point-set-topology

[rancid mantle]

How? It has been mentioned at some point but it didn't make much sense to state it that way and just leave it there,

[quartz horizon]

They are a composition of an inclusion with a projection

You have an inclusion from P to A x B

[rancid mantle]

Hmm what I understood is that it packages two spaces into one but with an extra set of instructions given by the morphisms to C. It will combine them but toss away anything that does not behave well with C (that is, the morphisms conflict). Is that right?

[quartz horizon]

And then projections from A x B -> A, A x B -> B

[quartz horizon]

That’s another way to think about it!

[rancid mantle]

So

Wait

[quartz horizon]

Take B -> C to be an inclusion map, and A -> C any continuous function

[rancid mantle]

So if C is the space with only one point

Then the pullback is the product

Because there are no conflicts

[quartz horizon]

Yep!

It’s also useful to set Z = one point space here

That tells you the points in the pullback

[rancid mantle]

It seems a bit unexpected that those two pop up here, inclusion and projection

But in some way, if it combines the two spaces, there must be some product going on, and if it filters stuff, there must be some inclusion

[rancid mantle]

Okay I'll try to work it out

So I'll have A and B together, and I will only keep the things on A that have image in B, and those will be related

[quartz horizon]

Yes, exactly

[rancid mantle]

So then

[quartz horizon]

Yep, but “the things in A with image in B” is exactly there preimage

[rancid mantle]

Yes

Oh yeah

Okay

So P seen as a subspace of AxB

[quartz horizon]

You don’t have to use the product inclusion construction

[rancid mantle]

Will have

[quartz horizon]

You just need to find something that “does” the correct thing

[rancid mantle]

For each b in B, add the preimage of b along with b to P

if the morphism from A to C is f, P is the union of f^-1(b)x{b} for each b in B

Is that correct?

[heady skiff]

why is f(x) = inf Q(x) <= r? we don't necessarily have x in U_r, so r is not necessarily in Q(x), right?

[rancid mantle]

To me that was the natural choice but I wouldn't be able to do this in some other context. When would doing something else be useful?

Should we move to a thread?

[quartz horizon]

Yes

[quartz horizon]

You can just see it as a subspace of A, too

Because that “does” the correct thing

[heady skiff]

moreover i don't really see how this is using the infimum property - obviously r is a lower bound for all such s, but since inf Q(x) is the greatest lower bound, shouldn't we have inf Q(x) >= r?

[tiny obsidian]

The sets $U$ are defined such that $\overline{U_r}\subseteq U_s$ for any rationals $r<s$, and $\bQ(x)$ is the set of rationals $s$ such that $x\in U_s$?

[gentle osprey]

Edward II

[heady skiff]

yeah

[tiny obsidian]

Suppose $s>r$, then $r<\frac{r+s}2<s$. The first inequality tells us $\frac{r+s}2\in\bQ(x)$ and hence so $s$ wasn't a lower bound for $\bQ(x)$ by the second. So the infimum (which is a lower bound) is not greater than $r$.

[gentle osprey]

Edward II

[tiny obsidian]

We don't know and r isn't a lower bound for such s, since if x were in U_r then there could be U_q ni x with q<x

[tiny obsidian]

The obviously here is wrong, since if $x$ were in $U_r$ we might have $x\in U_q$ for $q<x$

[gentle osprey]

Edward II

[sonic crane]

Edward the 2nd is the Point Set Topology King

[heady skiff]

i meant r is a lower bound for all such s with r < s

[tiny obsidian]

Ok, then as I replied that doesn't necessarily mean r is a lower bound for Q(x)

[thorny agate]

Ok so this is my first time looking at sheaf stuff

[gentle osprey]

Spamakin🎷

Spamakin🎷

[thorny agate]

And upon more thinking I am still not sure about the stalks either 💀

[grave solstice]

the stalk at x is C

[thorny agate]

Yea I could have guessed that but I'm not sure how to prove that

[grave solstice]

just use the definition

[thorny agate]

hm ok came back from eating and the stalks make more sense now

still unsure about the gluing

[thorny agate]

I think the whole pairwise overlap thing solves my confusion here but I'm not seeing it

[grave solstice]

maybe do the case where all the U_i contain x (it is trivial that you can always reduce it to this case anyway)

[red folio]

For the definition of a subspace topology $\tau_{A}$ of $A \subseteq X$, since $A \in \tau_{A}$ does this mean A is also open in X?

[gentle osprey]

snus

[thorny agate]

I don't see the reduction 💀

I have figured out that case but I don't see the reduction

[robust drum]

No

It means A = A intersect a set open in X. But this is trivial because we can always take the latter set to be X itself regardless of what A is

[prime elbow]

I want to show {m + na | m,n in Z}, where a is irrational numbers, the set is dense in R.

I am trying to construct a surjective continuous function from R to R such that {m + na | m,n in Z } is an image of a dense set .

Any hint?

[red yoke]

This is a subgroup of R

Show that if some subgroup of R isn't dense, it must be generated by a single element

And thus a would be rational

[prime elbow]

Seems interesting

[prime elbow]

I have no idea

Can you please make this problem in sub-problems?

[wise mist]

think of it as a subring Z[a] of R
if na >= 1 u can always swallow the integer part into m
to make it eventually <= 1

if its 1 the set is discrete in R
if its less than 1 (na)^k will make it dense

[jagged perch]

Guys it is correct?

If I take complement of both the subset inequality will change?

[tender halo]

yes

[prime elbow]

Can you explain more ?

[granite crane]

does it look good?

[vital river]

point-set topology so far feels like 80% memorising the definitions so that I know wtf is going on

i hate this so much

[wise mist]

k is natural

never mind u dont have to even take the sequence there

as R\Q is already dense in R

[vital river]

is it true that a set is open if every point in it is an interior point?

So contrapositively, if there's a point in A that's not an interior point, then A cannot be open

[mellow basalt]

Yes

[vital river]

can a similar characterisation of closed sets be done with boundary points some how?

[mellow basalt]

Yeah one common definition of a closed set is that it contains all of its boundary points

[vital river]

so A is closed iff it contains all boundary points + interior points?

or is it unidirectional only

A is closed **if **it contains all boundary + interior points

[dire dove]

Any A always contains its interior points so that condition is irrelevant

[mellow basalt]

If A contains all of its interior points then it's open Boundary points cannot be interior points

[vital river]

why is this set not closed then?

[dire dove]

Shouldnt it be if all points of A are interior then its open?

[granite crane]

If A contains all of its boundaries points and interiors then it is closed yes.

[mellow basalt]

Yes you're right

[opaque scroll]

1 and -1 are both boundary points

[dire dove]

Doesnt A always contain its interior points? I don't see how you would need that condition (or how having it would change anything)

[vital river]

±1 aren't part of the set doe

[opaque scroll]

Exactly

[vital river]

they're the accumulation points

[opaque scroll]

Hence it's not closed

[vital river]

wot

ohhhoh

oops

I meant open

why is it not open

[opaque scroll]

A set is open iff all it's points are interior points

Here none of them are

[vital river]

how so?

take n = 3

then we have -1 + 1/3

how is that not an interior point

[opaque scroll]

So a point is in the interior if the set contains a neighborhood of that point.

So some interval (a, b) with a<-1+1/3 < b

The set does not contain any such set (it's not uncountable, it doesn't contain any irrational numbers, whatever justification you want to use)

[vital river]

what

for $a_k$, why can't we just take $a_{k\pm 1}$ to be the bounds?

[gentle osprey]

Kakaka

Kakaka

[vital river]

for the interval (a,b) that you mentioned

[opaque scroll]

I'm not entirely sure what you're suggesting, what would that be for n=3

[vital river]

ohh

wait

[opaque scroll]

In either case, it's clear that the set doesn't contain any intervals at all. So it doesn't really matter what you choose for a and b

[vital river]

we need to find an open ball

so a symmetric interval about -1 + 1/3

but no such radius >0 exists in the set

is that what you're saying?

[opaque scroll]

Yup

[vital river]

thanxx

lemme backtrack to see if I get it now

[vital river]

actualy why is it "clear"

how would you prove it

that so such $\epsilon >0$ exists such that $B(x,\epsilon)\subseteq A$

[opaque scroll]

Two possible arguments in parenthesis here

[gentle osprey]

Kakaka

[vital river]

why do we need irratoinal numbers?

[opaque scroll]

Intervals contain irrational numbers

[prime elbow]

Thank you ❤️

[wise mist]

i think i have rather proved its not a discrete set, where the limit points are the integers

[gritty widget]

Well consider a small ball about any point, then it falls outside the set

a set is open if we can cover it by open balls which is a strict subset of it

since we can't do that here, it rules out the possibility of it not being open

[heady skiff]

can i have a hint for showing that every subset of R under the cofinite topoolgy is compact

[tiny obsidian]

Pick a set in your cover. What elements are left over and how can you cover them

[low flame]

So all of the subbases are nhbds around zero

Thus the intersection of subbases is a base set around zero?

[fervent steeple]

i believe loossemble knows. but im not sure

if i elaborate the government will end me

ttyl

[low flame]

Thanks for the help

[uneven bronze]

Here's a classic. I have a silly doubt. Am I right that f^{-1}({y}) contains only a single x by definition of a function?

[unreal stratus]

No

Functions needn't be injective or surjective

(As a silly example take the constant map X -> {y} for any set X)

[uneven bronze]

ah right ok, thanks

[uneven bronze]

I feel like this proof is so convoluted. How do we know f^{-1}(U) equals that union?

One inclusion has been shown, the \supset inclusion. But seeing the \subset inclusion eludes me.

If x in f^{-1}(U), how do we know there is a ball around it with radius \delta_x?

I'm even doubting the \supset inclusion. It's only true if x\in B(\delta,x)\cap f^{-1}({y}). Meeh.

[tiny obsidian]

every ball with centre x will contain x, so since x in f^-1(y) by definition it's clearly in that intersection so there's no issue for supset

[uneven bronze]

ok, true, that makes sense 😄

[tiny obsidian]

for the subset we're taking the union over all x in f^-1(U). So for any x in f^-1(U) the relevant ball B(x,delta_x) will give x in the union

(x in f^-1(U) means x in f^-1(y) for some y in U by definition so we do have delta_x from the proof before, if that's a concern)

[uneven bronze]

ok 👍 makes sense, I think I'm beginning stabilize (in my understanding) 😄

[low flame]

can anyone tell me if i am thinking correctly about this?

i think that the bassis element is the intersection of all those inverse projection of nhbds around zero

because f(c) = 0

and its setup such that each interval (a,b) is a nbhd around 0

and a basis consists of the intersection of sub-basees right?

[tiny obsidian]

I think so

I don't really get the question since it basically becomes "a neighbourhood of 0 contains a basis element with 0 in it" and just saying what that basis element could be

which as you've said is finite intersection of preimages of intervals containing 0

[low flame]

okay thank you 🙂

[sonic crane]

I dont understand. Sorry if my handwriting is bad.

f is a continuous function from X to Y

A is a subset of X

I just dont understand that implication where its coming from

The images preimage and complements are all confusing me

This was in a proof about f(A bar) < f(A) bar

For f cts

[sonic crane]

Nevermind all good

[vital river]

So to generalize, any set "spanned" by a sequence, e.g ${x_n: n\in \mathbb N}$ cannot be open right? Since it would only contain discrete points, and the open ball must be "dense", i.e $A\subseteq \mathbb R$

[gentle osprey]

Kakaka

[unreal stratus]

It can't be open, but it needn't be discrete and it can be dense!

For example just like list the elements of Q

[gentle osprey]

𒀭

𒀭

[robust fable]

Wait nevermind I forgot how bases work...

[trim spade]

how you doing

[robust fable]

Apparently I’m stupid lmao

[trim spade]

☹️

[robust fable]

Well, at least I didn’t need a hint for this

Love it when topology questions are literally just “have you considered what your basis is”?

[low flame]

for seven, I was thinking wouldnt i just create some bijection. for example H: F -> N such that for f_i in F we can find H(f_i) = k.

but also is this collection formed like this lets say C= (1,2,3) we get 1, 2, 3, {1,2},{1,3},{2,3},{1,2,3}

So could i create a sum_k=0 ^n C choose k, which is a countable sum of sets, and is countable?

[sterile jungle]

u can just write the collection as a union of sets of the form A_n = {F | F subset C has cardinality n} where n >= 0. each of these sets is atmost countable (A_n has the same cardinality as C^n which is countable). and you have a countable union. so {F | F subset C is finite} is countable

[low flame]

wow thanks

[gritty widget]

I mean there is stuff to think about. In general the set of points in a sequence could be closed. For example, take the constant sequence. The set spanned by it is closed as the one point set is closed in R.

[gritty widget]

I thought about it more and my arguement of it being closed is wrong. I hence deleted the messages

[low flame]

oh yeah this is for like an infinate countable set

If its a finite set theyre the same i think?

[granite crane]

how

[alpine nest]

From definition.

Every neighborhood of 2 contains all but fintely many elements of your sequence; also every neighborhood of 3 contains all but finitely many elements of your sequence.

[granite crane]

its strangee

[gritty widget]

https://math.stackexchange.com/questions/4974832/are-convergent-sequences-in-mathbbr-neccesarily-closed-sets

[granite crane]

if $p,q\in X$ with $p\neq q$ there exists two functions $f_p,f_q:X\to \Bbb R$ such that $f_p^{-1}(0)={p}$ and $f_q^{-1}(0)={q}$. Now my idea is to find some open sets say $U,V$ of $\Bbb R$ such that $0\in U,V$ and $f_p^{-1}(U)\cap f_q^{-1}(V)=\emptyset$.

[gentle osprey]

Afzal

[granite crane]

Is it correct idea?

[gritty widget]

reminds me of the urysohn elmma

[quick delta]

I wouldn’t consider two separate $f_p, f_q$ (because I don’t think there’s a nice way to force the resulting opens to be disjoint)
I would only consider $f_p$

[gentle osprey]

Micose

[gritty widget]

nice question, where did you get it from? @granite crane

[granite crane]

idk about it

[limpid fern]

ITM

lee

[granite crane]

i mean why? according to the problem for each p there is some function f, so the choice of f depends upon p right

[gritty widget]

your claim is correct

but I think micose is saying something else

she is talking about the approach in itself rather than correctness of the statement made

[granite crane]

oh yeah DerpZ told the name of the book (its ITM by lee)

[quick delta]

Your claim is right, I just don’t feel that (half of) it is useful

[quick delta]

*she

[gritty widget]

oh oops, I am used to everyone I talk to ebing male 😹

[granite crane]

I see, let me think more about it

[alpine nest]

Yeah, I think that's a better strategy

[quick delta]

Maths moment

[gritty widget]

Well, R is hausdorf. That gives immediately that point sets are closed, and hence point sets are closed in your beginning space

[alpine nest]

Internet moment, really; but fortunately we have the pronoun indicators:

[gritty widget]

oh that's what it is for

[low flame]

so the product topology right vs box topology.

With the box youre taking the product of every open set within the list of topoligies right.

While with the product topology were taking the inverse projection of a finite number of sets and crossing that with the rest of the topologies right? Also, because were in a way smearing out the image across the various other topolgical spaces, and taking the intersection of them (sub-bases)

were in a way then kinda making things smaller with the product topology vs the box then. because of the intersection of the sub-bases right?

Am i thinking about this correctly? i find it super confusing tbf.
oh yeah this is for like an infinate countable set

If its a finite set theyre the same i think?

[granite crane]

yes

[gritty widget]

as pre image of closed is closed

[granite crane]

yeah cuz of continuity of f

[gritty widget]

there is a characterization of hausdorfness through the self product

a space X is hausdorf iff the diagnol line is closed in XxX

diagnol line = {(x,x) } i.e: pairs with same x and y component

I am thinking maybe that is helpful here somehow because that immediately says when something is hausdorf by using when something is closed

and the continou function tells us when something is closed

[gentle osprey]

Micose

[granite crane]

Let me work on it

[gritty widget]

do you understand why your first approach wasn ot a good idea?

[granite crane]

let me on new idea before looking at this spoiler

[gritty widget]

or is potentially not a good idea

[alpine nest]

Yeah, that should also be a valid approach

[low flame]

I guess im very wrong then

[granite crane]

i think cuz open sets are making things difficult here

[gritty widget]

wdym?

[alpine nest]

I'm honestly not sure I understand your doubts properly, but what i can say is that 1. every set that's open in the product topology, is also open in the box topology, but not necessarily vice versa; and 2. in case of the product of finitely many spaces the two topologies are the same.

[granite crane]

existence of two open sets U and V such that there inverse images are disjoint. It not some characterization of cont funcitons ( or not trivial or easy) i think thats why

[gritty widget]

well I think something else to say is

you'd need an intuition of how the preimage changes when you add two such functions

for instance, if I had a simple function f(x) = x, and I consider the unit open ball about the point x=0 in the codomain, the preimage is the same.

If I turn my function to f(x) = x+1, then the preimage of the unit open ball about x=0 in the codomain becomes shifted to the left

[low flame]

I just am looking for the reason why open sets in the box topology arent open in th product. And i think its because of the use of the sub-basis right.

So we if we have a countably infinite set right, we will take a finite number of these intersections with inverse projections that forms a subbasis, and then cross it with the rest of the spaces.

this is why open sets in the box may not be open in the product right?

[gritty widget]

this buisness is quite easy to understand with simple functions like that, but when given functions with the property in the question, you'd need to think a bit more on that

[low flame]

its like the product doesnt use every open set, while the box does

[gritty widget]

did that make sense

[alpine nest]

I mean, the definition of a basis set in the box topology is straightforwardly broader/easier to satisfy than in the product topology (in box it's "product of open sets", and in product it's "product of open sets, of which only finitely many are non-trivial"), so there's at least as many sets in the basis for the box topology as there are for the product topology.

And to show that there can be strictly more, you can just demonstrate a set that's open in the box topology, but not in the product topology.

[alpine nest]

The product topology does involve all the open sets in all the component spaces, just not in every possible combination of them

[low flame]

okay i think i get it 🙂 sorry i have found this to be very confusing 😛 thanks

[granite crane]

yes.

[gritty widget]

well maybe I should explain why I said add

I think for each individual function you can squeeze out that point sets are closed

but right after that, what would you do? I guess the natural thought process would be to do some algebra and try come up with a new function

[granite crane]

I got it, now things make sense

thank you Micose and something

[quick delta]

Very slight nitpick on the first line: such that f^-1(0) = {p}
As that’s needed to conclude f(q) =/= 0

[granite crane]

ah yes if q\neq p

[fair idol]

If I have an indexed family of functions on topological spaces $f_i:X\to Y_i$ and consider the initial topology of $X$ given by this family, are there other functions $g:X\to Y$ that are continuous with respect to this topology that are not from the family of functions?

[gentle osprey]

HausdorffT1

[fair idol]

Oh a constant function i guess

[thorny agate]

I am still stuck on the gluing condition for this sheaf in one particular case:

[gentle osprey]

Spamakin🎷

[opaque scroll]

I mean you just define zU to be zi for one of the i's where Ui contains x

[thorny agate]

wut

How is that well defined

Ohhhh

[opaque scroll]

For all the Ui that contain x, the zi are equal. And for the other it's always 0

[thorny agate]

Cause of the overlap being identity

Yea

[opaque scroll]

So that uniquely specifies zU

[thorny agate]

Ok cool

Thanks

[sonic crane]

Is f(A^c) subset f(A)^c not always true?

Is it potentially not true when f is not injective ?

Ik not topology but its in my topology stuff that

Im struggling with

Chat gpt?

I am correct right

[junior granite]

hello folks, how would i go about proving the first "axiom" of the definition of a basis? This is the problem i'm working with but i'm not sure how to approach it.

Left is the setup, right is the notes I'm referring to.

[prime elbow]

So can you prove 1? For any n in X can you find B_k such that n in B_k ?

[granite crane]

Wdym by true?

[prime elbow]

And 2, let B_k_1 intersection B_k_2 be non-empty. Say x in B_k_1 and B_k_2 then x is a divisor of k_1 and k_2.

Now think about what will be B_x ? And prove that || B_x contained in the intersection of B_k_1 and B_k_2 ||

[rancid umbra]

they are asking if f(A^c) is a subset of f(A)^c for all subsets A of the domain of f

[prime elbow]

Is a number divisor of itself?

[granite crane]

Oh, damn I didn't read that subset

[junior granite]

yes

does 1 hold for B_0 because B_0 = X? or am i making stuff up lol

[prime elbow]

So for n in X what will you choose k such that n in B_k ?

[prime elbow]

B_0 ?

I see

But I think take n in N and 0 not in N

[junior granite]

hmm

i'm not sure then

[prime elbow]

Then for x in X pick k = x thus x in B_x

[junior granite]

Would this suffice then?

I think I am confused on what exactly you are asking, sorry.

[prime elbow]

You have to show that for any x in X there exists a basis element in B such that x belongs to that element, right ?

[junior granite]

Yes? I think so.

[prime elbow]

So pick x in X then since x in B_x

[junior granite]

Is that not what I did? Or did I just say that's what I needed to do? I'm confusing myself now I think D:

[prime elbow]

You assumed 0 in N then B_0 = X but what if 0 not in N

[junior granite]

What about B_1 = empty set, so B_1 is in the topology?

[prime elbow]

Yes

And you do not need to empty set in your basis set

[junior granite]

hm?

[prime elbow]

Yes

[junior granite]

I don't need the empty set in the basis ?

[prime elbow]

Yes

[prime elbow]

Which book?

[prime elbow]

Just think why do you need an empty set in the basis set ?

[junior granite]

Sorry I was just confirming that's what you meant, the wording confused me.

[prime elbow]

If you have an basis and T generated by that basis set then how does the open set look?

[junior granite]

Topology, by James Munkres

[prime elbow]

Okay

[junior granite]

Though the book is rarely used in our class. Just vaguely referenced here and there.

[prime elbow]

Okay

[junior granite]

Not sure, I think that's where I'm confused.

I don't really understand the concept of a basis so proving something is one is challenging.

[prime elbow]

I can understand

Just work it out and if you can look at topology without tears

It is interesting book " topology without tears "

[low flame]

its the smallest collection of open sets that forms a topology i thought?

and then for your question, youre just showing that each bn is in the topology

Then if b3 = b1 cap b2 then b3 is a basis in X i think

The proof is right in munkres book

[opaque scroll]

It's any collection such that every open set in your topology can be written as a union of basis sets.

No condition of it being "the smallest" or anything like that

[low flame]

im learning myself im sorry

[low flame]

i guess i took the smaller thing since B is a subset of the open sets.

thank you for correcting me 🙂

[junior granite]

isn't that intersection empty? since 1 is strictly not in the set X.

so B_1 would be empty

[alpine nest]

They meant B_1, B_2 and B_3 as three arbitrary sets from your basis, not related to the B_n's as the specific sets defined in your original problem

Unfortunate terminology collision

[junior granite]

oh oh i see

[heady skiff]

1a. Let $(f_i(x)){i \in I} \in \prod{i \in I} X_i$. Let $\prod_{i \in I} U_i$ be any open set containing $(f_i(x)){i \in I}$, so that $U_i = X_i = I$ for all but finitely many $i$. Fix $x \in I$. Since open sets of $I$ are of the form $[0, 1]$, $[0, a)$, $(b, 1]$, or $(c, d)$ (where $0 < c < d < 1$) and $(0, 1) \subset [0, 1]$, $(0, a) \subset [0, a)$, $(b, 1) \subset (b, 1]$, we see that for a given $U_x$, we can find $\epsilon > 0$ such that [B{\epsilon}(f_i(x)) \subset U_x] By the definition of pointwise convergence, there exists $N \in \mathbb{N}$ such that if $n \geq N$, $|f_n(x) - f(x)| < \epsilon$ and thus $f_n(x) \in B_{\epsilon}(f_i(x)) \subset U_x$ for all $n \geq N$.
\newline
\
Would it make sense to take the supremum of all such $N$ for each $x \in I$? Or could this be possibly unbounded and hence not exist

[gentle osprey]

okeyokay

[heady skiff]

Because the definition of convergence in a topological space has no analogue of pointwise convergence

It requires a single N for the sequence

[heady skiff]

assuming b is true, why is c true? if F(I) is dense in B(I) then each point in B(I) should have a sequence of F(I) converging to it, right? or is this only true if B(I) is metrizable?

[robust drum]

Suppose X is a normal space. I can separate any two closed sets by disjoint open sets, and therefore I can separate finitely many closed sets by disjoint open sets. Does this hold for arbitrarily many closed sets? (for my purpose, I only need countably many).

[unreal stratus]

No e.g. try separating all the rational points of R from one another

[robust drum]

damn

is there some conditions on the closed sets or open sets I can give

in particular for this case I have disjoint closed sets of a smooth manifold

and I want to find disjoint open sets containing each

[unreal stratus]

Well R is as nice a smooth manifold as you can want and is a subspace of any (dim >= 1) smooth manifold

[robust drum]

damn you're right

oh wait

I'm also told closed sets are compact

wait

that also holds here

ok im being dumb

lol

[unreal stratus]

well

the closed sets i mentioned compact sure

but also you could do like S^1 and consider the rational points on that

[robust drum]

Wait is it true that a compact set has finitely many connected components

[unreal stratus]

if you want a compact smooth manifold version

[opaque scroll]

It can have infinitely many

[unreal stratus]

In nice cases yes, not in general

Consider Cantor set

totally disconnected (so connected components are points), cardinality R, and compact

[robust drum]

damn

[opaque scroll]

Union_n [1/(2n+1), 1/2n] union {0} for example

[unreal stratus]

Lol honestly I didn't know the answer (since it will be "yes" for nice spaces) and then jagr said the answer and that gave me an example aha

[unreal stratus]

Nice

Is the easiest way to see this is compact like

Pick an open in your cover containing 0 and that covers all but finitely many of the closed intervals

[opaque scroll]

What are these "nice" spaces though?

[unreal stratus]

Actually my idea of nice may be a tautology lol

Something like a locally finite CW complex for example

[opaque scroll]

I guess the easiest way is that a subset of R is compact iff it is closed and bounded

[unreal stratus]

But the locally finite is then doing all the work so it is a tautology

[unreal stratus]

True though then you have to argue this is closed which I think is basically the same

But yeah sure it is "by inspection" the closure of union of (1/2n, 1/n) for example, or consider sequences (they either -> 0, or they are eventually in one of the intervals) ig

[robust drum]

what I'm actually trying to prove is that if f is an embedding when restricted to a compact subset of a smooth manifold, there's a neighborhood of that compact subset on which f is still an embedding. I'm trying to prove it by passing to the connected case, and arguing that the set of points on which it's not injective is clopen

but I can't figure out the connected case

probably goes in #diff-geo-diff-top at this point though

cause it looks like I wont be able to just point set my way to the connected case

[opaque scroll]

Or see that the complement is the union of open intervals

[unreal stratus]

What do you mean by "embedding" here

Topological embedding?

I'm a bit confused cause you say smooth manifold but then compact subspace

[robust drum]

local diffeomorphism + injective

ur right

[unreal stratus]

Or perhaps like just underlying space of a smooth manifold

[robust drum]

the wording of the q is

local diffeo + one-one

not embedding

and then you're supposed to find an open set containing the compact thing where it's actually a diffeomorphism (with the induced smooth manifold structure on the open set)

[unreal stratus]

I guess I would think in terms of the differential

[robust drum]

yeah, it seems true to me that you can cover it with an open set on which the differential is nonsingular at least, but maybe that introduces points where the map itself is non injective

[unreal stratus]

Well if it's a local diffeo then the differential is non-singular

[robust drum]

sure, but also on the open set is my point

oh wait

ur right

im dumb

lol

[unreal stratus]

Yeah this happens everywhere on that compact thing

Now do you see how to extend to an open subset?

[robust drum]

here's the actual question

I see that the differential is nonsingular on the open subset fs

but the way I was thinking of this in the connected case is to argue that we can take the open set to be connected, and that set of points which get identified is clopen in that open set

[unreal stratus]

Well now we are done up to injectivity right

[robust drum]

yeah

[unreal stratus]

Like this is a local diffeo on U and injective on Z

[robust drum]

yeah

[unreal stratus]

now need to show injective on U sure

Well this will be where we use compactness of Z I suppose

Well also we will need to shrink U possibly

[robust drum]

Because f is a local diffeo, we can take our open set to be a finite (by compactness) union of open sets on each of which f is injective

[unreal stratus]

I mean I have a funny slight overkill

[robust drum]

but im not sure how to finish there

[unreal stratus]

Actually no it is not even an overkill because can adapt

[unreal stratus]

Yeah so now consider f^{-1}(x) for x in U

What can you say about it

[robust drum]

(btw I just realized that I read the problem slightly wrong and actually can reduce to the connected case bc Z is a submanifold)

[robust drum]

but responding to this, I think we can say that f^{-1} of each point is a discrete union of points in each of the sets covering Z

[unreal stratus]

Eh tbh maybe this odens't work

[robust drum]

well it's ok

for my purposes

bc we can actually reduce to connected case

(components of a manfiold are open, therefore compact manifolds have finitely many)

and thus by normality we can actually separate them all

[unreal stratus]

I don't see how that helps

[robust drum]

well normality means I can separate any two

and then because finite unions of closed sets are closed

[unreal stratus]

Yeach but there are infinitely many x possibly

[robust drum]

I can take a big finite intersection to separate any

wait a minute which part are u saying breaks

the reduction to Z connected

[unreal stratus]

Well tbh I dno't see what you are doing

[robust drum]

well to reduce to the case where Z is connected

[unreal stratus]

Well that seems unnecessary but ok

[robust drum]

well then my original strategy works

[unreal stratus]

But yeah okay sure you can definitely do that

Oh what is your strategy then

[robust drum]

that is, you can show that the set of points on which each f is not injective is clopen

in the bigger U covering each Z

[unreal stratus]

Sure

[robust drum]

then you can argue from there that it can't be all of that U

[unreal stratus]

Yeah actually that is similar to a thing I had nice

[robust drum]

nice

I was just tripping abt how to reduce

but I didnt see "submanifold"

I'm kinda curious bc I think that's literally the only place where I used submanifold

np lol

[unreal stratus]

Lol

[robust drum]

this pset has taken me legit like 40 hours

it's insane

worst pset of my life

that problem is like the easiest one

[unreal stratus]

I'm sorry 😭

[robust drum]

oh as an example of how bad this pset is, my friend and I just realized that this is literally just false in the non connected case

because you can have two copies of the same manifold mapping in the same way

my reduction is bad bc I was assuming disjoint unions of diffeos are diffeos

and that's not true

but this is just evidence of how this prof doesn't even try the problems before assigning them

[unreal stratus]

Oh wait wdym

Like a disjoint union?

[robust drum]

yeah like

[unreal stratus]

But then each copy is open so you just take U = Z

[robust drum]

also closed

wait

ur right

wait im dumb

and confused

nvm ignore me

it's probably still provable

[unreal stratus]

No ur not dumb lol

[robust drum]

but the reduction isnt as obvious as I thought it was

im just malding over how much time ive spent im ngl

ah it's fine you just take neighborhoods in the codomain instead of the domain

bc the f(Z_i) are all disjoint

bc f is injective on Z

and then you take disjoint neighborhoods of all of those things in the codomain (f is a local homeomorphism too, so the f(Z_i) are all closed and finitely many of them), take preimages, and then intersect those preimages with the neighborhoods of Z_i on which f is an embedding

breh

[hoary sphinx]

How would you prove (in the case it's true) that a connected open subset of R^n is homeomorphic to R^n?

If the set is convex I suppose you can choose any point in the subset and draw all lines that go through that point

Is it false?

What would be a counterexample

Let me think for a sec

Oh yeah it can have holes I see

How would you prove then that say {x in R^n: |x| in (1,2)} is NOT homeomorphic to R^n?

Hmm

Hmm ok yeah that makes sense

I guess you would generalize the loop then

Like a function from S^n to R^n

That can be deformed

I see yeah makes sense I guess

Ok I see

Thanks!!

Wait but what does deforming the loop to a point mean?

(If it's not too hard)

Like formally I mean

Ok I'm going to look into that

Thank you

[sonic crane]

Profs notes ….

Showing that f(bar(A)) subset bar(f(A)) for every subset A implies F continuous

Not only do i not understand the intuition but i also dont even understand the logical structure here

Right now I am not understanding how we showed X\finv(B) is open

With this stuff i find it hard to see which parts are just set theory and which parts are topology

Oh because the elements of X\finv(B) are precisely the inverse image of Y\B

Ok, i still just do not see the intuition.

I get the structure of the proof now tho (i think)

[tiny obsidian]

Intuition for the result?

[sonic crane]

Yea i suppose

[tiny obsidian]

The closure of A can intuitively be thought of as "the points near (or in) A"

So the result is saying f is continuous if and only if points near A get mapped to points near f(A)

(without all points near f(A) necessarily originally being near A as continuous maps can bring things close, hence why only one inclusion)

Also this proof is quite hard to parse I agree

I also don't quite understand what is happening there

[alpine nest]

I'm having trouble with this one, because it needn't be true in general if f isn't surjective.

The preimage of B might be empty, and what then?

Unless the sign isn't equality but inclusion

But then it's an inclusion in the opposite direction to the next one in the chain, which would be highly unorthodox

[tiny obsidian]

"w/ Y\B o = (Y\B)" ??? that's making me think some parts got erased, though I don't think they'd help too much given it'd be a few words at best

[plush folio]

I dunno if this helps with the intuition, but it might be useful to notice how this proposition relates to sequential continuity: if $x_n$ is a convergent sequence in A, then $\lim x_n \in \overline{A}$ and $f(\lim x_n) \in f(\overline{A})$. Then by continuity $f(\lim x_n) = \lim f(x_n)$ and $\lim f(x_n)$ is in $\overline{f(A)}$

[gentle osprey]

sheddow

[real granite]

Not sure where best to post this, but I have question about Hilbert spaces arising from quantum mechanics.

1. Is a Hilbert space necessarily complete (i.e. Cauchy sequences are convergent)?
2. When we say a Hilbert space is "separable", do we mean that it is Hausdorff (as stated in the screenshot), or do we mean separable in the sense of there being a dense countable subset? A friend at a different university suggested that the lecturer was confusing these different things.

yes the bit in brackets is what i thought

yes

he's mismatched the definition with the term

silly physicist not knowing topology

i concur

[unreal stratus]

Oh lol

Please don't multipost - i wrote a near identical answer in another channel lol (just delete the post in the other channel next time ig)

😭

But yeah best is probably #real-complex-analysis (or #advanced-analysis for harder questions about hilbert spaces etc)

[real granite]

i did strikethrough it

[unreal stratus]

Sure but I would just delete lol and you didn't say you'd actually posted elsewhere ig

but dw

[heady skiff]

anyone know if this is right?

is the converse to a) true?

i feel like it's intuitively true - for each i in I, we can set U_i to be any ball of radius e > 0

which is open in the product topology since it's only one open set

then |f_n(i) - f(i)| < e

or we can just do something like B_(1/n)(f(i)) and use archimedian property

[opaque scroll]

Convergence in product topology is exactly pointwise conference yes, if that's what you're asking.

[heady skiff]

can i have a hint for 1c? say f(a) =/= 0. then there's a delta neighborhood of a such that f(a) =/= 0 for all i in B_d(a) and hence (f(i))_(i in I) has infinitely many points nonzero, but i don't really know how to derive a contradiction here - because if we have an neighborhood of (f(i)), then only finitely many X_i = U_i, and so my original idea of forcing one of the f_ns to have infinitely many nonzero points goes to waste...

[opaque scroll]

Think about a sequence of functions in F(I), gn -> g. How big can the support of g be? What is the relationship between where the gn are 0 and where g is 0?

[heady skiff]

well we know that f doesn't have finite support since there are uncountably many points where f is nonzero... so maybe something stating that a sequence of functions with finite support must converge to a function with finite support, but that doesn't look to be the case because of b)

hmm

[opaque scroll]

Remember being in the closure of something is different from there being a sequence which converges to that point.

That's only true for first countable spaces

Sequences can only capture a "countable amount of data"

[sonic crane]

Bruh

[heady skiff]

maybe some countability argument then? like given any basis element U containing (f(i)) then if we did have (f_n(i)) in U we would have to infinitely many points where f_n(i) is nonzero

i'm bad at countability arguments

[sonic crane]

Well B = f(finv(B)) right?

[opaque scroll]

So if we think about this sequence f_n. Where each f_n has finite support.

Consider the set where at least one f_n is nonzero. How big is this set?

[sonic crane]

And B is closed so B = closure of f(finv(B)) @alpine nest

So just take complement and you are good no?

[heady skiff]

well it can be the empty set, since each f_n can be zero, or it can be all of the f_n, i.e. they are all nonzero at a single point (since they're not necessarily continuous, nonzero at a point wouldn't imply nonzero at uncountably many points)

[opaque scroll]

Right, so the points where at least one is non-zero is just the union of where each of them is non-zero.

So that's a countable union of finite sets

How big can that be?

[alpine nest]

Let f(x) = x^2 from R into R, and let B = [-3,-2]

What's f(f^(-1)(B))?

[heady skiff]

so that means ${a \in I \mid f_n(a) \neq 0 \text{ for some } n \in \mathbb{N}} = \bigcup_{n \in \mathbb{N}} {a \in I \mid f_n(a) \neq 0}$?

[gentle osprey]

okeyokay

[heady skiff]

and that would be countably infinite

i can kind of see the argument from here? something like f(i) has uncountably many points where it's nonzero, but any neighborhood will contain only countably many points where the f_n are nonzero, so it can't converge? i'm not sure how to make this precise though

[sonic crane]

Damn!

So my prof straight up wrong

[opaque scroll]

Well, you already have the idea with pointwise convergence.

So if f_n(a) = 0 for all n, then f(a) = 0.

So f(x) is 0 everywhere outside this countable set.

But you know f is nonzero on an uncountable set

[heady skiff]

i think this makes sense. i'm having trouble seeing how this countable set give points a such that f_n(a) = 0 for all n tho

[opaque scroll]

I mean by definition.

You have the set where at least on f_n is non-zero, then outside that none are non-zero, aka they are all 0

[heady skiff]

ah okay i see now

if a is in this set then f_n(a) is for at least one n in N

so the points b outside of this set have f_n(b) = 0 for all n in N

got it tysm! that rlly helped

[uncut fulcrum]

Does the order topology have a universal property?

[low flame]

does this apply to collections of finite subsets of countable sets as well?

[unreal stratus]

Well with some modifications yes

[low flame]

im trying to use it for this here

[unreal stratus]

Oh lol do you mean the result

Yes, finite union of countable is countable

Indeed (given some mild form of choice) countable unions of countable sets are countable

[heady skiff]

what do they mean by this? (proof of Tietze extension theorem) here they took a function f: A --> [-1, 1] and extended it to a function g: X --> [-1, 1], but it didn't seem to be an accident - anyways, they've shown that we do have such an extension, and why don't they stop there?

sure it's dependent upon f mapping into [-1, 1] but this theorem is up to homeomorphism so

[low flame]

Lovely. If i could also inquire about problem seven. can i use this to show that {N} U {N X N } U .... is countable?????

[unreal stratus]

Yes

[low flame]

BROTHER, youve take a lot of stress of my shoulders

thank you

[fierce lily]

Question: 1. does this question want to show that f(X) is open? i am somewhat confused about what local homeomorphism is open mean? 2. since X is topological space, can I just write X= union of A_i as A_i is open basis elements here?3. May I write f(U A_i)=Y? I really have some confusion about these expressions

[quartz horizon]

So the definition of open map here means “image of open is open” (contrast this with continuous meaning “preimage of open is open”)

Your task is to show that if f : X -> Y is a local homeomorphism, then the image of any open subset U, so f(U), is open in Y

[fierce lily]

so given f:X->Y is local homeomorphism, we can know f is bijective continuous with f^-1 continuous right?

[quartz horizon]

I think that would imply f is a homeomorphism

But there are definitely local homeomorphisms that are not homeomorphisms

[fierce lily]

I may need to know what information does f local diffeomorphism give( like continuity, bijective etc?) Or you can just share how you will get started?

[quartz horizon]

The screenshot you have gives all the details of what a local homeomorphism is

[novel acorn]

1. you need to show that for any open subset U of X, f(U) is open in Y
2. I mean you can but I don't see why
3. A local homeomorphism isn't generally surjective

also a homeomorphism is different from a diffeomorphism

[fierce lily]

ok, so local homeomorphism is different from homeomorphism. and for part3, if we just say f:X->Y is homeomorphism, then we can let f(U A_i)=Y right?

[novel acorn]

you can say the last thing is true without f being a homeomorphism
It just needs to be surjective
Because that's the same thing as f(X) = Y

[fierce lily]

I am struggling which information can I use as it given above? for example: continuity of f? and If I can use the condition f|Ux->f(Ux) is homeomorphism directly?

[fierce lily]

so in order to prove this question, which kind of information can I apply here? I think I can apply continuity of f and f|Ux is a homeomorphism, but not sure if it is true?

[novel acorn]

use the fact that a homeomorphism is an open map

[fierce lily]

so when i start with this question, first prove that f|Ux is a homeomorphism, then apply this fact?

and the condition that I can apply is f is continuous( but bijective is unavailble?)

[novel acorn]

It's in the set up of the problem that f|Ux is a homeomorphism so you don't have to prove that

[fierce lily]

so continuous and f|Ux is a homeomorphism are both given ?

[novel acorn]

yes that's what's written in the problem

[fierce lily]

but it seems that for every point x has open neighbor U, f(U) is open is also given?

and it still wants me to prove that

and also it seems that I can just say f|Ux is homeomorphism-> this map is open, and the proof is done?

[novel acorn]

okay yeah so
It doesn't want you to prove that

it wants you to prove that for EVERY open set possible

the image is open

the statement "for every x there is an open x \in U_x such that" is very different from "for every open U"

think of it like this

around every x there's lots of open sets containing it

the problem says

well we know this holds for one of these open sets

but prove that for all of them at least the image is open (we can't guarantee that it's a homeomorphism tho)

[fierce lily]

Then because X is topological space, suppose there is a topology T generated by B and pick every open U in T, U must contain point x in X if X is non-empty. Then because U is open, we get x in Bx subset of U. and because open set is union of basis elements, then U=Union of Bx such that f( U Bx) is open because of f(Bx) open, do you think this proof is valid?

My purpose is prove for every U in T, we have basis Bx such that f(U Bx) =U f(Bx) is open

[fierce lily]

the above statement is the proof I gave for this question, but I am confused about if I am allowed to just assume X is topological space and assume T is topology on set X generated by Bx?

[unreal stratus]

What do you mean by Bx

You are given topological spaces (which comes with the data of topologies)

[fierce lily]

for Bx, i mean X topological space-> have topology T on X generated by basis B. So Bx is basis elements such that for every U in T, we have x in Bx subset of U

[unreal stratus]

Topological spaces don't have a given basis though

[fierce lily]

so X topological space without given basis-> we can not assume there has a topology generated by some basis B?

[unreal stratus]

Well you can always take the basis to be the topology, but I would recommend just not thinking in terms of bases

[fierce lily]

I mean if my proof given above is correct? Or you can also share how you will get started if not thinking of bases?

[prime elbow]

Let f_n : X -> Y be a sequence of bounded functions from one metric space ( X,d_X) to another metric space ( Y, d_Y). Suppose that f_n converges uniformly to another function f : X -> Y. Suppose that f is a bounded function.

Now I have to prove that the sequence f_n is uniformly bounded, there exists a ball B_d_Y(y_0, r) in Y such that f_n(x) in B_d_Y(y_0,r) for all x and all positive integers n.

Now since it is uniformly continuous so for all e > 0 there exists m in N such that d ( f_n(x), f(x) ) < e for all x in X and for all n ≥ m.

Since f is bounded so there exists a ball such that f(x) in B(y,r) for all x in X

Now we have d(f_n(x), y) ≤ d(f_n, f(x)) + d(f(x), y) implies d(f_n(x),y) ≤ e + r, for all n≥m

Now we have to bound f_1,...,f_m-1

But they already are bounded by B(y_i,r_i) corresponding to f_i

Now I have to show that if f_1 is bounded by B(y_1,r_1) and f_2 is bounded by B(y_2,r_2) then f_1 and f_2 is uniformly bounded

We can show that by f_1 and f_2 is uniformly bounded by B(y_1, r_2 + r_1 + d(y_1) + d_(y_2) )

d(f_2(x), y_1) ≤ d(f_2(x), y_2) + d(y_2,y_1) ≤ r_2 + d(y_2,y_1) ≤ r_1 +r_2+ d(y_2,y_1).

d(f_1(x),y_1) ≤ r_1

So finite bounded sets can be uniformly bounded

So f_1,..,f_(m-1) is uniformly bounded and f_n for all n≥m is uniformly bounded

Thus the sequence is uniformly bounded

Is it correct?

[rancid umbra]

uniformly convergent, not uniformly continuous

[rancid umbra]

the outline seems correct

[gritty widget]

Can $\mathbb{R}$ be embedded in $2^\mathbb{N}$?

[gentle osprey]

bucketbot

[rancid umbra]

viewing 2^N as an ordinal with its order topology?

or just as sets?

[alpine nest]

Or as the space of binary sequences with product topology?

[gritty widget]

cantor space

well seems it doesn't embed tho

[alpine nest]

Well no, there's an immediate argument if you invoke ||connectedness||

Way to ruin my spoiler

I was joking!

[rancid umbra]

spoiler was the way to go anyway lol

[tender halo]

yeah cantor space is 0 dimensional, and any of its subspaces is therefore also 0 dimensional

[prime elbow]

Okay thank you ❤️

[unreal stratus]

I love the Cantor

[prime elbow]

Is there any source where Cantor is described in a better way? I face a problem when Cantor comes

[bronze wadi]

What are exactly the subspaces of R^n homeomorphic to R^n?

[sonic crane]

The only “clopen subspaces” are singletons? 🧐

I understand how the only connected subspaces of Rlower are singletons

[quartz horizon]

Ok that seems sus

The empty set and the whole space are always clopen

[sonic crane]

Hya. Thats what i was thinking

[quartz horizon]

Also finite unions/intersections of clopen should be clopen too

[alpine nest]

I'm almost certain it should have been "connected"

[sonic crane]

Ya that would make sense

[unreal stratus]

I am no Cantor-space expert tbh Idk

[quartz horizon]

Perhaps a mix-up occurred because like

You’re disconnected iff you have nontrivial clopen subsets

[unreal stratus]

Ye

[quartz horizon]

So it’s not like connectedness and clopenness are completely unrelated

[unreal stratus]

||modulo considerations and conventions about empty sets||

[alpine nest]

Yeah, probably

[quartz horizon]

Though I don’t think that particular relation is what’s being used here

[tender halo]

better than 2^N?

[prime elbow]

Sorry I don't get it

I want to study Cantor set properties but i don't get any intuition for it

[heady skiff]

could I have a hint for d please? My idea was to take any basis element of f and then construct a continuous function based off the basis element containing f. However I've run into some problems, such as making the distance between |f(x) - f(c)| < e for points x_i, x_j arbitrarily close to each other by say x_j in U_j = (1/2, 1] and x_i in [a, .0000001); i guess i could combat this by choosing delta which excludes that distance between each of the x_j which belong to U_j =/= X_j, but then what about the case that we have X_j = the entire space and some U_j next to it? I don't know how to define a continuous function f that satisfies the criterion for all U_j if that makes sense

here's the picture i had in mind

[dire dove]

They must be open by invariance of domain, and contractible because R^n is contractible

Apparently this is not sufficient though, googling i found this

https://math.stackexchange.com/questions/55114/are-contractible-open-sets-in-mathbbrn-homeomorphic-to-mathbb-rn

[bronze wadi]

Damn this question seems harder to answer than I expected

I thought there’d be a pretty simple characterization

[heady skiff]

can somebody check this answer? it feels wrong somehow

[rancid umbra]

i mean, in R^2 it’s true that every open bounded simply connected subset is homeomorphic to R^2

it fails in R^n by looking at B^n - 0 for n > 2

it doesn’t seem unreasonable that every open bounded subset of R^n with pi_n trivial is homeomorphic to R^n but i haven’t given this any thought

[unreal stratus]

Well, add connected

[heady skiff]

anyone?

[bronze wadi]

I noted this but I feel like it’s missing continuity?

I’m not 100% sure

[tender halo]

yeah it needs to be continuous

[bronze wadi]

alright thx

[unreal stratus]

Eh often people assume any map you talk in topology should be continuous

Though I guess yeah fair enough if this is pointset

[tender halo]

its not like there are any interesting theorems where the maps arent continuous

i cant think of anything really

[heady skiff]

can somebody check my d please?

[tender halo]

sure it works

[heady skiff]

oh damn okay

i'm surprised

[tender halo]

uhh

hmm

let me read it again

[heady skiff]

okay

[tender halo]

idk why are you doing all this work

[heady skiff]

yeah my classmates said show F(I) is contained in the closure of C(I)

but by then i was too deep into the proof lmao

[tender halo]

what is A

isnt it a single point

by that point you won right?

it extends to a continuous function that is trivially inside U

why are you doing all this arcane stuff with C

[heady skiff]

i guess A should technically be lambda, but they're all distinct points

[tender halo]

ah i see

[heady skiff]

yeah sorry if notation is bad

[tender halo]

A is not a subspace of B(I) though

[heady skiff]

oh i meant A is a closed subspace of I which is a closed subspace of R

[tender halo]

ohh sorry now i get it

i now understand whats happening, yeah that works

[heady skiff]

all g all g

[tender halo]

i dont know what you do in the middle but the core idea is correct

[heady skiff]

basically, i'm eliminating the concern that if you have two proper spaces U_i and U_j, (say disjoint) close enough to each other and f(x_i) in U_i and f(x_j) in U_j, you have some distance between them which cannot be arbitrarily bounded

to fix this, i'm just choosing delta so that we don't have to worry about that

for example say we have U_i = (1/2, 1] and U_j = [0, 1/4) with f(x_i) = 3/4 and f(x_j) = 1/8

then we can't bound their difference if |x_i - x_j| < d if that makes sense

[tender halo]

you have g, g is continuous function from a finite discrete space to R

it can be extended to f: R -> R, either by tietze or lagrange or just piecewise linear functions

f is obviously inside U

[heady skiff]

okay thanks

i'll double check my proof based off that and edit

i feel like it would be very nice to have an approximation theorem for continuous functions (maybe on a compact set) for e)

wait

wait this has something to do with series

which i completely forgot

[tender halo]

it has to do with stone-weierstrass theorem

[heady skiff]

oh yeah weierstrass theorem is literally this lol

funnily enough ive never encountered that theorem in any of my analysis courses

[fierce lily]

Can someone help me with this question? I am trying to argue set {x| f(x)>g(x)} is open set, and for every neighborhood containing x, there must be an open set U' containing x such that f(U')>g(x) for all x, because f is continuous. Do you think this is correct?

[prime elbow]

Order topology on Y is Hausdorff space?

[tender halo]

yeah i am not great at reading today apparently

[rancid umbra]

put the product topology dictionary order on Y x Y
consider the map x |-> (f(x),g(x))
show that {(y,y'): y <= y'} is closed in Y x Y and that its preimage under the above map is the set you are interested in

[heady skiff]

lmmao i remember this problem

kickedm y ass

don't we need M to be complete in order to involve Corollary 4.62?

[low flame]

hey im trying to do the second countablity of the product topology

Can i write under all the A1, A2... An

that each of these A are the length |N|^k = |N|

Then transition into the countable union?

[fierce lily]

can you share how you will prove [(f,g): f<=g} is closed in Y x Y here?

[fierce lily]

I construct cont map h: x->Y x Y but I have no idea how can I show {(f,g): f<=g} is closed in Y x Y

[rigid heart]

Hello, when you are for example removing a open disk from a torus, what is the most natural way to define the topology on your new surface with boundary plz?

[opaque scroll]

Subspace topology I guess

[rigid heart]

yeah but to define a manifold on it it must be with diffeomorphism to open sets of the closure of the upper half plane. Im troubled by how you would describe open sets that covers the disk

[opaque scroll]

I mean you can just cover the boundary by two open sets, just like you would put a manifold structure on a circle.

[rigid heart]

Then u wouldnt need to add anything to prove that it is second countable?

[opaque scroll]

No.

Like your questions are sort of orthogonal. The topology you can just take from the subspace topology and then it becomes second countable automatically.

Then you also want to put a smooth structure on it, but the smooth structure doesn't override the topology.

[rigid heart]

for what im interested in, i dont need smooth structure, ill come back, if i can ask, on my writing

thanks

[rigid heart]

im blocked when i want to say that a small enough neighbourhood of a surface without a disk is homeomorphic to an open set of the upper half plane intersecting the x-line, i feel i need to prove it

[opaque scroll]

Well, you're able to pick a neighborhood that contains the hole right, so it's enough to show it for R^2

[rigid heart]

a small enough neigbhourhood of the boundary my bad

[bronze wadi]

Im trying to wrap my head around this funky uniform topology on R^omega, is this statement true?

[rigid heart]

like I feel the urge to build homeomorphism else i dont find it very rigorous. But when you consider random open set of a boundary point, then it's not obvious to me why it is homeomorphic to an open set of the closure of the upper half plane intersecting the x line

[fierce lily]

I now start with a cont map h: x->Y*Y and want to prove {(f,g): f>g} is open, but I have no idea about how to prove this set is open?

[unreal stratus]

Preimage of a certain set

[fierce lily]

so I am trying to prove {(f,g): f>g) is open in order to apply its preimage of a continuous map

but I don't know how to argue that {(f,g): f>g) is open

because it seems to be different from what we operate like {x: x>2} etc

[unreal stratus]

No it is p much the same

So by using another "preimage argument" lol you reduce to showing that { (a,b) | a > b } is open in R^2

But you can now show this is the preimage of another open

(or do it by hand)

[fierce lily]

you mean trying to show { (a,b) | a > b } is open in R^2 is same as {(f,g);f>g} open in Y x Y?

[fierce lily]

I am actually considering that both f(x) and g(x) might be the subset of R^n, so if for every pair of (f,g), I don't know whether it is still correct to make epsilon ball around(f(x), g(x))

[unreal stratus]

Wait sorry I misread smth

[unreal stratus]

This doesn't really make sense like you're talking about f,g in which space

I thought you meant {(x,y) | f(x) <= g(y)}

sorry for not reading carefully enough

[fierce lily]

so is it not correct to construct a continuous map h: X->Y x Y defined by h(x)=(f(x), g(x))?

[unreal stratus]

Yes you should do that

Now the set is the preimage of { (y,y') in Y^2 | y <= y'}

and you can show this is closed from the definitions

[fierce lily]

then I am somewhat confused about how can I show { (y,y') in Y^2 | y <= y'} is closed?

since I think y is not necessarily belong to R but R^k?

[unreal stratus]

well Y right

needn't have anything to do with R^N

[fierce lily]

then how you will prove that set is a closed set?

[tender halo]

just by definition really

[fierce lily]

which one? because y seems not necessarily in R but in R^k, can I just use graphical proof in this case( I know it contains all boundary points means closure)?

[tender halo]

Y is just some ordered set

nothing to do with reals

which one
the definition of order topology

[fierce lily]

I know that order topology can be defined on any ordered set, so it should be the same as justifying like a subset of R^2 is closed right?

[tender halo]

well uhh sure (modulo what you mean by "same")

a graphical proof is not a proof though

[fierce lily]

one question: why p(x,y)<=d(x,y)( at the bottom of screenshot) can let us conclude that B_L^2(x, epsilon) subset of B_p(x,y)?

[fierce lily]

and also for part(b), I know it can be done if I can find an subset of R^inf that is open in finer topology but not in coarser topology, but I don't know how to find that set in these four topology?

I mean for ex: between Uniform topology and box topology etc.

[heady skiff]

can I have a hint for f? I know that f is in the intersection of an open set from B(I) and D(I) if and only if f(j) = 1 and f(i) = 0 for all i =/= j if and only if U_i contains 0 for all i =/= j and U_j contains 1, but i don't really know a finer characterization'

[tender halo]

hm? D(i) is just the discrete space with cardinality of the continuum

[slate bane]

How can I see that a continuous surjection from the disk to a space X such that when restricted separately to either the interior or the border is an embedding (homeomorphism into the image) is actually a homeomorphism from the whole disk to X?

[tender halo]

you just need to prove it to be a bijection cuz disk is compact

if there is a map f: X -> Y such that for a dense subset A of X the restriction of f to A is an embedding, then f(A) is disjoint from f(X \ A)

X needs to be Hausdorff

[slate bane]

My problem is that X is not assumed to be Hasudorff

[unreal stratus]

What happens if you take like the natural map D^2 -> D^2/((0,1) ~ (0,0))

That seems to be a counterexample but maybe it fails for some reason

[dim meadow]

Take a point on the interior and the boundary which map to the same point in X and draw a line segment between them. If you restrict the line segment to the open disk then you get the image is a loop in X. Which is mapped to by the half open line segment in the open disk. But this violates the homeomorphism property?

[slate bane]

Actually I think it works too. Probably they forgot Hausdorff in the text for X then

Wait sorry let me reread your message autumn, I thought you were on Potato side

[unreal stratus]

lol

i don't fully get autumnghost's message

[unreal stratus]

why should the image be a loop in X after you restrict?

[dim meadow]

The point is suppose the map is not injective. Then you take a closed line segment in the disk which maps to a loop in X. Then if you restrict it to the interior it still maps to a loop. But the map is a homeo when restricted to the interior

Because it's the image of a line segment which identifies the endpoints

[unreal stratus]

but you restricted it to the open disk

oh sure sorry

[dim meadow]

But you still have the same image

[unreal stratus]

i'm silly aha

yes okay

I was thinking loop in the sense of map [0,1] -> X

not like the set itself

[dim meadow]

Anyway do you agree with this argument?

[unreal stratus]

I don't see a contradiction yet

[dim meadow]

You can even just say the image is compact while the half open line is not compact

[unreal stratus]

that i agree with

nice

but now i'm confused as to how this interacts with my purported counterexample hmm

[tender halo]

X is the domain in the theorem i mentioned

so the disk

[dim meadow]

Lol

[slate bane]

Oh now I see it too wow, very nice. I think the problem is that when you fold the open disk onto itself on a single point you lose local connectedness maybe

[slate bane]

Ops

[tender halo]

so yeah by the lemma i posted its bijective and therefore a homeomorphism

[slate bane]

Oh wait that's where I wasn't convinced though

Aren't you applying compact hausdorff theorem?

[tender halo]

uh sure

[slate bane]

So we would need Y T2 as well

Or not?

[tender halo]

yea

its obviously Hausdorff because uhh

uhh

there should be an easy argument why im sure

ok i cant come up with an easy argument actually

how about this, Y is second countable, so Hausdorfness is equivalent to sequences only having one limit, and that obviously holds in Y

because if we have a limit inside the disk we cant have any limits on the border because we can separate the border and any point in the interior

[slate bane]

I can't see it easily

Why can't we have a sequence in Y in the image of the border the disk that converges to the image of an element of the interior of the disk?

[tender halo]

because if it converges to a point in the interior then look at the image of a slightly smaller inner disk

from some N all the points in the sequence are inside the smaller disk yet the closure of that smaller disk does not intersect the border

so it cant converge to a point on the border

[slate bane]

But when we look the image O of the small open disk, this is only an open in the subspace in Y which is the image of the interior of the disk, but open sets in Y neighbourhoods of that point could be larger

[tender halo]

it cant

[fierce lily]

If there are spaces X and Y with metric d and d'. and we know d>d', can I just argue that metric topology on X is finer? for every y in Y with y belongs to B_d'(x, epsilon), we can always find B_d'(x, epsilon) subset of B_d(x, epsilon)?

[fierce lily]

question part(a) when I want to know why l^2 topology is finer than uniform topology, I don't know why p<d can let us conclude B_L^2(x, epsilon) subset of B_p(x,epsilon)

[tender halo]

sorry i cant write anymore im gonna go sleep glhf

[slate bane]

I think we should try to prove the image of the interior is open

Thanks! Rest well

I could only say its closure is all Y for now

[slate bane]

Well, this problem is too hard for me. If someone comes up with a solution, please ping me

[slate bane]

Original problem

[plush folio]

My idea for this problem is that $g \mapsto g\cdot s$ is a homeomorphism for each s, so $Os = {o\cdot s | o \in O}$ is open for each s, and therefore $OS = \bigcup_{s \in S} Os$ is open. Does this work?

[gentle osprey]

sheddow

[plush folio]

it doesn't use the Hausdorff assumption, so I'm not sure

[novel acorn]

what does inclusion with the circle mean?

[plush folio]

O is open in G

just a weird piece of notation our lecturer uses, never seen it before

wow, there's even a unicode symbol for it: ⟃ maybe it's not so obscure after all

[tiny obsidian]

I mean the fact it's in unicode doesn't mean it's not obscure

like these brackets are in "Miscellaneous Mathematical Symbols-B Block"

⦕ ⦖

[ocean canyon]

If a continous surjective map f:X ---> Y takes saturated open sets to open sets, does it take saturated closed sets to closed sets, and vice versa?

[fringe thorn]

yes, a continuous surjection takes saturated open/closed sets to open/closed sets iff it is a quotient map

so saturated open sets to open sets <=> quotient map <=> saturated closed sets to closed sets

[tender halo]

what does saturated mean?

[fringe thorn]

choose your favourite characterization

these are all the ones I know

[ocean canyon]

yeah but it says "or", not "and" in Lee. So I was wondering whether one implies the other and vice-versa (prop. 3.60)

[fringe thorn]

unless I'm mistaken, I interpreted the proposition to mean "q is a quotient map iff it takes saturated open sets to open sets, or iff it takes saturated closed sets to closed sets"

it seems like that is indeed what Lee meant
https://math.stackexchange.com/questions/2068296/a-map-f-maps-saturated-open-subsets-of-x-to-open-subsets-of-y-if-and-only-if

proving the closed case is pretty similar to proving the open case anyways

[boreal sequoia]

is topology closer to algebra or calculus?

also any book recoms?

[tiny obsidian]

neither really, though if I was forced to give an answer it feels more like analysis than algebra to me

[tender halo]

topology of the reals is closer to calculus, topology of the integers is closer to algebra

[sonic crane]

So in the product topology X x Y, open sets are the collection of all unions of open in X cross open in Y right?

Im trying to remember what the weird rules were for products tho in terms of intersection or unions

Like U x V intersect M x N is not necessary U intersect M cross V cap N?

Or was that for unions?

[fringe thorn]

yes

[fringe thorn]

this one is true

the union one is false, yes

[grave solstice]

Many algebraic structures carry an interesting natural topology. For example, Galois groups

[paper wedge]

yo

[paper wedge]

we know that if f : X-->Y is continous , X is compact, and Y is hausdorff then f is closed

can i take X to be the discrete topology on R, Y to be the std topology

then X is not compat and literally any f ( identitiy ) would work as acounterxample for when X is non-compact

does this work?

[tiny obsidian]

not literally any f

[fringe thorn]

a constant f would not work, I don’t think

the identity seems fine though

[fringe thorn]

cause every closed subset would be sent to a singleton, which would be closed

[paper wedge]

yeah

thank u guys

and girls

[fierce lily]

i actually feel somewhat confused if this two p has the real difference: p and p_bar?

both seems to find the max of |x_i-y_i|? The only thing that seems different for me is that d_bar will discuss min{ d, 1}

[rigid heart]

u want to define a distance for two sequences, for example with (x_n := 1/n)_n and (y_n := n)_n p(x,y) = + inf, u dont want that so you change the metric to force a boundary to prevent urself from getting those illcases

[fierce lily]

so p_bar here just want to keep distance between each pair of x_i and y_i less than 1?

[rigid heart]

not for each, but for the set of pairs

[fierce lily]

why not for each? is it not correct for each i?

[rigid heart]

I mean ur checking the sup of the distance between each, so u dont keep the information about each i, u have a bound for the distance for each if it is not 1 but thats all u have

[fierce lily]

then do you think the uniform topology on R^j is the same as product topology. Since uniform topology induced by metric p on R^n is the same as product topology( thm 20.3 on munkres)

but here it only writes uniform topology finer than product

[tiny obsidian]

iirc Munkres uses 'finer' to include the case where they are the same

[rigid heart]

same for coarser, so 20.4 just proves that for J infinite we have the uniform topology is finer than prod top and coarser than box top

[bronze wadi]

not sure i see the equivalence in definition here

are open sets satured?

because what if we have a open set in Y that is not saturated and neither is its preimage

or does surjectivity imply that the preimage of a set is saturated?

[craggy cedar]

Yes, the preimage of a subset of Y must be saturated

[bronze wadi]

ah i see why the rest follows then

wait is the preimage saturated because f(f^-1(A)) = A

i forgot that is true only for surjective maps

so in some sense, satured sets are sets on which the function acts in an injective matter right

[craggy cedar]

Ah, not quite

Consider X={0,1} and Y={2} and the map f(x)=2. This is surjective, and the only saturated subset of X is the one for which f is not injective when restricted to it

[bronze wadi]

right cause the preimage is many elements

the sets just happen to have a property that injective maps give

if every subset of X is saturated, do we have that f is injective?

well, we only need to consider singletons being saturated

[craggy cedar]

Sorry, back. The preimage of a subset of Y is saturated because if we have x in f^{-1}(A), then we have f(x) in A and f^{-1}(f(x)) is a subset of f^{-1}(A)

[bronze wadi]

i mean if we apply f^-1 to both sides we get what we need right

both justifications work tho youre right

[gentle osprey]

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

[sonic crane]

Very basic question but, subsets of X x Y in general cant always be represented by U x V where U subset X and V subset Y right

[fringe thorn]

no

[sonic crane]

yea it seems counterintuitive if u havent thought about it properly before

to me anyway

[fringe thorn]

even for open sets in X x Y, they are in general unions of open sets of the form you describe

[sonic crane]

yeah

thx

[fringe thorn]

just a explicit counterexample I’ll speak to: let X = Y = R, and take A \subseteq X x Y to be the open ball of radius 1 in R^2

it’s fairly intuitive that there’s no way to write this ball as the Cartesian product of two subsets of R

[sonic crane]

yea i guess cause subsets like (subset R) x (subset R) are necessarily some rectangle type of thing .... but maybe with some missing pieces

This is weirdly subtle to me lol

[drowsy iron]

Can someone give me an intuitive reason why this is the case? Cause I would think that it would only have 2 elements, one equivalence class of loops that go around the origin, and one equivalence class of loops that don't go around the origin

[quick delta]

What about loops that go around the origin twice?

[drowsy iron]

Ah of course, hadn't thought about those

So is the fundamental group of R2-{0} isomorphic to the integers?

[quick delta]

Yes

Although proving that is slightly fiddly

[drowsy iron]

Ah alright, thanks a lot

[sonic crane]

A subset X, B subset Y, showing that product topology on A x B (generated by subspace topologies on A and B) is the same as the subspace topology on A x B

This should be kind of trivial when u set it up right?

Basis element of product topology on A x B is U x V with U open in A V open in B, this is equal to the basis element for subspace on A x B, (U x V) cap (A x B)

[drowsy iron]

Oh yeah, I forgot about deformation retracts. Haven't done topolgy since last year

[sonic crane]

Oh i think something is off here

U open in A doesnt mean U is open in X

I think I know what to do though

[sonic crane]

So is the part that makes this work is that intersections work well with cartesian products?

[quartz horizon]

Yeah essentially

[sonic crane]

The main part of this question is justifying that open sets in gamma are like {(x,f(x)) | x in an open interval in R} right

[gritty widget]

I mean maybe you can start with constructing a basis for the subspace topology on the curve

then you define a new function,
g: x \to (x,f(x) ) from R to graph

and you also have the projection from graph to R, $\tilde{g} : (x,f(x) ) \to x$

[gentle osprey]

something

[gritty widget]

if you a continous surjective function in two directions then the spaces are homeomorphic

[gritty widget]

it's a difficult question to be honest

to the first order, your observation is all there is to it. There are two main categories of homotopy classes , one which wind around the removed point, and ones which don't

but within the stuff of loops which wind the removed point, the loop that winds around n times is not homotopic to the one that windws around m times (m neq n)