point-set-topology

[real granite]

It is sequentially but not topologically

[alpine nest]

You're correct, the subspace topology on (0,1] consists of {1}, the emptyset and all of (0,1]

So it's connected (and weird)

[real granite]

Yes

How about (ii)?

I think it's just the trivial toplogy right? The subspace topology consists of ø and the entire subset, nothing else

[alpine nest]

Yeah, that's what I was typing basically.

in (ii) you've got the trivial (antidiscrete) topology, so it's connected

[real granite]

It feels weird understanding topology. A few months ago I was finding this stuff impossibly difficult

[keen seal]

It's useful

For linear codes especially

[alpine nest]

Yeah, absolutely, I actually knew of that but I forgot in the moment.

I genuinely don't think of non-topological context when I hear "metric", a lamentable omission on my part

[gritty widget]

you can also do geometry with metric spaces

stuff like isometries

which is not recoverable from the topology

[real granite]

Does $\bZ=\bigcup_{n=0} ^{\infty} \qty({-n} \cup {n})$ work as an example of a non-compact subset?

[gentle osprey]

Douglas

[real granite]

As the above shows, I sometimes post questions even though I've got the right answer

[gritty widget]

Have you found an open cover with no finite subcover?

[real granite]

yes

cos any finite "subcover" won't actually cover Z

[alpine nest]

Yep, a space with the discrete topology is compact if and only if it's finite

[gritty widget]

depends on the original cover

if the cover contained the open R, then there will be a finite subcover {R}

[alpine nest]

Fair point

But any finite subcover of the one implicitly described by Douglas in the union.

[real granite]

no i meant like

of the cover i suggested

you only need to show that there exists one cover with no finite subcover right

[gritty widget]

i didn't know you meant a cover by your equality, i thought you was just defining the non-compact set

[gritty widget]

yes

[alpine nest]

Yeah, it wasn't clear, which is why I mentioned the general fact about the compactness (or lack thereof) of spaces with the discrete topology.

[gritty widget]

In retrospect it seems clear because theres no reason to write Z in two different ways there if not to show an open cover of it

But I just didn't even notice that

[sonic crane]

Su the box and product topologies has as basis all sets of the form U1 x U2 x … for Ui open in Xi (product topology only has finitely many Ui that are not Xi), but this does NOT mean that every open set in these topologies have the form “open set x open set x open set …” right?

Maybe im getting confused by smth

[gritty widget]

Do you know an example of an open in RxR that is not the product of two opens in R?

[alpine nest]

Yep, those are the basis

[gritty widget]

R being the real numbers

[alpine nest]

So there are open sets that aren't in the basis

[sonic crane]

So an arbitrary union of some “open set x open set ..” thing we cant write as a like componentwise union ?

But we can do that for intersections?

[quartz horizon]

Yeah a union of products isn’t nice in general

But an intersection of products is

[sonic crane]

Idk how i didnt know this lol

[gritty widget]

For example, is the interior of a circle the product of two opens?

[sonic crane]

Oh yeah, its not

Thanks for that example

[gritty widget]

Yes, I believe so

If (x,y) is in AxB and in A'xB' then x is in A and in A', and y is in B and in B' so (x,y) is in (A\cap A')x(B\cap B'). Conversely if (x,y) is in (A\cap A')x(B\cap B') then x is in A and A' and y is in B and B' so (x,y) is in AxB and in A'xB'

This can probably be better and more succintly written

[real granite]

(From an open book past paper.)

7. (i) Not sure how to do this besides considering a Cauchy sequence.

(ii) $Tf$ is obviously differentiable, so it is also continuous. The integral is obviously less than or equal to $\int_0 ^x y^3 \dd{y}$, so this ensures $Tf$ is in the range.

(iii) Not sure if my method was sufficiently rigorous. $$d(Tf, Tg)=\frac{3}{2} \sup|\int_0 ^x y^3 (f^2(y)-g^2(y)) \dd{y}| \leq 3\sup|\int_0 ^x y^3 (f(y)-g(y))\dd{y}|$$ by using the range of $f,g$ to bound $f+g$. "Factoring out" $f-g$ gives $$d(Tf,Tg) \leq 3\sup|\int_0 ^x y^3 \dd{y}| d(f,g) =\frac{3}{4} d(f,g).$$ Hence $T$ is a contraction.

(iv) $[-1,1]$ is complex and $T$ is a contraction, so the fixed point theorem implies the existence of a unique solution to the given IVP.

[gentle osprey]

Douglas

[alpine nest]

7. (i) - yes, consider a Cauchy sequence and prove directly that it converges

7. (ii) - probably fine, I agree with your reasoning that Tf is continuous so it remains to show that Tf(x) is always in [-1,1] if x is in [-1,1]

7. (iii) - I think is fine;

7. (iv) - you'd probably need to properly write down that f being a fixed point of T implies that it satisfies the differential equation. More importantly, the important thing is the completeness of X, rather than of [-1,1]

[real granite]

Uh yes and also I said complex instead of complete

But I did mean complete

And yes X's completeness is the main bit

[alpine nest]

Anyway, on the whole I don't see any glaring errors

[real granite]

Lovely

[alpine nest]

Comedy option for (i) based on this; if the course had the result that the set of all bounded functions on some set (or the set of all continuous function on a metric space, or something of the kind), with this metric, is a complete metric space, then all you need is to prove that X is a closed subset of C([-1,1])

Which will imply that it's complete.

[rancid umbra]

have you tried anything?

[rancid umbra]

in case you haven't gotten anything, maybe consider what R^3 - B^3 is homotopic to or strongly deformation retracts onto

[west brook]

Nice pun

[rancid umbra]

for van kampen, you should split the space into two parts:

U = {(x,y,z) in R^3 - B^3 : z > -1/2}
V = {(x,y,z) in R^3 - B^3 : z < 1/2}

U and V are both contractible

splitting it into a part containing a large sphere centered at the origin and a part that looks like R^3 - some other sphere centered at the origin makes you “go in circles”

[rancid umbra]

in general, you should try to show that if you can write a space as the union of two simply connected open sets whose intersection is path connected, then the entire space is simply connected

[livid narwhal]

How I have a very basic question in topology. Can anyone please answer about my confusion.

Is $\beta={[a,b): a<b, a,b\in\mathbb{R}}$ a basis for a topology?

[gentle osprey]

Vishnu das

[livid narwhal]

My confusion is I have to inclue $\Phi$ empty set with this collection then it is a basis otherwise how it could be a basis.

[gentle osprey]

Vishnu das

[opaque scroll]

It's a basis. I'm not sure what leads you to think a basis needs to contain the empty set, but it doesn't

[alpine nest]

The emptyset is the union of an empty subcollection of basis sets

So you don't have to explicitly include it in the basis

[livid narwhal]

Then the definition of munkres said take any two $B_1$ and $B_2$ from $\beta$ now if $x\in B_1\cap B_2$ then there exists $B_3\in\beta$ such that $x\in B_3\subseteq B_1\cap B_2$ then how this definition is satisfied?

[gentle osprey]

Vishnu das

[light herald]

isn't intersection of two [a, b) sets either empty or also an [a, b) set?

[livid narwhal]

Yes what happen if the intersection empty in that case what you choose?

[light herald]

then there is no element x that would be in B_1 intersection B_2

so the implication is true in vacuum or whatever it's called

vacuously satisfied

[sonic crane]

For a subbasis for a topology, the subbasis elements themselves are basis elements right

[tender halo]

yea

[sonic crane]

Finite intersection of subbasis elements with n=1 i guess

Thx

Subbasis was pretty weird when i first read about it but i guess it turns out to be useful

The definition was just weird at first understanding why its helpful to have such a thing

[tender halo]

i dont like how subbasis is usually introduced

its just a family of sets whose union is X

which in turn generates some topology on X

so if it generates yours its a subbasis for your topology

[sonic crane]

Yeah and i just understood better now why its called SUBbasis

[tender halo]

instead of just saying that the textbooks are doing a weird song and dance

[sonic crane]

Yea i feel like in munkres it was just so random and came out of nowhere

I dont think he explained at all why it was even introduced

He just gave defns

[rancid umbra]

i think it came up somewhere with continuity for me

as a way of checking continuity on some simpler sets

at least that’s how i first learned about it

[sonic crane]

Yeah i learned that now too and that shows me how its useful

[tender halo]

also you can check regularity only on subbasis sets

[rancid umbra]

oh cool

[tender halo]

tychonoff-ness too

[rancid umbra]

i wonder what properties in general only depend on subbasis elts

[tender halo]

uhh i dont know any more besides compactness i dont think

[gritty widget]

is this part obvious?

https://www.researchgate.net/publication/303344462_A_proof_of_the_nonretractibility_of_a_cell_onto_its_boundary

this is the original paper

[umbral panther]

What do you know about regular values?

[gritty widget]

just the definition

[umbral panther]

The entire point of a regular value is that it feeds into the inverse function theorem to conclude that the inverse image is a manifold. It’s not obvious, but you’re supposed to have seen it before

[gritty widget]

in a diffgeo book?

or a calculus book?

I haven't read a diffgeo book yet

[umbral panther]

Anywhere that introduces regular values should immediately address this

[gritty widget]

thanks, will try to see in relevant sources

[gritty widget]

so I should know that the preimage is a manifold, but what about the 1 dimensional part and the part saying boundary is on S^{n-1}

[real granite]

If anyone can help in #diff-geo-diff-top message that would be much appreciated

[sharp forge]

The codimension of r^-1(a) is equal to the dimension of S^{n-1}, so that the preimage manifold is 1-dimensional.

Thinking about the rank of the derivative is a good way to make sense of this.

Ignore my second comment, we can't say much about that with this map.

[gritty widget]

can you recommend a book that talks about this because I'm not familiar with it

[sharp forge]

Milnor's Topology from the Differentiable Viewpoint is good

idk though, I'm not the best at this stuff

[gritty widget]

thanks

[sharp forge]

See lemma 4

[restive flicker]

Question:
How do I show for any pair of sets B subset of C with B,C homeomorphic to the unit ball in Rⁿ for some n (with B,C subsets of some topological space X)-
There exists a homeomorpism from C to the unit ball that sends the points in B to a smaller ball?
The problem is I know only the definition of homotopy and maybe the fundamental group of a circle, tell me if it's enough to show the long statement above.(If not ig which materials do teach it)

[tiny obsidian]

I'm not sure this is true

Not without extra constraints on what B is allowed to be

Like, let C already be a unit ball and pick B so that C\B is not connected, and B homeomorphic to the unit ball

e.g. just squish C in one direction

[rancid umbra]

C\B is homotopy equivalent to S^{n-1} by pulling everything radially to the boundary of B, if C is the open unit ball and B is an open ball in C.

the only time when C\B is disconnected is when they are subsets of R.

[gaunt linden]

B doesn't have to be an open ball in C, just homeomorphic to one.

So we could, as Edward proposes, take C = { (x,y) | x²+y² < 1 } and B = { (x,y) | x² + 2y² < 1 }. Then C\B is disconnected.

[tough sluice]

I just completed reading the second chapter of Rudin (Basic Topology) and became really interested for no reason. Can I have some book recommendations to continue learning topology?

[gritty widget]

I suggest Willard's General Topology

[tough sluice]

i will try it out thx

Is it difficult compared to Rudin?

[tender halo]

Munkres' Topology is also a good book

if you want to do hardcore pure point-set topology, then Engelking is the standard reference

[sonic crane]

Im reading munkres and i like it

[tough sluice]

thx, I'll try all these three out

altho i probably won't be able to handle the engelking one (I still have moments where the rudin is challenging)

[sonic crane]

Its all hard math is hard 😁

But yeah idk i havent read other books but Munkres is pretty friendly

[tough sluice]

fair point

Wow, it has diagrams

interesting

[sonic crane]

Oh yea for sure

[tough sluice]

Also I'm starting to feel like every math book assumes that the reader doesn't understand elemantrary set theory

[sonic crane]

Lol yeah whenever its an introduction to a subject they always start with the set theory background

Im pretty sure future textbooks you would study on like algebraic topology wouldnt have that set theory again

[tough sluice]

algebraic topology is, like, completely different from point set topology right?

[sonic crane]

Well no its still topology

But now you use algebra structures to study properties of topological spaces

You’re still studying topology but now with a new toolbox at your disposal

[tough sluice]

ah, topology combined with abstract algbera give algebraic topology?

[sonic crane]

Yeah, like the idea is that structures from algebra would be encoding properties of topological spaces

And then you can use results from algebra and interpret them as results about the topological space

[tough sluice]

Oh wow, I see

[sonic crane]

Yeah its pretty cool. I have not learned it properly yet but ive seen a few of the big ideas

[tender halo]

half the books on advanced set theory start with overview of basic set theory

which i find amusing

[tough sluice]

btw topology isn't grad level right?

[sonic crane]

Ive seen it either senior undergrad or first yr grad level

[tough sluice]

well, that's scary

but ehhh i'll figure something out

[sonic crane]

Yeah its mostly because its definitely more abstract than the math you learn earlier on in uni

[tender halo]

i had it in my first semester

[sonic crane]

Yea i bet schools are quite different, i am speaking from what ive seen at canadian universities

[tender halo]

anyway i wouldnt say its a difficult subject at the basic level

[sonic crane]

Its just a mathematical maturity thing that matters i think

[tender halo]

yea

[tough sluice]

Yeah, lemme try it out, and if I can't handle it... I have many more years to get to uni so I'll be fine

[tender halo]

the abstraction pileup is the difficult part

[sonic crane]

Yeah

[tough sluice]

Ya I don't think I've gone to that levels of abstractions yet

[sonic crane]

What math have you learned so far

[tough sluice]

Uh, I'm currently in the 3rd chapter in D&F (abstract algebra), and just completed the 2nd chapter in Rudin

so, not much experience

[sonic crane]

Cool

I dont really know what its trying to say when it says all three topologies are “”different”” when J is infinite

Also is that implying that the first sentence there is when J is finite ?

And then in that case with J finite isnt box and product topology on R^J would be the same

[gritty widget]

yes

[sonic crane]

Yea when J finite box and product is same

Im just a bit lost on what this thm is saying

[gritty widget]

Its not true when J is infinite

[sonic crane]

Yes

Oh wtf

Yea ok we already know that but the thm is actung like we dont

The only new thing it introduced was this uniform topology

Nvm ill read this over again

[granite crane]

Uniform Topology, hearing for the first time, interesting

[warm hedge]

i am trying to do thing exercise and i am not sure if i am doing it the right way if anyone could help

i was thinking to take these A collections of sets

but instead of $\mathcal{A}_0$ being a countable base i would take $\mathcal{A}_0 = \mathcal{B} +$ the complements of the elements of $\mathcal{B}$

[gentle osprey]

whitefang

[warm hedge]

to make \mathcal{A}_0 a boolean algebra

but i dont know if every $\mathcal{A}_n$ will be boolean algebras

[gentle osprey]

whitefang

[warm hedge]

they are closed under complements i think

but i dont know about finite intersections

(btw the triangle symbol is the vaught transform)

[sonic crane]

The connection between metric spaces and topology has been the most interesting thing so far

Im looking forward to learning those metrization theorems

[sonic crane]

Hahaha who put that reaction

are those theorems not as interesting as i think ?

[tender halo]

I wouldn't say they are particularly interesting

Its just a bunch of different variations on the same theme of if the space is, in some sense, locally "metrizable" (i dont mean actual local metrizability, just something akin to it), then its metrizable

[sonic crane]

@tender halo what would you say are the interesting parts of topology then?

Maybe what do u think are the interesting parts one would learn in a first course, and then just in general what parts do you find interesting?

[tender halo]

I am not qualified to talk about algebraic topology and stuff, so if we are talking about just point-set topology, and just parts I find interesting as opposed to important (cuz metrization theorems are of course hugely important)

I would say for the first course the interesting parts are

the diagonal theorem and the fact that all Tychonoff spaces are embeddable into I^m for some cardinal m (which doubles as Uryhson metrization theorem)

the fact that locally compact spaces are precisely open subspaces of compact spaces

Cantor space, Baire space and their isomorphisms to subsets of the real line

Tychonoff theorem proof via ultrafilters

sequential spaces and Frechet-Uryhson spaces

just in general uhh

Q is the unique countable dense in itself metrizable space

Stone-Cech compactification, Cech-complete spaces and Baire category theorem, Hewitt compactification

Stone spaces

Stone topology on the ring of continious functions of a compact space, C and C*-embedability, z-filters and z-ideals

admissible and acceptable topologies, compactly generated spaces, uniform spaces

complete metrizability and its heritability, completion of a space

a lil bit of non-algebraic dimension theory maybe

[wispy veldt]

Stone-Cech ✨

[keen seal]

wait what?

I don't understand what you mean

[tender halo]

any countable metric space with no isolated points is homeomorphic to Q

[real granite]

#diff-geo-diff-top message any help would be appreciated thx

[keen seal]

ahhh but what did you mean dense in itself every space is, no?

[tender halo]

no

dense in itself means every point is a limit point

i.e. the derived set is again the whole space

if your space has isolated points there are no limits (that dont contain that point) that converge to it

[hexed steppe]

do people really use this terminology

ive never seen it

[light herald]

it's also called crowded set iirc

dense in itself is a bit confusing as it could refer to set being dense as a subset of itself

[hexed steppe]

never heard that either

i would always just say “no isolated points”

unless this language is really commonly used in some area

[hexed steppe]

yeah as evidenced by the above conversation

[light herald]

ig that also works
but being able to say that a cafe is dense-in-itself opens new possibilities of being a nerd in front of your friends

[hexed steppe]

it doesnt just “also work” i think it is standard?

unless the audience is like

exclusively point-set topologists

i think saying anything other than “no isolated points” would cause confusion

[tender halo]

yeah i see it often in point set topology texts

or at least often enough

[tender halo]

well the problem with that is that its not an adjective

language is much nicer if you can express a property as an adjective

[hexed steppe]

i think language is nicest when it conveys meaning with minimal friction

[dusky patrol]

Why does f(U) need to be open?

[hexed steppe]

it is a definition?

[dusky patrol]

Ik but why

Why does it matter

Whether f(U) is open or not

[hexed steppe]

idk

it doesnt matter to me personally

[rancid umbra]

they are defining what an open map is.
it is a map f : X —> Y that takes open subsets of X to open subsets of Y

[dusky patrol]

Not that part

In the def of local homeomorphism

[rancid umbra]

ah

[rancid umbra]

otherwise, maps like the inclusion of R^k into R^n for k < n would be considered ‘local homeomorphisms’ when they are not

since no open subset of R^k is homeomorphic to an open subset of R^n unless k = n

[dusky patrol]

What would this inclusion map be? On an unrelated note, do y'all have any hints for 2.8(a)?

Ah ok

[dusky patrol]

Why?

[dusky patrol]

I just started topology and don't know what homology groups or invariance of domain are :p

Ig i just have to accept it for now?

[rancid umbra]

take an open subset U of X. look at y in f(U). there is some x in U such that y = f(x). use the local homeomorphism property. you may need to shrink some sets

[dusky patrol]

. Let $V_x$ be the open set containing x such that the local homeomorphism property holds. Then $U=\bigcup_{x\in U}V_x \cap U$, with U and $V_x$ open. Hence $f(U)=\bigcup_{x\in U}f|{V_x}(V_x \cap U)$. Because a homeomorphism is open, each $f|{V_x}(V_x \cap U)$ is open and so is their union

[gentle osprey]

oxil764

[dusky patrol]

Does this make sense?

[tender halo]

yea

[gentle girder]

is your latex font comic sans? or just something close to it?

@dusky patrol

[dusky patrol]

Yes

[alpine nest]

I'm just disappointed the math environment isn't in comic sans

[dusky patrol]

I don't know how to change it😭

My papers are gonna be in comic sans

If i ever write them

[alpine nest]

Sadly, when you publish them they're going to be in whatever font the journal mandates

[gritty widget]

sounds like a good reason to start a journal

[rapid osprey]

Translated to english: 8. Remark about compact metric spaces. A metric space is compact if it has the Borel-Lebesgue property, that is to say, if, for every open cover, you can extract a finite subcover

This is from a paper from 1951

Kinda cool that compactness was still considered primarly sequential up to 1951 even

[echo oyster]

Why completeness in R is an axiom ?

Can it not be proved from already established axioms ?

[gritty widget]

It can

[gritty widget]

If whatever you are reading uses it as an axiom it's probably because they don't want to construct R

The construction is complicated and so some people just mention it as an axiom rather than a consequence of its construction

[echo oyster]

Oh

[dire dove]

I wouldnt say it's a matter of the construction being complicated

[quartz horizon]

Depends what construction you use

[gritty widget]

For most purposes you don't need to use an actual construction of R, you just need to know it exists

[quartz horizon]

The two main ones are dedekind cuts and cauchy sequences

[gritty widget]

The construction just proves it exists

[quartz horizon]

But there are many many ways to construct the reals

[echo oyster]

Can I get a reference pls to read about it

[quartz horizon]

My personal favourite is neither of these

Eudoxus is also very fun

[dire dove]

It's more of a universal property kind of thing (where you define the object as the thing that satisfies some set of axioms and then construct it)

[quartz horizon]

Mmmmm even then

I wouldn’t say that the usual axiomatisation of R is a universal property

It’s usually expressed as the “unique complete ordered Archimedean field” (hopefully I got all the adjectives there correct)

And that isn’t a universal property, strictly

There are ways to phrase R as an object satisfying a universal property, though

[dire dove]

Yeah i was just drawing a parallel to other things which are defined using universal properties

[echo oyster]

If possible please

[dire dove]

You dont need archimedean btw

[quartz horizon]

I see

[gritty widget]

Depends on what is meant by complete. If you mean Cauchy complete then you need Archmedean. If you mean Dedekind complete then you don't need it.

[quartz horizon]

Interesting

[dire dove]

Right

[quartz horizon]

https://projecteuclid.org/journals/rocky-mountain-journal-of-mathematics/volume-45/issue-3/Survey-Article-The-real-numbersA-survey-of-constructions/10.1216/RMJ-2015-45-3-737.full

That’s the most comprehensive list of constructions I know

(Though my favourite is not on there)

[dire dove]

Baby rudin chapter 1 is about these things

[echo oyster]

Ok thank you both of you

[quick crane]

The one variable book here covers the construction of $\mathbb{Z}, \mathbb{Q}, \mathbb{R}$ from $\mathbb{N}$ using Cauchy sequences: https://mtaylor.web.unc.edu/notes/math-521-522-basic-undergraduate-analysis-advanced-calculus/

[gentle osprey]

L

[indigo mural]

I am just starting out learning topology, can I get some hints? Which of the following subset(s) of ( M_{n}(\mathbb{R}) ) is(are) closed in standard topology?
(a) ( O_{n}(\mathbb{R})=\left{A \in M_{n}(\mathbb{R}) \mid A^{t} A=I d\right} )
(b) ( S L_{n}(\mathbb{R})=\left{A \in M_{n}(\mathbb{R}) \mid \operatorname{det}(A)=1\right} )
(c) ( G L_{n}(\mathbb{R})=\left{A \in M_{n}(\mathbb{R}) \mid \operatorname{det}(A) \neq 0\right} )
(d) ( G L_{n}(\mathbb{R})^{+}=\left{A \in M_{n}(\mathbb{R}) \mid \operatorname{det}(A) > 0\right} )

[gentle osprey]

dThirteen

[indigo mural]

I think a and b are closed, c and d are not closed

[restive flicker]

I think requiring the closure of B being a strict subset to C is enough to rectify that

[restive flicker]

Can you prove it is?

[dusky patrol]

Why do you always link stuff from mtaylor?

[prime elbow]

To show that if E is disconnected in M then there exists non-empty set A and B in M such that A intersection cl(B) is empty and B intersection cl(A) is empty and E = A U B.

Let E is disconnected so there exists non-empty set clopen U and V such that E = U union V and U intersection V is empty but on mathstack they argue that let A = U and B = V, but how A intersection cl(B) is empty?

They argue that cl(B) = B because B is closed in E, but B is closed in E not in M so how cl(B) = B?

M metric space

[echo oyster]

Ok thank you

[plush folio]

Yep 👍 the easiest way to prove it is to write the sets as the preimage of a continuous function

[tiny obsidian]

B is the complement of A, an open set

[tiny obsidian]

C = R^2 is homeomorphic to the (open) disc, as is an infinite strip B = R x (0,1) which has closure a strict subset of C. My argument continues to hold

(in fact my original counterexample also has cl(B) strict subset C, where closure is taken w.r.t. C)

[prime elbow]

But How A is open ?

[tiny obsidian]

because A is U, which is open?

oh, they even tell you B=V is clopen so in particular closed directly

[prime elbow]

But it is open in E

[tiny obsidian]

oh

[rancid umbra]

this was already resolved and pointed out to be wrong in the context of the convo

[tiny obsidian]

good point, I missed that. You can write $A=\operatorname{cl}(A)\cap E$ and $B=\operatorname{cl}(B)\cap E$ (basic-ish property of closures), and then using that show [A\cap\operatorname{cl}(B) = A\cap B] by using properties of intersections

[gentle osprey]

Edward II

[tacit basin]

yes

For a), the map from M_n(R) -> M_n(R) given by A^TA is continuous since it's a product of the transpose and identity functions which are continuous. Then note that O_n(R) is the inverse image of Id, a singleton (closed), and is closed

For b, c, and d, just note the determinant function is continuous and then use inverse images

[prime elbow]

Got it thank you

[indigo mural]

thanks, i will write it out and see that i understand

[tacit basin]

Do you know that for cts functions, inverse images of open sets are open and same for closed?

Thats what Im using here

[rapid osprey]

This one is fun

It's the slopes one right

[quartz horizon]

Yeah

[rapid osprey]

That one's great

I read a paper on it a month ago

Quite elementary too

[quartz horizon]

Mhm it is

Still not my favourite but I love how you don’t even need to go through Q

[rapid osprey]

What's ur fav

[quartz horizon]

It’s a variation on Bachmann’s construction

You encode real numbers as a pair of rational sequences a_n, b_n

a_n is strictly increasing and acts as a sequence of lower bounds

b_n is strictly decreasing and acts as a sequence of upper bounds

with a_n < b_n for all n, and b_n - a_n tending to 0

Two such pairs are identified if they overlap for all n

You do arithmetic with interval arithmetic

You do order by waiting for inequivalent sequences to become disjoint, and then using the order on Q

And you do completeness by proving a version of the nested interval theorem for sequences of open intervals

[rapid osprey]

Huh

Cool

I get a spirit of dedekind in this construction

[quartz horizon]

It’s inspired by the method of exhaustion

[quartz horizon]

I really like this proof btw

[quartz horizon]

You just consider such sequences but where a_n and b_n are real

And to prove completeness, you interleave a sequence with rational endpoints between your sequence with real endpoints

[quick crane]

It's something I read that covers the topic he requested. I link books by other authors too.

[quartz horizon]

After proving Q is dense in R

[rapid osprey]

Wait, so you make 4 sequences a_n', a_n'', b_n', b_n'' where a_n'<a_n<a_n'' and b_n'<b_n<b_n'', where the primed sequences are rational

[quartz horizon]

You start with a real endpoint sequence

And you only need a single rational endpoint sequence

It’s like, you want a_n < a_n’ < a_{n + 1}

Where primed is rational

Similarly b_{n + 1} < b_n’ < b_n

This is exactly what the density of Q in R lets you do

By your construction, the primed sequence will define a real number

It’s not hard to show that this real number is between a_n and b_n for all n

That’s your point in the intersection, and this proves the nested interval theorem (for open intervals, which is actually sufficient)

[rapid osprey]

I see

[rapid osprey]

If I have a metric space $(X,d)$, and a metric $d^\ast$ defined on a closed subset $A$ of $X$, and $d^\ast$ is topologically equivalent to $d\restriction_A$, can I extend $d^\ast$ to $X$, where the extension is also equivalent to $d$?

[gentle osprey]

aNDY

[rapid osprey]

My guess is almost definitely

Along the lines of Tietze-Urysohn

But how hard is this fact to prove

[prime elbow]

If M is metric space and connected, M has at least two points, then M is uncountable.

Any hint?

It is connected so any continuous map from M to discrete space is constant

[left eagle]

Is being closed or open a topological property? Im confused... because being open or closed only makes sense for subspaces of a topology?

[prime elbow]

I don't know much but since (0,1) is not closed in standard topology on R, but (0,1) is closed in its own subspace topology

[granite crane]

Suppose $x\neq y$ are two distinct points in $M$, so $d(x,y)>0$. Now you have to make partitions.

[gentle osprey]

Afzal

[prime elbow]

How?

[left eagle]

yeah thats what i was thinking but I think if we are only talking about sub-spaces of the same topology then it would be a topological property (i think)

[granite crane]

I mean there are uncountably many numbers in between 0 and d(x,y).

[prime elbow]

Yes then?

[granite crane]

Notice $A:={q\in M:d(p,q)< \delta}$ and $B:={q\in M:d(p,q)> \delta}$ are separated sets for any delta

Using this idea form a partition

[gentle osprey]

Afzal

[prime elbow]

We can map x to d(x,a), where a is fixed

So M maps to R

In non-constant continuous mapping

But how uncountable related to it?

[prime elbow]

What do you mean by topological property?

[red yoke]

Note that image of connected space is connected

[prime elbow]

And in R only intervals are connected

[red yoke]

Yea

So its image is uncountable

[granite crane]

oh cool thats an easy idea

[left eagle]

a property that is shared by all spaces homeomorphic to it

[prime elbow]

Yes

[granite crane]

This is how I did this problem

[robust drum]

No - it’s relative to an ambient space

[prime elbow]

So if A is countable then f(A) is countable?

Yes I think so

[tidal lynx]

You can also use the function provided by Urysohn's Lemma, and this extends the result to normal spaces

Another method: if $X$ is not uncountable, then for two different points $p, q \in X$ with distance $r = d(p, q)$, we have for the set
$$ S = {d(x, p)~|~x \in X} $$
that
$$ |S| \leq |X| < |(0, r)| ,$$
meaning there exists an $s \in (0, r) \setminus S$,
for which then $B_s(p)$ is clopen and proper, i.e. that $X$ is not connected.

[gentle osprey]

T Stepped

[granite crane]

wow there are bunch of ways

[paper wedge]

quick bad question

topologies of the form (a set is open iff it contains U) for some certain U

are always connected right?

as in no clopen sets?

just a quickie

[tender halo]

yes

[indigo mural]

Yes I had read that. So, this is what I think now: $O_{n}(\mathbb{R})={A \in M_{n} (\mathbb{R})\ |\ A^{t} A=Id}$
Let $f(A)=A^T\cdot A$, $f:\mathbb{R}^{n^2} \to \mathbb{R}^{n^2}$. Then $X \in f(A)$ has entries which are polynomials of entries of A. Each polynomial, viewed as a function from $\mathbb{R}^{n^2} \to \mathbb{R}$ , $f_{i,j}$ is continuous as for each open U in $\mathbb{R}$, $f^{-1}{i,j}[U]$ is open in $\mathbb{R}^{n^2}$ (why? because product and sum of reals is a real), and then as $f$ is made up of $f{i,j}$ then $f$ is continuous(...because product?). Again, $O_{n}(\mathbb{R})=f^{-1}[Id]$. So now if ${Id}$ is closed in $M_{n} (\mathbb{R})$ then $f^{-1}[M_{n} (\mathbb{R})-Id]$ is open in $M_{n} (\mathbb{R})$ and thus $O_{n}(\mathbb{R})$ is closed.

[gentle osprey]

dThirteen

[indigo mural]

You know, I suck at math, but I like it. So I picked up what I thought was easiest, Topology Without Tears, but I am getting confused many times when doing problems. This isn't my uni coursework, so I don't have anyone to talk about problems irl.

[gritty widget]

why open sets can't be compact?
like why is (0,1) not compact?
while (0,1) covers (0,1)

[quartz horizon]

Compact means that every open cover has a finite subcover

It’s not sufficient to just give an example of an open cover with a finite subcover

Concretely, $\bigcup_{n = 1}^\infty (0, 1 - \frac 1n)$ is an open cover of $(0, 1)$ with no finite subcover

[gentle osprey]

Pseudonium

[quartz horizon]

In R^n, the only compact set that is open is the empty set

[left eagle]

Is there a way to prove that 4 cannot be true (without using that weird theorem that states it?)

[quick delta]

There is an example such that 4 holds in n = 1

Hint: ||think about tan||

[left eagle]

what?

[quick delta]

There is a closed $X \subset \bR$ and non-closed $Y \subset \bR$ such that $X$ and $Y$ are homeomorphic

[gentle osprey]

Micose

[left eagle]

ye ik but i was sure that 4 was false (https://en.wikipedia.org/wiki/Invariance_of_domain)

[quick delta]

Neither Y nor X have to be open

[left eagle]

oh

[quick delta]

Concretely, the pair I’m thinking of are ||Y = [0, pi/2) and X = [0, \infty)||

[left eagle]

BRUHHhhhhhh

I forgot that closed sets arent just [a, b]

[quick delta]

Yeah, if X is bounded then you’ve (presumably) already proven this can’t happen in 3

(Because the proof I’m thinking of for 3 doesn’t need that Y is bounded)

[left eagle]

yeah when i was trying to think of examples I was thinking of like [0,1] U [2,3] U [4,5] U .... but i was like theres no way thats homeomorphic to a non-closed set

[prime elbow]

How?

In the book they gave a hint that find non-constant real valued continuous function on M

[tender halo]

dont need normality, Tychonoff-ness is enough

[restive flicker]

Good point

But there is a theorem about manifolds that says if you have a regular disk- a disk which has a neighborhood homeomorphic to a 2-ball (iirc) then the subtraction of the disk from the neighborhood is homeopathic to an anulus

Which is weird given the example you gave me so that probably isn't an actual theorem

Can you prove without this "fact" that the connected sum of two manifolds is also a manifold?
And are all possible connected sums of connected manifolds homeomorpic?

[restive flicker]

Actually maybe if we require cl(B) to be sub to C and the boundaries of B and C to be homeomophic to spheres

Unless you're first example disproved that

[robust drum]

Do you have specific questions? People here are happy to help

[robust drum]

I actually looked this up on stack exchange and I think you need the theorem you mentioned - the annulus theorem - to prove that connect sum is well defined in general

It’s apparently pretty hard to prove

That said, someone also said that if instead of taking general connect sums you restrict yourself to connect sums where you only allow yourself to remove coordinate balls in differential manifolds, it’s supposed to be much easier

[restive flicker]

What are coordinate balls?

What I called "regular disks"?

[robust drum]

A coordinate ball is a subset of an n manifold homeomorphic to a Euclidean n ball

[rapid osprey]

bumpies :)

[tender halo]

uhh sure i think

[restive flicker]

Wasn't that what I was talking about (up to the non negligible "subset of a manifold")

[robust drum]

Hmm good point. I think maybe I’m misremembering the stack exchange then

https://mathoverflow.net/a/121642

Ok this is what I was thinking of

The detail is that you already assume you have some fixed charts on your manifolds and then a coordinate ball is a ball contained inside some chart who’s image is a standard ball under that map

[tender halo]

http://matwbn.icm.edu.pl/ksiazki/fm/fm77/fm77118.pdf

there are several links in the references for other proofs

[restive flicker]

I was asking in the non differentiable case

But would be looking for a theorem that says there's some automorphism mapping every ball on a manifold to any other ball (ball on a manifold being something homeomophic to a euclidean ball)

But originally was asking about there being a homeomorpism from the subtraction of balls (with probably a few extra conditions) to the anulus

[restive flicker]

Basically the thing you said was a theorem but for connected manifolds instead of differentiable ones

[rapid osprey]

Saw this. Was hoping there was a more elementary proof, but it's okay

I'll just sti down and go thorugh it

[robust drum]

it looks like the author of that stack overflow is working in Top not Diff so should be what you're looking for

[restive flicker]

Cool thx

Wonder if the anulus theorem proves the stuff I want to show

[warm hedge]

Lets say we have an equivalence relation in a topological space X and [x] are the equivalent classes, which are G_delta sets. what exists in the closure of [x] - [x] ?

like there are some y which arent equivalent to x but what about [y] ? $[y] \cap [x] = \emptyset$ but $[y] \cap \overline{[x]} \neq \emptyset $ ?

[gentle osprey]

whitefang

[red oxide]

are we given specifically what the equivalence relation is?

[warm hedge]

in the context i am seing it the e.r is an orbit one from the action of a polish group to a polish space X

[umbral panther]

An arbitrary equivalence relation? Anything could happen
Say R and x ~ y if they’re both zero or both nonzero

Which is the orbits of R* on R

But what if the group is Q with the discrete topology acting by addition on R. Those equivalence classes are awful

[warm hedge]

I am asking this because for example i have, for given $x \in X$, this set $Y = { y \in X : \overline{[y]}_G = \overline{[x]}_G }$ and i wonder, wouldnt this set jist contain all the elements of [x] ?

[gentle osprey]

whitefang

[warm hedge]

so Y = [x]_G ? or i am missing something

[umbral panther]

I gave you a counter example

[bronze wadi]

just to verify something, if f is continuous and f(A) is a subet of B, then is the image of the closure of A a subset of the closure of B?

[tiny obsidian]

yes, because $f(\overline A)\subseteq\overline{f(A)}$ if (and only if) $f$ is continuous

[gentle osprey]

Edward II

[bronze wadi]

yeah thats an equivalen statement basically but i forgot why its true

[tiny obsidian]

$f^{-1}\left(\overline{f(A)}\right)$ is closed and contains $A$. Other details left to you

[gentle osprey]

Edward II

[bronze wadi]

ah ok yeah makes sense

thx

[prime elbow]

A subset of a metric space (X,d) is said to be totally bounded if for each e >0 there exists x_1,.., x_n such that S \subset \Cup_{i = 1}{n} B(x_i, e).

How to show in R, [a,b] is totally bounded?

My work:

If 2e> b-a, then take any point in [a,b] such that around its ball contains the entire [a,b].

Geometrically I understand how it works but I don't know how to write it.

[paper wedge]

how do u understand it geometrically

this kind of argument is

like u see alot

okay firstly i would guess ur definition of compactness is sequential compactness

that would make more sense in for ur "geometric" intuition no?

@prime elbow

[prime elbow]

Like if you give me an epsilon then I have the choice to choose finite x_i such that union of all open ball of x_i of radius epsilon covers [a,b]

[prime elbow]

They haven't introduced yet

[paper wedge]

okay

[paper wedge]

okay then whats the problem

[prime elbow]

How to write?

[paper wedge]

suppose A is compact

what do u want to show

[prime elbow]

Sorry but they haven't introduced compact yet, but I know compact stuff little bit

[prime elbow]

A is any set of R? If yes then it is bounded and closed

[paper wedge]

okay do u know how [a,b] has like this property where

every sequence of points in [a,b] has a convergent subsequent

kinda bolzano-weirestrass?

do u know that

[prime elbow]

Yes

[paper wedge]

okay

[prime elbow]

[a,b] is compact

But I don't want to use that

[paper wedge]

okay forget about compactness now

[prime elbow]

Yes

[paper wedge]

this is called sequential compactness but like

whatever

so now

we will go for the contradiction

[prime elbow]

Sequential compactness? Definition?

[paper wedge]

a subset is sequentially compact iff it every sequence has a convergent subsequence

but we are just using bolzano-weirestrass from basic real analsyis no?

[prime elbow]

Yes

[paper wedge]

okay

[prime elbow]

Thank you

[paper wedge]

okay like geometrically its like

if u know every sequence has a convergent subsequence

really imo when u try to like find out why something is true u wanna draw a picutre where the shit doesn't work

so it's like someway

being not totally bounded would fuck up the bozlano weirestrass theorem

in ur opinion, how can we do that like

[prime elbow]

So [a,b] is sequential compactness

[paper wedge]

in general what do we have to do to contradict bolzano weirestrass theorem

what do we have to show exists

[prime elbow]

To show there is a sequence which has no convergent subsequence

[paper wedge]

yes exactly bro

so for now we will assume the contradiction

[prime elbow]

Okay

[paper wedge]

that is [a,b] is not totally bounded

and from this we must somehow CONSTURCT a sequence

that in no fucking way has a convergent subsequence

like just no way

can u try to do that

[prime elbow]

So if A is sequential compact => A is totally bounded?

[paper wedge]

this kind fo argument like keeps showing like i remember seeing it in functional analysis proving the like S^n is never compact in inf dim space

literaly same argument but u use riez lemma

[paper wedge]

yeah in

R

[paper wedge]

.

[prime elbow]

I need to prove this, because if we have this result then [a,b] is totally bounded and any bounded set is totally bounded in R

[paper wedge]

now that u have a plan just try to fuck around with it

like unfold definitions see what u have

[prime elbow]

Okay

[paper wedge]

bro isnt this what we are trying to do

[prime elbow]

Yes

[paper wedge]

ok

now u want to try tro construct a sequence

that surely never has a convergent subsequence

by assuming that [a,b] is not totally bounded

and hence contradicting bolzanto

[prime elbow]

Why [a,b], we can generalize it

[paper wedge]

u can try to think about how such a sequence should look like by just unfolding the defintion of convergence

[paper wedge]

yeah

[paper wedge]

bolzanto lmao

[prime elbow]

I am stuck at constructing the sequence 😭

Yes A is not totally bounded so there exists epsilon such that for all n in N there is no x_i such that A contained in union of open balls x_i of radius epsilon

[paper wedge]

okay

[prime elbow]

But how to construct sequence

[paper wedge]

we know that

there exists an epislon > 0 such that A can not be covered by finitely many epislon balls

so if x is in A then there surely must be some epislon ball around some other element x_i such that x is not in this x_i , hence d(x,x_i) >= epislon

so for example suppose that A is not covered by B_e(x_1) U B_e(x_2) U B_e(x_3) U ...

[prime elbow]

And this epsilon is fixed, right?

[paper wedge]

yes

it is the epislon choice from the contradiction assumption

[prime elbow]

Yes

[paper wedge]

the fact that A is not contained means that there are elements in A that are not in any of the balls

choose these elements

the fact that they are not in the balls will give us d(x_i,x) >= epi

so can u finish

it

[prime elbow]

Wait

[prime elbow]

I am not sure about it, what if there exists x in A such that it is always contained in epsilon ball of every element y

[paper wedge]

yeah i worded out correctly but we really dont care about those

u can find points x_i such that x_i are not in any of the balls

if u define x_1 to be the point in A but not in the first ball

and x_2 to be the point in A that is not in the first ball nor the second ball'

and so on

u get ur sequence

the balls being centereda round x_is

so more clearly

or yeah i fucked the indices up

x_1 is in A

x_2 is in A-B_e(x_1)

x_3 is in A-(B_e(x_1) U B_e(x_2))

and so on

these exist cuz i know A cannot be covered by the union of these balls

so there is a set of points that are not in any of them

[prime elbow]

Yes

[prime elbow]

So we have sequence x_1,x_2,...,x_n,..

[rancid umbra]

hint: [a,b] is compact. choose a suitable open cover and apply compactness

[prime elbow]

Yes compactness implies totally bounded

But I don't want to use compactness

[prime elbow]

Got it, thank you ❤️

[rancid umbra]

you can kinda make this constructive without appealing to any compactness arguments (sequential/limit point compact, etc)

for simplicity, wlog [a,b] = [0,1] (otherwise just dilate and translate, the same argument will go through)

for a natural number n, partition [a,b] uniformly by
0 < 1/2n < 2/2n < … < (2n-1)/2n < 1

now B(k/2n, 1/n) for k = 0,…,2n covers [0,1]

given ep > 0, choose n such that 1/n < ep

for the corresponding partition above, B(k/2n, 1/n) subset B(k/2n, ep)
@prime elbow

[prime elbow]

Thank you ❤️

[prime elbow]

Totally bounded implies bounded right?

They define compactness in metric space as sequential compactness, yes both are equivalent but they haven't been introduced yet.

I need to prove that let (X,d) be a compact space then it is bounded.

Since it is compact so it is totally bounded and totally bounded implies bounded.

The problem book hasn't been introduced totally bounded so can I use that or go through bounded?

[quick delta]

I mean just doing the proofs "Compact => Totally bounded" and "Totally bounded => bounded" in sequence is probably what I'd have done anything

Even without saying the words "totally bounded"

[prime elbow]

Yes

[opaque scroll]

Or just unbounded => not compact directly

[prime elbow]

But compactness=> totally bounded by contradiction

[opaque scroll]

Like balls of increasing size will give an infinite cover that can't be reduced to a finite one.

[prime elbow]

Sequential compactness of A => A is complete, A is metric space

Right?

[quartz horizon]

I guess the way I might do this now is

If A is compact, so is A x A

[prime elbow]

Oh

[quartz horizon]

Then use continuity of the metric

[prime elbow]

I need to prove that result

[quartz horizon]

I think you can prove it with sequential compactness, no?

You take a subsequence twice

[prime elbow]

In general metric space if (x_n,y_n) converges to (x,y) then x_n converges to x and (y_n) converges to y, ?

In R^n it is true

[quartz horizon]

Yes, and vice versa

The various product metrics all let you “package” and “unpackage” convergent sequences

That’s one thing they have in common

[prime elbow]

It is somehow related to Bolzano Weierstrass theorem

[quartz horizon]

Mhm

[prime elbow]

In R^n it is true

But for arbitrary metric space?

[quartz horizon]

Yes

With any standard notion of product metric

[prime elbow]

And this is what we want to prove

[quartz horizon]

Ofc there are metrics you can put on the Cartesian product which make this false

But the standard ones are, like

[prime elbow]

A is compact then A×A is compact

[quartz horizon]

d_1, d_2, d_infty

[gaunt linden]

Beware though that it is not necessarily true for infinite products.

[quartz horizon]

Mhm

Though countable products of metrizable spaces are metrizable

And in that the case the countable product still has this packaging property

E.g. this is why the Hilbert cube is metrizable

[prime elbow]

Is this if and only case?

[quartz horizon]

Hm, interesting question

The answer is yes

[prime elbow]

I think we can because since A is compact now if you take sequence in A×A then component wise we have sequence x_n,y_n then it has subsequence which is convergent, so (x_n,y_n) has subsequence which is convergent.

Yes this is not so good proof but is it correct?

[prime elbow]

Because we can treat x_n sequence in A as (x_n, y) in A×A ?

[quartz horizon]

Some variation of that shows that A compact => A x A compact

[quartz horizon]

That’s one way to do it, yeah

Another way to do it is use that the continuous image of a compact space is compact

And use one of the projections A x A -> A

[prime elbow]

I see

[quartz horizon]

Either are fine, honestly I like your way because it’s more elementary

[prime elbow]

But I like when continuity comes

[quartz horizon]

Yeah cause then it’s like

You use the fact that d : A x A -> R is continuous

So if A x A is compact

Then d must be bounded

[uneven bronze]

I have tried googling this result, but haven't actually found a decent proof out there. I was wondering how one goes about showing that the closure of an open ball in Euclidean space is the closed ball? Or does this follow from separable-ness/second-countable-ness? I have read a hint that one could consider convex functions on a normed vector space (i.e $f(x)=\lVert y-x\rVert$ and $\mathbb R^n$ ). Then $$\mathrm{cl},{x\in X\mid f(x)<c}={x\in X\mid f(x)\le c}.$$I still do not see though why the above equality should hold?

[gentle osprey]

psie

[ionic bay]

This actually holds for all normed vector space

What have you tried?

The nice thing about vector spaces is that you could just draw a line connecting two points and rescale them and shift them around

[prime elbow]

Because in R compact set is bounded

[quartz horizon]

I was more using the result that continuous functions on compact spaces are bounded

But sure, that works too

I like the way you phrase it though

[gaunt linden]

As an elementary approach, I would just show that 1) every vector with |v|=r is in the closure of the open ball, 2) every vector with |v|>r is not in the closure of the open ball.

[rancid umbra]

another approach:
cl(B) = int(B) U bd(B) = closed ball so all you have to do is figure out the interior and the boundary

[prime elbow]

Thank you, you helped me a lot ❤️

[ionic bay]

Also a sequential approach:
For x in the closure, take a sequence in the ball converging to x and use triangle inequality. In the other direction, if |f(x)|=r, draw a line connecting y and x and take a sequence of point along that line converging to x. This implies the inclusion in the other direction. If |f(x)|<r the inclusion is trivial

[gaunt linden]

If |v|>r, then the open ball around v of radius |v|-r is disjoint from the original ball (because of the triangle inequality). This shows that v is not in the closure of the original ball.

[prime elbow]

I am interested, is there any way to prove d: M × M -> R is continuous by sequence?

[quartz horizon]

Yes, once you fix a metric on M x M

[uneven bronze]

Thanks for the hints I think I have proved it now

[quartz horizon]

Oh hmm, actually

For that particular case it might be a bit fiddly

But I’m sure it’s possible

[prime elbow]

Fix?

[quartz horizon]

Given a metric on M

There are many choices for metrics on M x M

[gaunt linden]

The metric is what he wanted to prove continuous in the first place.

[quartz horizon]

Right, but what I mean is

To say that d : M x M -> R is continuous

You need to make M x M a metric space

Unless you’re viewing them as topological spaces

In which case I suppose you can use the product topology

[quartz horizon]

The standard ones I know are the d_1 metric, the d_2 metric, and the d_infty metric

Any d_p metric for p in [1, infty] essentially

[prime elbow]

Yes

[quartz horizon]

These all give different, but homeomorphic, metric spaces

[gaunt linden]

More primitively, a metric is continuous (with respect to itself) in each of the two arguments separately.

[quartz horizon]

Right, sure, currying and all

Separately continuous

Being jointly continuous is a little stronger

[gaunt linden]

Right. But separate is all we need for eg showing that a closed ball is indeed closed, being the preimage of the closed interval [0,r].

[quartz horizon]

Wait was that the original goal

I came in the middle

[gaunt linden]

I'm not sure, that's more of a guess on my part.

[quartz horizon]

Sure sure

Yeah separately continuous is a lot easier

I think you just show it’s 1-lipschitz

You can also generalise it to “distance from a subset” I think

Which is 1-lipschitz by a similar argument

d(S, -)

[gaunt linden]

We can then show in general that a separately continuous function on a finite product is metrically continuous wrt either of the usual product metrics.

[quartz horizon]

Ooh, interesting

Huh, so a separately continuous function on R^2 is jointly continuous

Did not expect that to be true

[gaunt linden]

Hmm, no that sounds wrong. I must be talking nonsense.

[quartz horizon]

I can’t exactly recall a counterexample rn

But yeah it does feel like a pretty strong result

Hence why I’m a little suspicious

Maybe you can take one of those functions where partials exist but is not differentiable

[gaunt linden]

xy/(x²+y²) with 0 at the origin is a counterexample.

[quartz horizon]

Or that yeah

Neat

I just find that “metric is continuous” is a cute result

[warm hedge]

sorry i dont get it

[tender halo]

a metric is continuous iff it generates a coarser topology

i think

[quartz horizon]

Wdym?

[tender halo]

a pseudometric d on the underlying set of the topological space X is a continuous function X x X -> R iff it generates a topology coarser than X

[quartz horizon]

Right but this is true for all functions

The preimage of a topology is a topology

[tender halo]

hmm

[quartz horizon]

So a function f : X -> Y is continuous iff it generates a coarser topology

[tender halo]

yeah but here you have a function X x X -> Y

and not X -> Y

[quartz horizon]

Same thing

X = M x M

Y = R

[tender halo]

yeah but it generates topology on X x X

not on X

[quartz horizon]

This is just a rephrasing of the continuity definition

Ugh ok look

If you have a function Z -> Y

It’s continuous iff it generates a coarser topology

Now set Y = R and Z = X x X

[tender halo]

yes

i know that

how do you then go from a topology on X x X to a topology on X

[quartz horizon]

I wasn’t suggesting that

[prime elbow]

If E is the set of all x in [0,1] such that whose decimal expansion contains only the digit 4 and 7.

But 0.48 is the limit point of E and it is not in E so how E is closed?

Yes 0.477777777777........ = 0.48

[tender halo]

"generates a topology coarser than X" here means generates as a pseudometric

[quartz horizon]

Oh that’s what you meant

I see

[tender halo]

i really hope 0.47777777... \neq 0.48

[prime elbow]

But every open ball of 0.48 intersects with E

Why E is closed?

[quick delta]

What about the one of radius 0.001

Everything there is either 0.48XXXX or 0.479XXXX
So doesn’t consist of only 4s and 7s

[tender halo]

0.47(9) = 0.48, not 0.4(7)

[prime elbow]

My mistake

[prime elbow]

Why?

[tender halo]

uhh

idk thats how numbers work

[prime elbow]

Got it

My mistake

Thank you ❤️

[rancid umbra]

check yourself using the formula for the geometric series

or note that 0.077777 is just 7/9 * 1/10 = 7/90 since x/9 = 0.(x) for 1 <= x <= 9

[prime elbow]

Yes

[unreal stratus]

0.477777... < 0.478

[uneven bronze]

I have to ask a similar question I asked recently. I'm reading Gamelin and Greene's book on topology, and in a solution to an exercise, they claim that the closure of open balls in R^n are closed balls. But this depends on which metric we equip R^n with, right? This statement is only true if the metric is derived from a norm I think.

[prime elbow]

Yes I think so

[tender halo]

yep

the closure of the open ball is contained in the closed ball of the same radius

but doesnt have to equal it

[prime elbow]

In normed vector space, closure of B(0,1) is closed ball around 0 of radius 1.

[prime elbow]

But in normed vector space?

[uneven bronze]

Ok, thanks. I can prove the claim, but I can not tell what property of a norm is key in proving this. Or put differently, why does it fail for some metric spaces equipped with certain metrics.

[prime elbow]

Norm has an interesting property that B(x,r) = x +rB(0,1)

[uneven bronze]

indeed

[tender halo]

take a metric bounded by 1 corresponding to any metric

the open ball of radius 1 will be well just some ball, and the closed ball will be the whole space

[uneven bronze]

ok

[prime elbow]

But how can I generalize to B(x,r) ?

[tender halo]

the property of the norm used is that uhh the distances are kind of uniform?

i guess

[uneven bronze]

I think the key property is that $\lVert kx\rVert=|k|\lVert x\rVert$, maybe that is what you meant

[gentle osprey]

psie

[tender halo]

the intuition that breaks in the general case is that the distances that are far away might look different when you are closer

in some sense, metrics are "local"

[alpine nest]

Yes; although as a reminder, if people talk of topological notions (such as openness, closure, continuity etc.) in R^n or C^n, and they don't specify the topology, then the topology is Euclidean.

In the antidiscrete topology the closure of any (nonempty) set is the whole space

[prime elbow]

How to show sequentially compact implies compactness in metric space?

I understand book proof but is there any other way which is an interesting one ?

[gritty widget]

What is the way your book proved it?

[tender halo]

sequentially compact spaces are countably compact (trivial)

countably compact metric spaces are lindelof

lindelof countably compact spaces are compact (definition)

[prime elbow]

They haven't introduced yet

[tender halo]

what

[prime elbow]

Tao II

[prime elbow]

They haven't introduce Lindelof space yet

[tender halo]

they won't probably

[prime elbow]

But I learnt Lindelof space in some blog

[tender halo]

lindelof spaces are spaces where every cover has a countable subcover

[prime elbow]

What is the definition?

Oh

And countably compact ?

[tender halo]

every countable cover has a finite subcover

so compact spaces are precisely lindelof countably compact spaces

[prime elbow]

Yes

[prime elbow]

How?

[tender halo]

uhh a metric space is second-countable iff every discrete subspace has at most countable cardinality

from countable compactness you can prove that every discrete subspace is finite

so its second-countable, and therefore trivivally lindelof

[tender halo]

you can try proving that one on your own

its a good exercise

[prime elbow]

Okay

So second countable definition?

[tender halo]

has a countable base

[prime elbow]

Okay thank you

And atmost countable means there exists injective function from Y to N

[tender halo]

sure

either finite or countable

[prime elbow]

Yes

[west brook]

Is atmost supposed to be at most or almost?

[prime elbow]

Infinite countable or finite

Actually countable used for infinite set

[tender halo]

its actually usually unclear

some use countable for "finite or countable" and denumerable for just infinitly countable

so really whatever the author decides

[uneven bronze]

I am working perhaps with an unusual definition of boundary. Let $E\subset X$, where $X$ is some metric space, then $$\partial E=\overline{E}\cap(\overline{X\setminus E}).$$ I am stuck at showing that $\partial E\subset E\implies E\text{ is closed}$. Naturally, I'd like to show somehow that $E=\overline{E}$, but how?

[gentle osprey]

psie

[tender halo]

thats one of the usual definitions

[uneven bronze]

good 🙂

anyway, I think I know how to prove it

by contradiction

[gritty widget]

You have to show $\bar E \subseteq E$. Let $x\in \bar E$. If $x \in \overline{X\setminus E}$ then use the hypothesis. Else there's a neighbourhood of $x$ that doesn't intersect $X\setminus E$ therefore..

[gentle osprey]

Bequi

[uneven bronze]

👍

[gritty widget]

I'm curious how you used contradiction though, if you'd like to share

[uneven bronze]

yes, I'll type it

[uneven bronze]

Ok. Suppose for a contradiction that we can find an $a \in \overline{E} \setminus E$. As $a \notin E$, $a \in X\setminus E \subseteq \overline{X \setminus E}$. On the other hand we know that $a \in \overline{E}$. So $a \in \partial E\subset E$, a contradiction.

[gentle osprey]

psie

[gritty widget]

Nice

[prime elbow]

Let (X,d) be metric space and K_1,..,k_n,.. are non-empty compact subset of X such that K_1 contains K_2, K_2 contains K_3,...
Let I be index set.
Now I want to prove that intersection of K_i over I is non-empty.

Now my idea is: Since all are compact spaces let open covering of K_1 then there exists a finite open covering of K_1, so that covering is also covering of k_i, i in I.

Intuitively, there exists open set such that it contains all K_i.

Is it correct?
I don't want a hint

[opaque zodiac]

What you said is true but I don't see how it leads to showing that the intersection is non-empty

[prime elbow]

I see

[quartz horizon]

I’m not sure why you introduce an auxiliary index set, isn’t it just the naturals?

[prime elbow]

Yes natural

[uneven bronze]

I have a very basic question. I know of the statement that if $A\subset B\subset X$ and $A$ is dense in $B$, and $B$ dense in $X$, then $A$ is dense in $X$. However, I am in a situation where I have $A$ dense in $X$ and we don't have any information about $B$ except that $A\subset B\subset X$. Is $B$ dense in $X$? It seems intuitively obvious, but I struggle showing it symbolically.

[gentle osprey]

psie

[quartz horizon]

Yes, this is because closure is monotonic

[alpine nest]

The closure of B is a superset of the closure of A

And the closure of A is the whole space

[uneven bronze]

ok 👍

[uneven bronze]

I am trying to understand the Baire Category theorem (BCT); I know the statement of the theorem and various other versions. Then it is stated in my book that

"Single-point subsets of the real numbers are nowhere dense, and the uncountability of the real numbers is just the fact that countable unions of such single-point sets cannot be the whole of the real numbers."

Why not? What would contradict the BCT if it were?

[keen seal]

Man, is the definition of continuity through preimages of open being open equivalent to preimages of closed sets being closed?

the direction => is known and usually used

But I suppose so

[gritty widget]

Yes, inverse images commute with complement

[keen seal]

noice, thanks

glad I didn't have to grab paper

[quartz horizon]

Yeah inverse image commutes with every subset operation

Union, intersection, complement, set difference, symmetric difference, …

[alpine nest]

The reals are a complete metric space, therefore they can't be a union of countably many nowhere-dense sets (such as singletons)

[uneven bronze]

I still do not quite understand. One version of the BCT states that if $(E_n)$ is a sequence of nowhere dense subsets of a complete metric space $X$, then $\bigcup_{n=1}^\infty E_n$ has empty interior. So in turn, we must show that $\mathbb R$ does not have empty interior. Maybe that is super obvious, but how does one show the interior of $\mathbb R$ is non-empty?

[gentle osprey]

psie

[tender halo]

what do you think the interior of R is

[uneven bronze]

not sure, but all of R perhaps? But that would mean the interior would be a closed set, which is bonkers

[tender halo]

why is it bonkers

it is both an open and a closed set

just like the empty set

but yes, the interior of R is in fact the whole of R

[uneven bronze]

ok, from what I've learnt, the interior is an open set. Didn't know it can be both open and closed.

[tender halo]

well the interior of any clopen set is itself

just like its closure

[uneven bronze]

ah ok, that makes sense

[alpine nest]

The interior of every open set is itself. If that open set happens to also be closed (as is the case when your open set is the whole space), then the interior will also happen to be closed.

Daily reminder that open vs closed is not a dichotomy; a set can be open, closed, both, or neither.

[uneven bronze]

ok 👍 good reminder!

[alpine nest]

For a more "low-level" argument that the interior of \bR is nonempty: 0 is an element of R, and the open ball (-1,1) is an open subset of \bR, which contains 0. Therefore 0 is in the interior of \bR (as is every other element of R, but I was going for a very explicit argument)

[gentle fjord]

How to prove that, every compact metric space has a countable base?

[fringe thorn]

are you allowed to use the result that a metric space is 2nd countable (has a countable base) iff it is Lindelof (every open cover has a countable subcover)?

[gentle fjord]

No

[red yoke]

A compact metric space is totally bounded

This gives you the countable base

[quartz horizon]

Ah right, you union all the finite epsilon-covers as epsilon ranges over, say, positive rationals?

[alpine nest]

Or even just reciprocals of natural numbers

[quartz horizon]

Mhm that’s sufficient too

[alpine nest]

I prefer that approach since it allows for a nice enumeration

[quartz horizon]

Though since positive rationals are countable, I was happy to be a little greedier

[red yoke]

Wait what is this replying to

[quartz horizon]

your comment?

maybe what I said doesn’t work

I thought it would

[red yoke]

This is if you know the space is separable

[quartz horizon]

Hm

Then I don’t immediately see how total boundedness gets you a countable base

I guess I’d have to think about it

[red yoke]

Wait no I misread it

Yes

[alpine nest]

The finite covers using epsilon balls are exactly what you get from total boundedness (or, in the same way, from compactness)

[quartz horizon]

Yeah exactly

[warm hedge]

i am trying to do this exercise

so i just need to so that the function in the hint is borel

Lets take the function $f : X \to X$ that maps $x \to \overline{[x]}_G$. To see that its Borel lets take $A\subseteq X$ to be open. We need to so that $f^{-1}(A)$ is borel. $f^{-1}(A)= { x \in X : f(x)\in A} = { x \in X : \overline{[x]}_G \in A}$. So if $A \subseteq \overline{[x]}G$ then $f^{-1}(A) = \emptyset$ which is borel. If else then $f^{-1}(A)$ will be union of equivalent classes, that means union of $G\delta$ sets. If the union is countable then its borel.

[gentle osprey]

whitefang

[warm hedge]

i guess until here its correct

how can i show that the unions are countable and not uncountable?

I guess it has to do with the fact that every orbit is Gδ

[prime elbow]

Let (X,d) be a metric space, let E be a non-empty compact subset of X and let x_0 be a point in X. Show that there exists a point x in E such that d(x_0,x) = inf{ d(x_0,y) | y in E }.

They give a hint : let R = inf{ d(x_0,y) | y in E }. Construct x_n in E such that d(x_0,x_n) ≤ R + 1/n.

Now use the compactness of E there exists subsequence of x_n which converges in E, say x is limit point of that convergent subsequence.

Now R ≤ d(x_0,x) because x in E.
For e> 0 such that d(x_n_k, x) < e so d(x_0,x) ≤R+ 1/n_k + e by using limit we have d(x_0,x) = R.

Is it correct?

[opaque zodiac]

Yes

[prime elbow]

I think there is a problem

For each e> 0 the x_n_k depends on e so d(x_0,x) ≤ R + 1/n_k + e how it imply that d(x_0,x) ≤ R?

[opaque zodiac]

Oh you have it for every positive e

[prime elbow]

So what do I now?

[opaque zodiac]

That tells you that d(x_0,x) \leq R + e for every positive e, so d(x_0,x) \leq R

[prime elbow]

So my proof is correct?

[opaque zodiac]

Well I would phrase the last line a little bit differently

to emphasize the fact that what's important it that it works for any e

[granite crane]

If you don't mind, can I ask which book is this?

[prime elbow]

Yes but how do I modify my proof ?

[opaque zodiac]

I mean it kind of depends on how precise you want to be it does need more. You're glossing over the details of how you know that these inequalities work

For starters you don't actually know how fast the sequence is converging to x.

I might say something like this

[prime elbow]

I am confuse because I showed that d(x_0,x)≤ R + 1/n_k + e, maybe I can take e > 1/n_k such that it will be d(x_0,x) ≤ R + e but I am not sure

[opaque zodiac]

Fix a sequence $(x_n)$ of elements of $E$ such that for each $n$, $d(x_0,x_n) \leq R + \frac{1}{n}$. By compactness we can find a convergent subsequence $(x_{n_k})$. Let $x$ be the point that this sequence converges to. Since $E$ is compact (and therefore closed) $x$ is in $E$.

Fix an $\varepsilon > 0$. Since $(x_{n_k})$ converges to $x$, we know that for sufficiently large $k$, $d(x_{n_k}) < \frac{1}{2}\varepsilon$. Furthermore, for any sufficient large $k$, $\frac{1}{n_k} < \frac{1}{2}\varepsilon$. Therefore, by the triangle inequality, we have for any sufficiently large $k$ that
[
d(x,x_0) \leq d(x,x_{n_k}) + d(x_{n_k},x_0) \leq \frac{1}{2}\varepsilon + R + \frac{1}{n_k} \leq R + \varepsilon.
]
Since we can do this for any $\varepsilon > 0$, we have that $d(x,x_0) \leq R$. Finally, since $x \in E$, we have that $d(x,x_0) \geq R$. Therefore $d(x,x_0) = R$.

[gentle osprey]

Exomnium

[opaque zodiac]

yeah that's the idea

the thing about these kinds of proofs is that there's a lot of different ways of writing them and there's a lot of different levels of rigor that someone might be looking for

[prime elbow]

Got it, thank you ❤️

Actually I thought the same way but you helped me a lot ❤️

[warm hedge]

Su Gao's invariant descriptive set theory

[uneven bronze]

Consider the metric space $(X,d)$ and let $S$ be the set of Cauchy sequences in $S$. Define the equivalence relation $\sim$ in $S$ by declaring $(s_n)\sim(t_n)$ to mean that $d(s_n,t_n)\to0$ as $n\to\infty$. \

Let $\tilde{X}$ denote the set of equivalence classes of $S$ and let $\tilde{s}$ denote the equivalence class of $s=(s_n)$. Then $\rho(\tilde{s},\tilde{t})=\lim_{n\to\infty} d(s_n,t_n)$ for $\tilde{s},\tilde{t}\in\tilde{X}$ is a metric. \

I need to show that, for $x\in X$ and $\tilde{x}$ denoting the equivalence class of the constant sequence $(x,x,\ldots)$, the function $x\mapsto \tilde{x}$ maps $X$ onto a dense subset of $\tilde{X}$. I can show that it maps $X$ to a dense subset, but how would I go about showing the mapping is surjective?

[gentle osprey]

psie

[uneven bronze]

If you're interested, I can show why the map maps X to a dense subset of tilde X, maybe it'll clarify things.

[tender halo]

huh

wdym how do you show the mapping is surjective

you need to show that the image is dense

a map is of course surjectrive onto its image

[uneven bronze]

ah, ok, that clears up the confusion I think. Thank you!

[gentle osprey]

psie

[uneven bronze]

I have received a hint to consider the map that takes an equivalence class of a sequence in tilde X and maps it to a limit of that sequence in Y. Why would this map be an isometry? I am not sure what the metric on Y is, hence I'm not sure how to verify that it is an isometry.

[tender halo]

the inclusion map has to be an isometry of course

(otherwise the statement is not true)

as X x X is a dense subspace of X tilda x X tilda, the metric d as a continuous function from X x X -> R has at most one extension to X tilda x X tilda

so the inclusion isometry has only one extension

[uneven bronze]

hmm, I have a hard time understanding this, why would the statement be false if the inclusion from X to Y were not an isometry?

[prime elbow]

I think this statement is not true because you can take f: R to R by x->0, but f^-1{0} is not finite

[tender halo]

well the metric on Y has to be connected to the metric on X somehow

otherwise you would get that any two metrics are isometric, which is patently not true

[tender halo]

besides generating the same topology, that is

take X to be already complete

[gritty widget]

I agree with you

It should be say "either finite or the whole space" instead, I believe

[prime elbow]

Yes

[uneven bronze]

Why does the metric d as a continuous function have only one extension? You seem to be using some theorem or proposition which perhaps is not known to me (yet).

[tender halo]

a continuous function from a dense subspace of X to Y has at most one extension to a function from X to Y

should be obvious why

[uneven bronze]

hmm, I still do not see how to verify that the map from tilde X to Y is an isometry (that is, the one that maps an equivalence class of a sequence to the limit of that sequence). It feels like one needs to know the metric on Y for this.

[tender halo]

the point is that once you fix a metric on a dense subspace there is at most one possible metric on Y

[uneven bronze]

ok, so why introduce Y if tilde X is already a complete space that contains X? I have worked a couple of exercises before this one, where I have shown that tilde X is complete and that X gets mapped to a dense subset of tilde X under the map that takes a point x in X and maps it to the equivalence class of the constant sequence (x,x,...). I feel like what you are saying is the very thing I need to prove 🙂

[tender halo]

because you need to prove a complete space that contains X as a dense subset exists

that was what the construction with the Cauchy sequences

the thing you are proving now is that there is only one completion up to isomorphism

[uneven bronze]

right, ok

[gentle fjord]

Suppose X is a metric space with a countable base.

How do I show the every open covering of X has a countable subcover(Lindelof)?

[gritty widget]

Let {B_n} be a countable base for X and {A_v} be an open cover. For each B_n, choose an A_v such that B_n is contained in A_v if possible, obtaining a subset of {A_v}. Clearly such a subset is countable. Can you see it is a cover?

[gentle fjord]

Yeah, I got it

Can someone please check my proof for problem 27

[gentle osprey]

psie

psie

[uneven bronze]

How can I verify that f is an isometry? (I feel like I am missing a key connection between f and the embedding map)

[tiny obsidian]

To verify that f is an isometry, all you need to do is check that for (equivalence classes of) sequences $s,t$ we have $d(f(s),f(t)) = \rho(s,t)$.

[gentle osprey]

Edward II

[tiny obsidian]

Or in other words, that $d(\lim_{n\to\infty}s_n,\lim_{n\to\infty} t_n)=\lim_{n\to\infty} d(s_n,t_n)$

[gentle osprey]

Edward II