that starting with an explicit construction for the one-dimensional case (an interval with endpoints identified is homeomorphic to a circle) will probably be much easier to handle.

[quartz horizon]

And the idea of universal property is to focus on how that gluing affects continuous functions out of the space, i suppose

Continuous functions can’t separate “glued” points, essentially

[alpine nest]

Well yeah, since topology can be characterized by what functions are continuous.

That is something that's often used whether or not you employ the language of universal properties etc.

Weak topology and product topologies are often defined as "the weakest topology in which a certain collection of maps is continuous"

[quartz horizon]

Mhm

[quartz horizon]

Specifically, which functions to the sierspinski space are continuous

But also in the sense you mean with initial/final topologies

[alpine nest]

I was thinking of the more specific cases, where in the weak topology you're looking at linear maps into R or C, and in product topology you're looking at projection maps into the component spaces.

[quartz horizon]

Mhm mhm

You can also phrase these as universal properties, which is often called their “characteristic property” for some reason

[candid wyvern]

i dont understand this example, starting from why h is a homeomorphism. isnt the map from M_f to N a deformation retract. or inclusion map? it isnt bijective right?

and whats the mapping cylinder? ive been thinking of it as a join of the domain and the image

[alpine nest]

TBH I prefer the term "characteristic property", since to me it conveys that it's a kind of property that uniquely characterizes your object. Meanwhile when I see "universal property", it sounds to me like it's some kind of property that all objects have.

[quartz horizon]

I usually read “universal property” these days as specifying how your object relates to everything else in the “universe”

I do see how “characteristic” property as uniquely characterising an object is good, though I think universal/characteristic properties are useful beyond defining an object up to isomorphism

[sonic crane]

Thanks guys!

when we talk about a circle or a sphere, what is the topology on it? Should I think of these objects on their own or as being embedded in R^n?

Do they take the subspace topology from R^n or smth?

[gritty widget]

They do

But they're still topological spaces on their own

And they can embed in nonequivalent ways (for some sense of nonequivalent) into R^n

Knot theory is about that

[sonic crane]

ok, but in a first course i should indeed think of these shapes as being embedded in R^n and having subspace top?

with usual topology on R^n

[craggy cedar]

Yes

[sonic crane]

Thx

[craggy cedar]

I mean topologically they are all equivalent though, no?

[gritty widget]

Their topologies are equivalent, but embeddings have more information than just the topology of the domain

[sonic crane]

To say that p is a quotient map is equivalent to saying that p is
continuous and p maps saturated open sets of X to open sets of Y (or saturated closed
sets of X to closed sets of Y )

I'm not really understanding this too well

The saturated open sets of X are a smaller collection than the open sets of X, im just not seeing how if the saturated open sets map to open sets in Y, then all open sets in X map to open in Y

[robust drum]

You’re right, in general quotient maps need not be open maps

An surjectivebopen map will be a quotient map but the converse is not true

[robust drum]

For a counterexample, consider identifying two sides of a square to a cyllinder. In the square, an open semicircle on one of those sides will be open (by the def of the subspace topology), but its image in the cylinder is a semicircle together with a closed line segment, and its image is neither open nor closed

[sonic crane]

Oh ok I was getting definition of quotient map confused!

One sec

Let X and Y be topological spaces; let p : X → Y be a surjective map.
The map p is said to be a quotient map provided a subset U of Y is open in Y if and
only if p−1(U) is open in X.

This does NOT mean that open in X gets mapped to open in Y

i thought it did but i think i see the difference now

[robust drum]

Yeah. The nuance here is that the semicircle I described isn’t saturated

Bc the line segment is on one of the edges of the square, so getting identified, it’s preimage isn’t just the open semicircle but the semicircle together with a line on the other side. The full preimage isn’t open

[sonic crane]

I think this example is a bit too much for me rn

[robust drum]

Lemme draw it

Shitty notes app drawing

Notice that the inside of the red semicircle, including the line on the right, is open in the drawing on top

But in the bottom, it gets mapped to this half open thing where half of its boundary is open and half isn’t

I really should’ve drawn the semi circle part with a dashed line

That would’ve made it clearer what I meant

[sonic crane]

For p bijective (or probably just injective), the saturated open sets are the same as the open sets right

Random question, [0,1] U [2,3] with the subspace topology from R, the set [0,1] is both open and closed?

[quick delta]

Yes

[sonic crane]

thx

[opaque zodiac]

#JustSayClopen

[alpine nest]

apparently some people are oposed to that

[quick delta]

And they are wrong

[fringe thorn]

like me, who prefers to say "oposed" instead

ah, I just realized you said the joke too

well done!

Some like the word clopen. Others are oposed.

[alpine nest]

what a terrible joke

[fringe thorn]

hey now, I didn't make it

I stole it from Carla a few months ago