point-set-topology

[hexed vault]

Please try to understand and internalise this way of burning and boiling mathematics to its core because this is what mathematics really is about.

Many times , the genius isnt in proving theorems but rather making the correct definitions such that they truly characterise the "exact" properties you want to model. And generalisation is just a consequence of this action.

@quartz horizon @storm wadi i hope this helped , again , these are my views , they could be wrong altogether aswell , please take with a grain of salt

[worn mortar]

This entire response was very well written

Well, one can make the case for some concepts and principles in engineering school.

But this doesn’t really add or take away from anyone’s points

[young stone]

An object c in a category C is called detecting if for every monomorphism f: a -> b that is not an isomorphism there exists a map h: c -> b that doesn't factor through f

[rancid umbra]

so in Set, 1 is detecting since for any strict injection f : a —> b, we can choose an element x of b not in the image of f as our map x : 1 —> b that doesn’t factor through f.

but 0 is not detecting

[young stone]

Yep

There is a related notion of a separating object, an object c of a category C is separating if for every pair of distinct morphisms f,g: a \to b there exists a morphisms h: c \to a such that fh ≠ gh

A separating object in a balanced category is detecting

Maybe this should be in #category-theory

[cosmic mirage]

hmmm how is it related to logic?

[hexed vault]

Topos theory , but if you want a much more directly related answer , open sets characterise the truthability of any statement in that open set

[hexed vault]

For further readin , read topology via logic book

[unreal wigeon]

Is there a nice reason that monics in Top are topological embeddings (i.e. homeomorphisms onto their image)?

[cosmic mirage]

thanks for the reference!

[gritty widget]

Monomorphisms aren't in general embeddings in Top

for example, letting DX be the discrete space with the same points as X, consider Id:DX to X sending every point to itself. This is a monomorphism independent of X.

[unreal wigeon]

Omgg, of course...! Thank youu!!

[gritty widget]

no problem

[unreal wigeon]

I actually have another quick question! How about continuous maps with left inverses??

[gritty widget]

Those are called continuous sections I think, you can probably find a good bit of info about them by searching for that

I don't think they're all embeddings but I can't think of a counterexample off the top of my head

[unreal wigeon]

I'll have a search! Thanks again haha :))

[rancid umbra]

the topological embeddings are precisely the regular monics in Top

[finite token]

If you restrict the domain your left inverse of f to im(f), don't you get a two-sided inverse just from set-theoretic properties?

[quartz horizon]

two-sided set-theoretic inverse sure

but it may not be a two-sided continuous inverse

[quartz horizon]

these are split monics

otherwise known as retracts

https://ncatlab.org/nlab/show/extremal+monomorphism

in Top, extremal monomorphisms are subspace embeddings

extremal is the weakest kind of mono/epi stronger than an ordinary mono/epi

[unreal wigeon]

Thank you guys!!

[unreal wigeon]

Just to be sure, if the left inverse is continuous, then it's an embedding?? So I guess the statement is that continuous maps with continuous left inverses are embeddings

[quartz horizon]

Yes this is true

[finite token]

I assumed continuity of the left inverse was implied

Otherwise it's just another way to say injective

Which is in turn just another way to say monic

[storm wadi]

This is a really beautiful way of describing it. I will certainly come back to this thought going forward as you're right it certainly captures the essence of abstract math. Category theory sounds really interesting too, hopefully I can take a course on it soon. And thank you for taking the time to help improve my conceptual understanding!

[rancid umbra]

couldn’t think of a counter-example either

i feel like there should be one tho

[dense laurel]

"What is closeness ?"

does it have to do with https://en.wikipedia.org/wiki/Separated_sets ?

also, what might "precisely" mean in the last sentence of "precisely separated by a continuous function"

wikipedia seems a little verbose (not always precise) and scattered (too many pages linking for one definition)

[quartz horizon]

is this a correct argument?

[unreal stratus]

Looks good to me

Though what decade was this book written in if it says "separated" aha

[quartz horizon]

this is tom dieck's algebraic topology

2008 i think...?

[unreal stratus]

Ok yeah maybe just a bit peculiar then

[quartz horizon]

also - is there a nice way to package "clutching datum" in a categorical way?

i guess a standard example for this would be the clutching datum of the charts of a manifold together with transition maps..

[tiny ridge]

That's effectively the same data as a TOP(n)-valued 1-cocycle on the manifold M where TOP(n) is the group of germs of homeomorphisms of (R^n, 0)

In general I suppose you could describe a groupoid where objects are homeomorphisms between open subsets of R^n

Then an atlas on the manifold is exactly a Cech 1-cocycle valued in this groupoid

But do you really want to think this way?

[quartz horizon]

idk, do i?

[tiny ridge]

"No"

[quartz horizon]

so how should i think of clutching datum

[tiny ridge]

On a case by case basis depending on context.

[unreal stratus]

I wouldn't massively worry about this anyway like I wouldn't say this is particularly standard terminology but related stuff comes up all the time like descent data

Just here it is kinda funny in that these are just sets (or later topological spaces ig)

[lament steppe]

If I have a map f: X -> Y (f is a quotient map in my case but I dont think that matters) where we are restricting f to U which is a subset of X, then do we assume the topology on U is just the subspace topology U of X?

[tender halo]

yes

[lament steppe]

Is this by convention? Or is this forced upon us in some way?

[kind marlin]

in general it's safe to assume a subset of a topological space is meant to be interpreted as a subspace

subspace topologies satisfy a special property: it's the coarsest topology on the subset such that the inclusion map from subset to set is continuous

one reason this is desirable is that since any map restriction can be made by composing the inclusion with the original map, then as long as the map is continuous, the restriction is also continuous

[rancid umbra]

on the flip side of that, if U is endowed with some other topology, the inclusion map U -> X may not be continuous, and now your "restricted" function may not be either.

[warped shore]

I'm having a little bit of trouble with understanding local trivializations of vector bundles. pi^(-1)(U) and U x R^n being homeomorphic makes sense but the other conditions are a little confusing intuitively even though I could state them

Something about linear isomorphism of fibers to R^n

Idk if this is the right channel btw

[kind marlin]

to add onto this

assume that we want the restriction to be continuous, and we equip some topology U' such that the inclusion is continuous to make this true. Since U is the coarsest topology on the subset such that the inclusion map U -> X is continuous, U is coarser than U', meaning the identity map U' -> U is continuous, so we can "factor" the inclusion U' -> X as U' -> U -> X

so, if we want a restriction to preserve continuity, then every topology we could pick for the domain to give a continuous inclusion can be factored through the subspace topology, making it the "minimal" topology to do this. (you could probably find discontinuous inclusions that give continuous restrictions, but that would likely depend on the codomain of the map, whereas continuous inclusions do not depend on the codomain to make restrictions continuous)

[tiny ridge]

You'd want the homeomorphism U x R^n <-> p^-1(U) to commute with the projections of U, at least

The says fibers are mapping to fibers. Since you're working with vector bundles you'd want this fiberwise map to be a linear isomorphism and not just a homeomorphism

The final point being the distinction between vector bundles of rank n and topological fiber bundles with fiber R^n

There are examples of vector bundles which are isomorphic as topological n-plane bundles but not as vector bundles due to Milnor but it's not quite trivial

[warped shore]

Yes of course

[warped shore]

What do you mean fibers mapping to fibers

[unreal stratus]

Point is you want the diagram with pi^-1(U) -> U x R^n and the projections to commute. In other words the map restricts to maps pi^-1({u}) -> {u} x R^n, and these two guys are called the fibres over u of the maps U x R^n -> U and pi^-1(U) -> U

[quartz horizon]

it might help to think of this in the "indexed" point of view as well?

[quartz horizon]

is the following proof correct:

[midnight umbra]

it's all correct although i think you can simplify a couple of steps by using the fact that the projection maps are continuous

[tiny obsidian]

Yes

[midnight umbra]

particularly, for $\prod_{j\in J}$ is closed $\Leftrightarrow$ for all $j\in J$, $A_j$ is closed, you can show the $\Leftarrow$ direction by

$$\prod_{j\in J}A_j=\bigcap_{j\in J}p_j^{-1}(A_j),$$

which are all closed

[gentle osprey]

Convergant

[tiny obsidian]

There's also a characterization of continuity in terms of closure which you could probably use to do the question in the provided order

[midnight umbra]

yeah that works for one direction of the double inclusion, although it doesn't really simplify that step because we seem to already have that the arbitrary product of closed sets is closed

i misunderstood what you said lol

yeah

[quartz horizon]

Oh damn that’s very nice

[quartz horizon]

Could you explain a bit more?

[midnight umbra]

it's the characterisation that $f:X\rightarrow Y$ is continuous iff, for all $A\subseteq Y$, $f^{-1}(\overline{A})\supseteq \overline{f^{-1}(A)}$

[gentle osprey]

Convergant

[midnight umbra]

so if you think about the double inclusion proof at the end, one direction can be done with this characterisation

[tiny obsidian]

[quartz horizon]

Interesting

Thanks for this, I’ll rewrite the solution to make it more efficient

I’m learning algtop but I want to shore up my point-set foundations, so I’m really grateful to get advice regarding that

[midnight umbra]

fits the name lol

[quartz horizon]

Yeah part of the reason I’m learning algtop is to get access to higher category theory

[midnight umbra]

since we're here, would anyone mind having a look at my proof that every metric space is regular and normal? my professor invited me to send it to him but i havent heard back after a week lol

[tender halo]

that works yea

for a cheat solution, consider the function f = d(x, F) / (d(x, F) + d(x, G))

its continuous from X to I, 0 on F and 1 on G, preimages of [0; 0.5) and (0.5, 1] are disjoint open sets containing F and G respectively

[midnight umbra]

cheat solution indeed lol but i like it

very slick

my original idea was to prove that $d(F,G)>0$ but this is actually false unless you assume they're also compact

[gentle osprey]

Convergant

[quartz horizon]

This even shows metric spaces are T6

I.e. perfectly normal

[quartz horizon]

ok I managed to shorten the proof considerably

[quartz horizon]

can i get a hint for this proposition

i know how to show that 1 implies 2 and 3, and that 2 and 3 imply 1

but i don't know how to show 2 implies 1 or 3 implies 1

if f is closed then i at least know that Y is T_1

[crisp lintel]

oh this is a fun one, I remember doing (3) implies (1) let me look and see if I can find a good hint

oh the way I did (3) implies (1) was by first establishing (2), so (2)\implies(1) should be easier

[quartz horizon]

yeah that i don't know how to do...

[crisp lintel]

you want to use some of the separation properties for a compact hausdorff space

[quartz horizon]

i know compact hausdorff spaces are normal

[crisp lintel]

in particular normality should help

[quartz horizon]

so if i take $y_0 \neq y_1$ then $f^{-1}(y_0)$ and $f^{-1}(y_1)$ are disjoint closed subsets of $X$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

thus i can find disjoint open neighbourhoods of them

[crisp lintel]

yea that's the first step, then there's one nontrivial step after that

you need to use these new open neighborhoods and basically make better ones

[quartz horizon]

hm...

[crisp lintel]

that interact better with the map

[quartz horizon]

so i have $U$ and $V$ disjoint open neighbourhoods of $f^{-1}(y_0)$ and $f^{-1}(y_1)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

the only way i can see to use f is closed is like

[crisp lintel]

or well not new open neighborhoods in X, but open neighborhoods in Y

[quartz horizon]

$f(U^c)$ is closed, and $f(V^c)$ is closed

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[crisp lintel]

the idea is that f(U) might not be open because you can't just take complements and use the fact that its closed

but you can define a neighborhood that is smaller than f(U)

and plays a bit nicer

[quartz horizon]

Which using f being a quotient map means $f^{-1}(f(U^c))$ is closed and $f^{-1}(f(V^c))$ is closed

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

so... $f^{-1}(f(U^c)^c))$ is open and $f^{-1}(f(V^c)^c)$ is open

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

maybe this is what i want?

[crisp lintel]

it might be, I defined them more explicitly in the thing I'm looking at but its possible it simplifies to that

[quartz horizon]

wait hang on

i recognise these operators

$f((-)^c)^c$ preserves intersections

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

so, since $U \cap V = \emptyset$, we have that $f(U^c)^c \cap f(V^c)^c = \emptyset$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

so hm, is it sufficient to use $f(U^c)^c$ and $f(V^c)^c$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[crisp lintel]

this I believe is correct

since the whole pre-image of y_0 belongs to U

I think gives you that y_0 is in f(U^c)^c

[quartz horizon]

ah, and then we can use $f^{-1}({y_0}) \subset U$, so that ${y_0} \subset f(U^c)^c$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

mfw right adjoint

ok so that shows f closed => Y hausdorff

[crisp lintel]

this is what I had written which is similar but not an abstract quotient map

[quartz horizon]

then, i want to show that R closed implies f closed

for f to be closed, i need that if $C \subset X$ closed, then $f(C)$ is closed, i.e. that $f^{-1}(f(C))$ is closed

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

now, $x \in f^{-1}(f(C)) \iff f(x) \in f(C) \iff \exists c \in C, f(x) = f(c)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

hm...

if $C$ is closed then $C \times X$ is too

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

meaning $C \times X \cap R$ is as well

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

since X x X is compact hausdorff, C x X intersect R is compact, meaning pi_2(C x X intersect R) is compact, thus closed

[crisp lintel]

yeah that sounds right

[quartz horizon]

and... i think that is f^{-1}(f(C))

[crisp lintel]

f^{-1}(f(C)) is the image of C x X \cap R under the right projection since its just the stuff equivalent to C

[quartz horizon]

yeah

ok so that should show R closed => f closed

f closed => Y hausdorff

and then the final

it's interesting that this Q required the right adjoint

[crisp lintel]

I did a lot of thinking ab this a while ago trying to get a nice classification of the closed *-subalgebras of C_0(X) for X locally compact hausdorff

if you assume that your subalgebra doesn't vanish anywhere (i.e., each point in X there's some function f\in A with f(x) nonzero)

then closed *-subalgebras correspond exactly to proper equivalence relations (or closed equivalence relations, if X is compact)

where a proper equivalence relation is one where the quotient map is proper (closed + pre-image of compact sets is compact)

but the locally compact case was a total headache

[quartz horizon]

i wrote it up

[crisp lintel]

Capitalize your Hausdorffs!

[warped shore]

Don't wanna drift too far off topic but I have noticed people writing Abelian as abelian...

I guess it's an honor when your name becomes recognized as an ordinary word?

I suppose boolean is Boole

[queen prism]

having one's name an uncapitalized adjective is the highest honor a mathematician can possibly get

https://mathoverflow.net/questions/44946/why-is-abelian-infrequently-capitalized damn I realized I was missing a word after the fact

[warped shore]

Oh right cartesian

That's another common one

[quartz horizon]

Is it bad that I’m having to look up a lot of point-set results

I’m getting stuck on lots of propositions in Dieck’s book

Like this

[hexed vault]

Could possibly be

[balmy briar]

Have you tried using this?

[quartz horizon]

i solved it already

[balmy briar]

What's the solution?

[opaque scroll]

Say K is not contained in any Xk, then for each i pick an xi in K that is not in Xi. Then K\{xi} gives an open cover of K

[balmy briar]

ah got it thanks

[quartz horizon]

i think you have to do a little more than this

[opaque scroll]

Like argue that points are closed in X you mean?

A set is closed in X iff its intersection is in Xi for all i

[quartz horizon]

Not that

Where’s the contradiction in your argument

[opaque scroll]

Oh yeah, derp. I meant
K \ {xi : i > n}

[quartz horizon]

And - how do you know that’s open?

[opaque scroll]

Because the complement is closed

Because in Xi it's a finite union of closed points

[quartz horizon]

why?

[opaque scroll]

Because the first i points are in Xi and the others are not

Well, the others are not anyway

[quartz horizon]

Hm…

How do you know x_{n + 1} is not in X_n

[opaque scroll]

It was chosen to not be in Xn+1

[quartz horizon]

all you know is that x_{n + 1} is in K \ X_{n + 1} right

[opaque scroll]

So then it's not in Xn

[quartz horizon]

ah I see, then that should work

[cosmic mirage]

huh i guess this is more or less the proof that topologically compact spaces are (a subcategory of the) categorically compact

[quartz horizon]

Oh I guess you’re right

Hom(K, -) preserves filtered colimits?

[cosmic mirage]

exactly

[quartz horizon]

That’s neat

[cosmic mirage]

in this case the inclusion K --> X has to factor through a finite stage because thats how unions work

[quartz horizon]

And for a more general map, the image is a compact subset of the codomain

And thus the inclusion of the image has to factor through a finite stage

[cosmic mirage]

yes

[quartz horizon]

Which I think means the map itself does

[cosmic mirage]

are you aware of the weird thing with categorical compact spaces containing more than topological compact spaces

[quartz horizon]

Yes

[tacit wren]

yeah exactly since the map factors through its image anyway

[quartz horizon]

Though the result here required each X_I to be regular

[cosmic mirage]

Yeah, you won't be able to say much categorically about spaces unless you restrict your notion of what a space is

any weakly hausdorff space will satisfy this condition though and that's not too much to ask for from a space

[quartz horizon]

oh wait not regular

mixed that up

it's just $T_1$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[cosmic mirage]

ye exactly

[cosmic mirage]

(maybe this is blasphemous in the point set channel but algebraic topology basically defines a space to be compactly generated weakly hausdorff)

[quartz horizon]

one question i have is that

is there any way to define that notion of space without first developing topological spaces?

[cosmic mirage]

not sure

[tranquil cosmos]

You can define spaces in general via locales

[cosmic mirage]

I need to get ready for class rn so can't think about this more but you might get some mileage out of CW complexes which can be defined more combinatorially, especially since attaching maps only matter up to homotopy and often boil down to degrees of sphere maps

[tranquil cosmos]

That is to say, instead of looking at a point set X with power algebra P(X) and a topology O(X) that is a subset of P(X)

You look at any poset K and a suitable subset L of K

With these "union of opens is open" etc. properties

I'm sure you can get slightly more abstract definitions of compactly generated and so on

[tranquil cosmos]

Or maybe K is a lattice or a complete lattice, something like that

And L (your locale) has all infinite joins and finite meets and the top and bottom elements

Idk if it's the right framework for compactly generated spaces in particular, but it's a way to think about point set topology without the point set

Sometimes called point-free topology or, very charmingly, pointless topology

[cosmic mirage]

>point-set topology channel
>no set of points

[tranquil cosmos]

😌

Yo these guys pop up in topos theory, where they're the lattice of subterminals

There's an even vaster generalization called a Grothendieck topology (you know shit gets real when it's named after Grothendieck), where you replace the lattice with a general category (which does not need to be posetal or anything) and you replace "open covers" (which survive into the picture of locales) with "horrible generalization of open covers called Grothendieck topology" and then you celebrate because it's so abstract

But the thing is, I'd bet that there's a good notion of compactly generated for those spaces

And it also makes bad spaces good

[rapid shuttle]

You're arrested due to charges of blasphemy for mentioning algebraic topology in this sacred channel

[rancid umbra]

what does categorically compact mean?

[cosmic mirage]

i mean in the sense of a compact object in a category

So preserving filtered colimits

Oh wait this is bad notation, I am saying that X is compact if Hom_C (X, -): C --> Set preserves filtered colimits

so sneakily viewing X as the functor it corepresents via Yoneda

[rancid umbra]

ah okay

[cosmic mirage]

and in a cruel twist of fate, there are more spaces that are compact in the category of (nice) spaces than just those which are compact topological spaces

[rancid umbra]

like categorically compact is a strict generalization over topologically compact?

[cosmic mirage]

yes, a topologically compact space is compact in the category

actually now that I'm writing this I am not sure which category, but it's either a suitable category of spaces or the homotopy category of such a thing

[rancid umbra]

Top is really poorly behaved lol

[cosmic mirage]

If I make a statement about spaces, it is most likely to be correct in the homotopy category of CGWH spaces

[tacit wren]

actually categorical compact objects in Top are just finite discrete spaces, so it's way more restrictive

[cosmic mirage]

ah cool

Oh I guess you can detect this in Set where this is obvious

nice

does the thing I am saying only work in the ∞-category of spaces/anima 🗿

[tacit wren]

yeah mostly an anima thing, it definitely simplifies the categorical definitions

[cosmic mirage]

damn, too ∞-pilled for this

@quartz horizon sorry I need to amend my statement to you. This is the claim that compact objects preserve these specific colimits but they don't preserve all filtered colimits (see above)

and this is a shadow of the fact that you can compute homotopy colimits of a directed system of inclusions as just usual colimits

so compact objects in the ∞-cat of spaces, which include compact reasonable spaces, will preserve these too

@tacit wren thanks for the correction

[tacit wren]

all good man, category theory notation is a minefield anyway

[steep wedge]

Isn't exercise 20 ii) (the dense set part) wrong?
consider any non-empty topology which has a disjoint singleton free basis.
What could they have been meaning to ask?

Or infact any topology in which there is a basis with a non-singleton element disjoint from all others

[rancid umbra]

if A is dense in X, A’ = X because every point of X is a limit point of A.

so A - A’ = A - X = 0

[tiny obsidian]

that's not what dense means though

The obvious subset of Q is dense in [0,1] u {2}

but 2 is not a limit point of it

[rancid umbra]

did i forget to subtract a point

in the definition of limit point

[tiny obsidian]

well "limit point" normally means accumulation point

[rancid umbra]

(A - {x}) \cap U is non-empty for any open U containing x

[tiny obsidian]

yes

[rancid umbra]

alr

my b

[tiny obsidian]

A is dense in X if cl(A) = X

[rancid umbra]

ye

[tiny obsidian]

So yeah I agree the exercise is wrong

[rancid umbra]

what’s the best way to fix it

[tiny obsidian]

Probably a separation property...

[rancid umbra]

yea, you don’t want any open points i think

[tiny obsidian]

well open points aren't the issue

[rancid umbra]

hold on, sorry

[tiny obsidian]

(in fact we would prefer them since such X is irrelevant)

[rancid umbra]

this is partly a granularity issue tho.

we don’t want our open sets to be too big

and just stop shrinking at some point

like, then you run into the issue that you have an open set which is disjoint from all others

[tiny obsidian]

I'm pretty sure you need hausdorff

[rancid umbra]

was literally just typing that

[tiny obsidian]

t0 isn't enough

[rancid umbra]

then i think my original argument goes through

like, if A is dense in a Hausdorff space X, and x is in X, if we take some open neighborhood W of x and any other point y of W, we can separate x and y with disjoint opens U and V, resp., and now A \cap (W - x) is non-empty because it contains A \cap (V \cap W).

i still think you can’t have open points either, because if W = {x}, it won’t work. maybe i’m overlooking something again.

[tiny obsidian]

well the assumption is that X has no isolated points

so

we are given no open points

[rancid umbra]

ah nice

thanks Edward II

[tiny obsidian]

this still feels like a strange correction to the question

[rancid umbra]

why?

[tiny obsidian]

idk it feels like the sort of thing you wouldn't forget to write down as an assumption

[rancid umbra]

does the case when A is open work without the Hausdorff property?

[tiny obsidian]

yes

[rancid umbra]

yea, that is kind of weird then

[tiny obsidian]

if U is open in A and contains a in A, then U is also open in X so contains another point (X has no isolated points), and that point is in A so a is not isolated in A either

[rancid umbra]

i think you can fix it if you require points to be closed.

let U be an open nbhd of x in X, A dense in X.

U must have another point different from x because X has no isolated points.

if points are closed in X, then A \cap (U - x) is non-empty by density.

and so A’ = X

the cofinite topology on an infinite set is an example of a non-hausdorff space where points are closed.

so X needs to be T1

[frozen acorn]

Could someone check these proofs to see if there are any problems?

Im more concerned with problem 11 as I wrote 10 much more quickly

[rancid umbra]

i don't think the reason why the two open balls are disjoint should be omitted in number 10. same with 11.

otherwise, it looks solid.

[frozen acorn]

Yeah, thats what I was concerned about. I get why that statement is true for the euclidean metric, but im unsure on how to prove that they are disjoint in some arbitrary metric

[rancid umbra]

you need to use the triangle inequality

[frozen acorn]

Ok I think I see it

[warped helm]

huh why?

it just follows from the definition of the sets

[rancid umbra]

to prove that for y with d(y,x1) < d(x1,x2)/2, you can't have d(y,x2) < d(x1,x2)/2, you need prove that d(y,x2) >= d(x1,x2)/2

[warped helm]

oh i see

[frozen acorn]

Could I stop you there

[warped helm]

i was tunnel visioned on the centers

[rancid umbra]

yea

[frozen acorn]

These are homework problems, so I'd like to get it done myself, just needed a little hint

[rancid umbra]

ofc

[frozen acorn]

Thanks for the help!

[steep wedge]

I have no clue what hausdorff is

[rancid umbra]

X is Hausdorff if given any two distinct points x and y of X, there are disjoint open sets U and V containing x and y, respectively.

most nice spaces are Hausdorff, for example, metric spaces are Hausdorff

[steep wedge]

So basically everything is distingiushable

all singletons are closed.

[steep wedge]

Stronger than this

[rancid umbra]

yes, Hausdorff implies all points are closed, and the converse is false by the example i gave here

[steep wedge]

Yeh

Ok, with that it's easy ig

[rancid umbra]

yea

[rapid shuttle]

That's why it's named point set

The other one is algebraic topo

It's continuous

Kinda

[ripe creek]

töpölögy

[cerulean yoke]

Hi, can anyone help me with a topology question, I know this should be obvious but idk how to prove it rigorously.

By Jordan curve theorem a closed curve splits the plane into two connected components when removed. So let's say I have two closed curves X and Y. Suppose that X separates the plane into A and B. Suppose Y separates the plane into C and D.

Now suppose also that the curves intersect at exactly points, lets call these points p and q. Then the whole thing looks like 4 paths between p and q, which are disjoint except for at p and q. So let's say we take the path from X that lies (apart from p and q) in C. Let's take also the path from Y that lies (apart from p and q) in A.

These paths together form another closed loop. So by Jordan Curve theorem I understand that it splits the plane into two regions when removed. But how do I actually prove that these regions are AnC and BuD?

I'm sure this is like a one-line observation that I just don't see 🙁

[tender halo]

that doesnt seem true in case the curves touch internally

an ellipse inscribed into a circle or something

but we can ask for that to not happen, in that case uhh

[cerulean yoke]

OK, lets add the condition that A not subset of C, A not subset of D, B not subset of C, B not subset of D

And vice versa

Ie C not subset A etc

[tender halo]

its shorter to ask for p-q paths to lie in different halves

[cerulean yoke]

Sorry , wdym?

You are saying its shorter than my condition?

OK anyway yeah if they DONT touch internally then how do I prove this rigorously?

[tender halo]

B \cup D is connected, its easy to see that it doesnt intersect this new path, A \cap C is also connected and doesnt intersect the new path

therefore its the separation that we are looking for

to see that they dont intersect the path we need to use the conditions we imposed on them not touching internally or externally

that is, we need to say that the four p-q paths lie in A, B, C, D respectively

(p-q paths of X lie in C and D, p-q paths of Y lie in A and B)

[cerulean yoke]

Sorry, I am not understanding why the proof follows from this

[tender halo]

B doesnt intersect the new path because one half of it is on B's boundary, and the other half lies in A

[cerulean yoke]

The new path being the closed loop?

[tender halo]

yea

call it Z or something

[cerulean yoke]

ok wait so B is disjoint from Z and since B is connected it must be in one side of Z

same with D

and AnC

OK makes sense lol thank you kindly

[prime elbow]

I am trying to find an counterexample such that C is closed subset of R under standard topology and C is not convex and dist(0,C) not attained in C

[tender halo]

dist(a, C) is attained in C \subseteq R^n for any closed C and any point a

in any LCH metric space even

[cerulean yoke]

Hi. Sorry I'm still confused 🙁

Why must AnC be connected?

And why must BuD and AnC lie on opposite sides?

I understand I am probably being very slow, I would appreciate it if you kindly clarify

[cerulean yoke]

So my thought is we dont actually need AnC connected. But we consider a vertex in AnC and we try and get from it to BuD without touching Z. We can escape AnC by either escaping A or escaping C. If we escpe A then still in C so the part of X we must have touched is the part in Z. Same for escaping C.

But there's an obvious hole in this: we could escape both AnC at exact same time. Idk even if this is the right approach or not, I really feel like there's a much better way to do it....

[robust drum]

Elementary thing that I am struggling with. Given a manifold, a compact subset K, and a finite open cover U_1, ..., U_p, find a refinement of the cover to U'_i such that U'_i is precompact

the hard part here is obtaining just one precompact set for each i and that still being enough

My thoughts:

(1) wlog assume the U_i are precompact, by intersecting with some bigger precompact open set U containing K.

(2) try to find some smaller balls that are still enough to cover U_i cap K and such that each ball is compactly contained in U_i. Then the closure of the union will be compact as a subset of the closure of U_i, but the problem is that it's no longer guaranteed to be contained in U_i (because U_i cap K is no longer compact so not guaranteed to require finitely many)

despite each ball being compactly contained in U_i

using something like paracompactness is not desired because I'm trying to prove it roughly

[rancid umbra]

any manifold has a basis of precompact coordinate balls.

[robust drum]

true, but how do I get enough of them to find a compact set contained in each U_i such that the unions of the compact sets still cover K?

U_i cap K isn't compact, which takes out what I would like to do

[rancid umbra]

each U_i is the union of precompact opens

[robust drum]

yes, but when I take their closure it might leave U_i

like the interplay is

(1) I need enough of them to still cover U_i cap K
(2) I need the closure of these to still be contained in U_i

[rancid umbra]

wait, why do you need (2)? i thought you only needed a refinement of precompact opens

[robust drum]

ah good point, my question didn't strictly require that. Anyway what I'm actually trying to do DOES require that. More specifically, in the functional analysis class I'm in I'm trying to prove existence of partitions of unity in R^n (that's why don't use too much manifold theory) and the statement of the problem is:

"given any open cover of a compact set K by U_1, ..., U_p, find Theta_1, ..., Theta_p with Theta_i compactly supported inside U_i summing to 1 on every point of K"

I've already proven that a bump 1 function exists for an arbitrary pair of a compact set contained inside an open. The thing that's bugging me out here is the constraints that I find exactly one function in the partition of unity for each original open and that it's compactly supported in U_i, and that the points where one of them are positive cover K still

[rancid umbra]

ah okay

[robust drum]

like my strategy is basically: just find functions where the points they're positive cover K, then I can divide by the sum to get something that's identically 1 on K

open to an easier way of doing it tho

bc that's the problem statement as written this time not hiding stuff lol

[rancid umbra]

and you only need to do it on R^n?

[robust drum]

yeah

it's the same thing tho lol

i saw the proof for manifolds first it's lowkey easier once you've already shown paracompactness

i suppose I could directly prove paracompactness for R^n

but I don't think that's what he's looking for/it shouldn't be necessary?

idk, I could mimic the manifold proof verbatim obviously but it feels like I should already have enough using just the structure of R^n and existence of bump functions & that I'm only being asked to do it for compacts

[rancid umbra]

i have an idea. but i don’t want to answer prematurely. going to write some stuff down; im a bit rusty with partitions of unity. maybe somebody else more familiar will be able to answer before me

[fallen canyon]

what's the q?

[rancid umbra]

@fallen canyon

[crisp lintel]

you can prove this for any locally compact sspace

let me find it in my notes and see iif I can give a hint

[crisp lintel]

this is the kind of thing you need right? just to be sure

my hint would be to use induction

[robust drum]

Yes

Ok excellent

thanks

[fallen canyon]

i mean why is this hard - -is this fancy q on refinement?

sry just curious if there was a trick here

[crisp lintel]

the question is to find a partition of unity for a compact set

I wouldn't say its crazy hard but it is a bit annoying there's moving parts

even with this lemma getting a partition of unity takes a bit of a trick to combine functions in the right way

I think there are a few ways to do it though

[fallen canyon]

ya shrink to refined cover and then normalise

[robust drum]

The question is to show that you can shrink with the same number of sets

[fallen canyon]

im so confused so yr askiung fr a proof of the shrink lemma?

i mean lee has that

[robust drum]

What's the "shrink lemma"

[crisp lintel]

here its only for finite partitions of unity for compact sets which makes things a bit simpler I believe

though I think the induction argument for finite covers can just be replaced with zorn's lemma and it works in general provided you put the right assumptions on

[fallen canyon]

^^

@robust drum sorry !!!! i do apologise to everyone i usually do not post

[robust drum]

nah it's fine im overreacting cause i was frustrated

by the problem

sorry lol

[fallen canyon]

let me really try to be precise here - -and not be an a-hole

[fallen canyon]

this is the q?

[robust drum]

sorry I knew I was overreacting as soon as I typed the comment. You're not being an a-hole (more likely I was). I was just lashing out. Yep that's it

[fallen canyon]

so youe mention para compactness so did ya define it yet

[robust drum]

Yeah - open covers have countable subcovers

[tranquil cosmos]

that's lindelöf

[robust drum]

trueee ok

ur right

it's the open cover has locally finite refinement ?

[fallen canyon]

that not paracompactifited refinement

[tranquil cosmos]

paracompact is the thing guaranteeing partitions of unity i think?

so locally finite refinement yeah

[fallen canyon]

but you're on the right track

@robust drum yr a person with a blue thingy

[robust drum]

Yeah it just means im terminally online and talk a lot lol

[fallen canyon]

so i have to be careful here because i am not a mathematician per say -- just a math ohd and phys post-doc

paracompact is every cover opwn has local finite refine; linfloff is the countable sub -- in this instances no paracompact stuff is needed

[crisp lintel]

if you only need partitions of unity for finite covers over compact sets you don't need paracompactness or anything

[robust drum]

The lemma Blake shared is enough for my purposes yeah

[fallen canyon]

^^

thank you @robust drum keeping me in check

[robust drum]

np, I apologize for lashing out and thanks for the help all

[viscid blade]

bro is being cooked by eth zurich

[viscid blade]

what course is this btw ?

[robust drum]

FA2

[prime elbow]

oh

how?

[tender halo]

uhh im not sure if i lied there

hmm

[crisp lintel]

Yeah I don't think it's true in a general LCH metric space

[tender halo]

maybe you need like sigma compactness or something

[crisp lintel]

it follows in R^n by the heine borel property

but yeah I think sigma compactness would be enough

maybe

actually I'm not even sure about that

[tender halo]

nah still not true

i think you can like

[prime elbow]

what is sigma compactness?

[tender halo]

countable union of compact spaces

[prime elbow]

oh

[tender halo]

take R with metric bounded by 1, adjoin a copy of R with metric bounded by 1 such that distances between points in different halves are 100, adjoin another copy with distances to it from outside being equal to 100 - 1/2, another with 100 - 1/2 - 1/4, etc

take C with points in each copy except the first one and a from the first copy

then dist(a, C) = 99 but its not realized

[prime elbow]

what do you mean?

[tender halo]

but the space itself is locally homeomorphic to R and sigma compact and locally compact and all that jazz

[cosmic valley]

i love you guys naming things - perfectly completely nothing-to-see-here normal space that Hausdorff would let into his Haus.

[round lagoon]

Who's even naming ts bruh

[steep wedge]

Section 5 3) is obviously false, consider for example the constant empty function, what extra assumptions should I make for it to be true?

[steep wedge]

And another way it is false is that the indiscrete topology gives a subset of the conditions of any other topology, so any function which works from some topology also works for the indistinct topology

[barren sable]

I need recommendations for introductory books on topology

[warped helm]

munkres

[barren sable]

Any other one? This book is so expensive in my country

[warped helm]

all books are free

[barren sable]

Yeah but I want a physical copy

[plush salmon]

is it just me or does urysohns lemma feel very "choice-y"

im somewhat skeptical of being able to pick out that many neighborhoods

[plush salmon]

engelking if you hate yourself

kelley is a bit more gentle

and as a bonus, it has a treatment of a set theory stronger than zfc in the appendix

[queen prism]

print it out

[barren sable]

any thoughts on willard?

[plush salmon]

seems like it has somewhat strange ordering

its somewhat nice that it does a little alg top, but i think its better to just read may's book for that

[barren sable]

right, I think I'm gonna buy kelley then

[plush salmon]

wise choice

also the exercises are quite fun

iirc he makes the reader develop the theory of filters

[barren sable]

that looks good!

[midnight umbra]

can you show me the proof in your book?

[tender halo]

it uses DC

it is independent from ZF + CC

[quick crane]

"Real Analysis: Modern Techniques and Their Applications" by G. Folland

[solemn iris]

I understand the idea perfectly, though I don't understand why continuity is needed here. I get the $g(V)\subset U$ and $g'(V)\subset U'$ is due to continuity, but even then aren't $g^{-1}(U)$ and $g'^{-1}(U')$ both neighborhoods of $x$? It seems like I can just take their intersection and have a common neighborhood of $x$, then produce $y\in A$ etc.

[gentle osprey]

bluepianist

[warped helm]

that's essentially what they're doing

[hidden abyss]

$g^{-1}(U)$ needn't be a neighbourhood of $x$ if $g$ is not continuous

[gentle osprey]

Jussari

[solemn iris]

oh right brah i implicitly used that and i didn't realize it

thanks

[warped helm]

thanks chatgpt

[mild cape]

“Gauth

Also i didnt fully use “gauth”

I elaborated it so I can get better grammar and reword it easier

[warped helm]

dont copy and paste llm outputs to answer questions here

[tired roost]

LLM answers are strictly against the rules.

[worn mortar]

Hu General Topology is good so far in my experience. Kasriel is good.

[civic wigeon]

why is the product topology not just defined as a product of the open sets in each component? i.e, let's say we have top spaces X_i, and we want to define a topology on prod X_i over i. why are the open sets here not just tuples (U_i)_i in I, where U_i is open for each X_i? (apparently what I defined here is the box topology? i'm so confused)

[rancid umbra]

it doesn’t satisfy the universal property of the product in general

[civic wigeon]

interesting

[crisp lintel]

the main thing you want in the product topology is for each of the projection maps to be continuous, the product topology is basically the minimal topology where this happens

If you take the pre-image of an open set under one of the projections, you get a big product where all but one factor are equal to the whole space

so then taking finite intersections of those sets

[civic wigeon]

oh huh this doesn't happen in the box topology?

[crisp lintel]

you get the product topology

the projections are still continuous in the box topology it's just far too strong

Another way to see this is that the product topology is the topology of pointwise convergence, so a sequence (or more generally, a net) in a product space converges if and only if each component converges

so for example if you take the set of all functions from R to R (this is an uncountable product of R with itself)

the product topology as mentioned gives you pointwise convergence

the box topology is far stronger, for a sequence of functions to converge in the box topology it must converge uniformly (and in fact, even uniform convergence isn't enough for the box topology)

[civic wigeon]

i see, thank you

[midnight umbra]

in addition to the above, an arbitrary product of compact spaces isn't guaranteed to be compact in the box topology (whereas it is in the product topology, that's Tychonoff's theorem)

[red sage]

maybe try out the proof that a function given by components onto the product space with product topology is continuous if and only if the components are measurable.

In the proof you will notice that considering a basis with a topology where we constrain on finitely many components allows us to use property that finite intersection of open sets is open (as you know in general arbitrary intersection of open sets may not be open)

[red sage]

but this isn't true in general right?, if projections are continous with respect to box topology, then product topology would be finer than box topology, making it equal (as box is always finer than product) , but this isn't true in general.?

[midnight umbra]

the product topology is coarser than the box topology

thats precisely why the projection maps are also continuous in the box topology

because the preimage of every open set is open in the product topology, and the box topology is finer (has more open sets,) so the projection is also cts in the box topology

[red sage]

I think I got confused between the wording, "coarser" and "finer"

[midnight umbra]

thats kinda what i thought might have happened, no wories

[slender glen]

This is a pretty common confusion

[robust drum]

I’m surprised no one has given you a counterexample . The function f(t) = (t, t, t, …) is continuous into the space of sequences with the product topology but not the box topology

f : R -> R^N

Obviously this map should morally be continuous so this is why the box topology is bad

[midnight umbra]

good example

[robust drum]

Thx it’s not mine tho just the standard one in most textbooks

[midnight umbra]

lol yw

[midnight umbra]

another one is that if you equip ${0,1}$ with the discrete topology, the box topology on ${0,1}^\mathbb{N}$ is discrete (so not compact b/c the set of points is infinite,) but the product topology is the Cantor space, which has a much richer topology and is compact, hausdorff, completely metrisable, etc

[gentle osprey]

Convergant

[robust drum]

Also a nice example

It’s nice that it’s the cantor space, it makes sense intuitively but it’s also nice

[brittle rapids]

in general you need dependent choice but for second countable spaces you don't

[plush salmon]

that makes sense

[lament steppe]

If we have a map f: R^2 -> f(R^2) such that f(x,y) = (x,y,g(x,y)). The topology on R^2 is the Euclidean topology and the topology on f(R^2) is the Euclidean topology in R^3.

Is f always an open map? I want to say that if g is not continuous, this cant be true, but I have some logic written down that seems to hint at the fact that g could be discontinuous and f could still be an open map.

[midnight umbra]

what are we supposing about g?

[lament steppe]

g : R^2 -> R and pretty much nothing else

[midnight umbra]

as a simple example, if g is the constant map 0, then f is not an open map b/c $f(\mathbb{R}^2) = \mathbb{R}^2 \times {0}$ is not an open subset of $\mathbb{R}^3$

[gentle osprey]

Convergant

[warped helm]

i said this too but i think they mean f(R^2) as a subspace of R^3

[lament steppe]

Yea, I specifically mean being open in f(R^2).

It could be that I dont have a firm grasp of what it means for an the image of f to use the euclidean metric in R^3...

[warped helm]

it would just be the metric on R^3 restricted to f(R^2)

aka the subspace topology

[lament steppe]

So assuming that f(R^2) is a subspace of R^3, if f always an open map, I dont see how Convergant's example disproves my thought.

[warped helm]

it does not

hm

[midnight umbra]

it doesn't, in fact obviously f(R^2) is never going to work because it's by definition open in the subspace topology

I've been thinking about it a bit more and honestly not gotten very far. my first thought was to choose a really pathological function like Conway's base 13, but that actually would make f an open map... lol

I do think with a horribly behaved enough g you would be able to find a counterexample, but I haven't found it yet

[lament steppe]

Well thanks for digging! I felt crazy thinking that any map would work, so if others out there arent immediately disproving me, I feel slightly better about my thought.

Event if its ultimately false.

[warped helm]

hm ok i have a weird idea

take the unit ball in R^3 and look at cross sections in the same plane

for each line y = c have f map that line to a space filling curve that fills up that cross section

hm idt this will work actually

[urban zinc]

Yes it will always be an open map; the image of an open set U under f is the intersection of U x R (which is open in R^3) and f(R^2), so f(U) has to be open in f(R^2) by definition of the subspace topology

g can be any function

[crisp lintel]

that doesn't quite seem right

the g part of the function is also only applied to U

however I believe it's still right because f is invertible and the inverse of f is just the projection onto the first two components, which is continuous

actually nvm you are right

because g is a function, so the intersection with U x R takes care of the fact that g is only applied to U

[warped helm]

ok im glad its this simple

[midnight umbra]

https://tenor.com/view/shrimp-as-that-clash-royale-hee-hee-hee-haw-gif-25054781

thank you

[willow stirrup]

how do i prove a set is open a gain

in Euclidean space

[velvet salmon]

Depends, but usually you have to show that for any element x of the set some small ball around x is contained completely in the set as well

[willow stirrup]

wait I use triangle inequality

[willow stirrup]

should i just send question

[velvet salmon]

Probably if you want help for the specific problem

[willow stirrup]

i think the C1 part is for the second part

do I use the definition of an open ball

[velvet salmon]

Let x in U. Because U is open there exists an open ball around x entirely contained in U

What happens to that ball when you subtract a from every element?

[willow stirrup]

i wanna say the center of the ball is shifted by a

bc x is the center of the open ball contained in U right

sry im literally just learning topology on top of this course

[velvet salmon]

Exactly if you can show this you're done

[willow stirrup]

oh nice

i like use euclidean metric/ triangle inequality right

[velvet salmon]

I don't think you need the triangle inequality

[willow stirrup]

idk i did a problem like this in the past

this is fro my analysis class

[velvet salmon]

Yes, so you know that open balls are open. But here you just need to show that the shifted ball is actually an open ball around x - a. To do this, take any value y in the ball around x and look at the distance between y - a and x - a, what do you get?

[willow stirrup]

a neighborhood of a?

wait idk how that is relevant

[velvet salmon]

Let me rephrase. Your goal is to show that there is an open ball around x - a completely contained in V. I claim that shifting a ball of radius r around x by - a gives a ball of the same radius around x - a

This ball will then be contained in V if we choose r sufficiently small, because U is open

[willow stirrup]

waitone second

[velvet salmon]

To show that my claim is true, you need to verify that: the distance between some y and x is the same as the distance between y - a und x - a

Can you verify that this is true?

[willow stirrup]

yea Br(a) = { x \in U| d(x-a,y)< r} ?

is what i want to show right

[velvet salmon]

No this is just the definition of an open ball around a

The edited equality is not well defined. What is y?

[willow stirrup]

i wanna show this contained in U right so do i just take an element in the ball and show it is also an element of V or is that not sufficient enough

[willow stirrup]

i think i made a mistake

[velvet salmon]

Can you show that this is true?

[willow stirrup]

ignore what i typed give me a sec

y is an arbitrary point in V right

[velvet salmon]

y can be literally any point in R^n

[willow stirrup]

ohhh

i need to draw a picture i think

[velvet salmon]

I'm literally just stating that shifting by a constant doesn't change distances

[willow stirrup]

like the metric is invariant

[velvet salmon]

Try to write down what the distances are

[willow stirrup]

like d(x-a, y-a)= d(x, y)

[velvet salmon]

Yes

[willow stirrup]

oh

so like

oh

[velvet salmon]

To see this is true just look at the absolute value expression that d(x-a, y-a) evaluates as

[willow stirrup]

yea like

|y - a - (x - a)| right

why it do that

[velvet salmon]

Yes that's right and if you simplify it becomes...

[willow stirrup]

|y - x|

thanks g

[velvet salmon]

Yes, okay. So know you know that d(x,y) < r is equivalent to d(x-a,y-a) < r so from there you can conclude

[willow stirrup]

both balls are the smae

and then V is open

[velvet salmon]

Yes

Wait, not the same

But the translated ball is also an open ball

Of the same radius

[willow stirrup]

oh

so really what we’re saying because they are if the same radii

then V must be open as well

[hearty vale]

not just radii, also centers match

[willow stirrup]

oh

bet

[warped helm]

this is somewhat of a dumb exercise imo

dont even prove things directly just write down the clear homeomorphism g : U -> V

[willow stirrup]

its my analysis class

the second part is this

ive never taken topo

lowk

[warped helm]

has homeomorphism been defined for you

[willow stirrup]

no

[warped helm]

o oke

then a direct proof will have to do

[willow stirrup]

im cooked for RA exam tmrrw

i miss intro single var RA

[velvet salmon]

Wdym? If i shift the ball of radius 1 centered at 0 by 3 the resulting balls are disjoint

[willow stirrup]

what book is good for topology in RA

i need to keep track of the basics

[warped helm]

pugh

munkres

[willow stirrup]

which title for Pugh

[warped helm]

real mathematical analysis

[quaint spruce]

W. Rudin, Principles of Mathematical Analysis Ch.2

[warped helm]

naur

rudin is not explicit enough about lots of smaller details

[quaint spruce]

Not sure what you mean. Can you give any examples? My course followed it and I didn't find any such problems. Proofs also make clear statements and take the optimal path.

Maybe the later parts are very bad, but I thought especially the topology section is written very well

[sour lily]

browder, amann escher

[willow stirrup]

oh yea ive heard of Aman Escher

totally forgot

[warped helm]

mainly lots of the things in "exercises not in rudin" featured in these notes

https://math.berkeley.edu/~gbergman/ug.hndts/m104_Rudin_exs.pdf

[quaint spruce]

Are you saying that the number of exercises in rudin is not sufficient? The question was asking for which books cover the basic theorems, and I thought rudin does so.

[warped helm]

well yes, the exercises make you check a lot of details that rudin doesnt

[willow stirrup]

is this good

[raw hornet]

how to continue the proof can someone give a hint or something regarding it

[ruby delta]

can you recall the definition of the product topology?

[raw hornet]

Sure its defined to be the topology generated by U \times V where U is open in X and V is open in Y

[raw hornet]

.

[ruby delta]

so what does it mean for M to be open

[raw hornet]

well the thing there is no basis for M_x this is the thing that making me stuck if we know that ther is a basis or its not mentioned

i cannot say that V is a basis

but if that is the case then the union of V will equal to M_x and thus M_x will be open

[raw hornet]

so this is what makes me stuck

[raw hornet]

the only the definiton that i know being open means that M is inside T_y

[ruby delta]

but that's not the definition

what does it mean for a set to be open in a topology

[raw hornet]

like this is the only definiton in my book

i mean there are other definiton sure

but those are basis releated

[ruby delta]

M is equal to the union of U x V

[raw hornet]

sure that is true i agree with that since U \times V are basis

for the topology generated by T_x\times y

[ruby delta]

so what can you now say about M_x?

[raw hornet]

M_x is also the union of V maybe

but not sure if that is a valid argument

but this seems to me the case

[raw hornet]

im not sure whre you want to get ):

[ruby delta]

maybe it is easier to think about M_x as instead p(M)

where p(S) = {y in Y| (x,y) in S}

what this really is is a function that we are applying to M

[raw hornet]

this seems to be the projection function

i guess

right

[ruby delta]

yes

[raw hornet]

sure but is there something relatble to the projection function regarding open sets that im missing

[ruby delta]

this is set theory

if a set S is the union of X_i

and I have a function f

what is f(S) equal to

[raw hornet]

ah i see the union of f(S)

for sure

and then we get the result needed

[raw hornet]

im not sure if what i said right

[ruby delta]

I don't understand what you mean here

[raw hornet]

Could you elaborate because I do not get it and it’s to late now

[raw hornet]

😃

[ruby delta]

$M = \bigcup_{\alpha \in \Lambda} U_\alpha \times V_\alpha \implies M_x = \bigcup_{\alpha \in \Lambda} (U_\alpha \times V_\alpha)_x$

[gentle osprey]

HChan

[raw hornet]

OMG this is so smart

You have just fixed the x

And get all others y

[steep wedge]

I have two disjoint countable intersections of open sets F and G, assume all closed sets are countable intersections of open sets, show there exist two countable intersections of open sets A and B, such that A U B is the whole space, A is a superset of F and B is a superset of G.

Does anyone have a weak hint?

I am not even sure if this statement is true

[zealous berry]

just let A and B both be the whole space?

am i missing something

maybe you wanted A and B to be disjoint as well?

[steep wedge]

Yes

I forgot to mention that

[steep wedge]

this is qn 5

showing a and b equivalent is pretty easy

[steep wedge]

this is equivalent to this

the qn

[cunning sleet]

What book is this?

[steep wedge]

james dugundji

can some pls give a hint? i have been stuck on this problem to a long enough extent that i am doubting the validity, sadly this is trivially true for all finite topologies, and i dont have a great intuition for arbitrary infinite ones

so constructing a counter example is a bit difficult

[fierce mesa]

Hi! Can anyone give a hint as to how I should approach part b)? I am currently stuck on the forward implication:
[ x_n \to x \implies x_{n, k} \to x_k ]
If $x_n \to x$ by the above metric, I assume this means that for all $\varepsilon > 0$, there exists an $n_0$ such that:
[
n > n_0 \implies d(x_n, x) < \varepsilon
]
Working with the above condition, then:
[
\sum_{k=1}^{\infty}\frac{1}{2^{k}}\frac{\left|x_{n,k}-x_{k}\right|}{1+\left|x_{n,k}-x_{k}\right|}<\varepsilon
]
I'm not sure how to prove that $x_{n, k} \to x_k$ pointwise.

[gentle osprey]

Average Math Student

[fringe thorn]

use the fact that if the entire series is less than e, then each term in the series must be less than e

[fierce mesa]

oh

Yeah didn't think about that 😭

Thanks, I'll let you know how it goes 👍

oh wait

Whoops

[fierce mesa]

So you end up with something like:
[ \left|x_{n,k}-x_{k}\right|<\frac{1}{1-2^{k}\varepsilon}-1 ]
Now note that [ f\left(x\right)=\frac{x}{1+x} ] is bounded from above by $1$, so immediately the metric should be bounded above by the minimum of $1$ or $2^k\varepsilon$. If the $2^k \varepsilon$ is greater than $1$ then the inequality (with the metric, in the previous image) is true for all values of $n$. \

So we have that:
[ \left|x_{n,k}-x_{k}\right|<\frac{1}{1-2^{k}\varepsilon}-1 ]

Now since we know $2^k \varepsilon < 1$, that means that the choice of a new epsilon:
[ \varepsilon_{1}=\frac{1}{1-2^{k}\varepsilon}-1 ]
Will always be positive. \

Since we only want to prove $x_{n,k} \to x_k$ pointwise, it doesn't matter that $\varepsilon_1$ depends on $k$. Therefore the limit is pointwise convergent, as $\varepsilon_{1} > 0$.

[gentle osprey]

Average Math Student

[steep wedge]

<@&286206848099549185> sry for ping

[low stone]

Hey, so I solved it.

I shall be focussing on (a) since both parts are basically the same.

I wanted to ask where you are getting stuck exactly. That will allow me to provide you a good hint

[steep wedge]

i am unable to make a construction

by any chance is your soln "consider a family of open set who intersection is H, and take the intersection of those open sets in this family which are disjoint from G"

?

if so this doesnt work

@low stone

[low stone]

Ok, I'm here.

[steep wedge]

so any hint?

[steep wedge]

or was that not your soln?

[low stone]

It wasn't

Sorry, I took a while to ponder if they're different or not

[steep wedge]

was it "take union of (interior of complement of G) and H"?

this also doesnt work (i mean if it does, i am unable to prove it is F_sigma, and i am pretty sure you cant, but dont have an explicit counter example)

[low stone]

Ok, so the hint I will provide you is to think about the complements of G and H. Then, modify those (which are clearly F _sigma) to be disjoint and follow the given conditions.

Is that good enough? Or too vague?

[steep wedge]

one sec

i dont get it, i have gone along that line of thought before, but it didnt lead me anywhere

[steep wedge]

something like this, but with F sigma instead of G delta, rite?

[low stone]

Actually yeah that's exactly what my solution is

[steep wedge]

yeh, i have gone through that thought process but couldnt get a soln

[low stone]

With F sigma sets tho.

[low stone]

Yeah this.

[steep wedge]

, yeh, i mean that was the obvious trick ig, to make it more symmetric and all

but what next

how will i decompose the space as such

difference of F sigma/Gdelta sets dont give anything nice

i was thinking about somehow "incrementally increasing the sets" but countable union of G delta isnt neccessarrily G delta

so the method i tried wasnt working

[low stone]

The way I did it is basically a neat trick to divide the F sigma sets into disjoint sets which also cover the total set

[steep wedge]

i mean that is what you need to do, but how did you get that trick?

while still ensuring both are Fsigma?

[low stone]

If M and N are unions of countable closed sets Mi and Ni, then think of the structure which these countable sets make. Then, try to express M union N as a series of unions of sets which are made up of the countable sets from before.

M and N are the F sigma complements of G and H

If I tell you the series I got it will just kind of give the answer away

[steep wedge]

By any chance did your solution assume that there exists a closed set Ni which doesnt intersect with G or there there is a closed set Mi which doesnt intersect with H?

[low stone]

That's not an assumption, it is clearly true

[steep wedge]

Like jsut M1 N1 M2 N2 ... works

[steep wedge]

its not clearky true

[steep wedge]

thats the reason this doesnt work

[steep wedge]

if it is clearly true prove it.

[low stone]

Actually nevermind that.

I didn't even use that assumption

[steep wedge]

are you just taking union of all pairwise unions?

[low stone]

Close.

[steep wedge]

said intersections by mistake

[low stone]

Ah. You meant differences?

[steep wedge]

i meant union

difference wont work, cause the differnce of closed sets isnt anything nice

[low stone]

Oh I m blind sorry

[low stone]

No it does work!

[steep wedge]

how?

the difference of closed sets isnt anything nice

?

[low stone]

Think about it for a second. Like imagine M_0 to be the first element.

Okay, should I just tell you the sequence?

[steep wedge]

you mean like symmetric difference, rite?

[low stone]

Wdym not anything nice? Difference of two closed sets are F sigma

[low stone]

No, I mean set minus

[steep wedge]

how?

[low stone]

Why wouldn't it be

[steep wedge]

why would it be F sigma?

if it is Fsigma, then i would be done

[low stone]

Difference of two closed sets can be written as the union of closed sets

[steep wedge]

doesnt have to be a countable union tho, rite?

[low stone]

Just think about this. For any two closed sets X and Y, X\Y = X intersect complement(Y)

[steep wedge]

yes

[low stone]

Which is closed set intersect open set

Which is closed

[steep wedge]

yes

[steep wedge]

no?

where did you get that from

[low stone]

Wait no.

It's definitely not closed, but it can be a union of countable closed sets

[steep wedge]

it can sure

but why are you saying it will be

[low stone]

I am pretty sure it's possible for all intersections of open and closed sets

[steep wedge]

i dont see any reason for that to be true

[low stone]

Umm... All open sets are unions of countable closed sets?

All open sets are F_sigma

[steep wedge]

ohhh, i forgot we are assuming that

[low stone]

I assumed we were in R^n

[steep wedge]

nah, we are in a general topology

but it was given in the first problem to assume that all closed sets are G delta, for the remaining problems

that was stupid

[low stone]

[steep wedge]

ok yeh, now i dont need help

thx alot

[low stone]

Well we were both wrong

[steep wedge]

its fine ig

[low stone]

I'm sorry for overlooking that

[steep wedge]

this shit will happen in life

its fine

i am just happy that now i can solve the qn and move on

5 days

for missing one line

[low stone]

Good luck with the next ones

[steep wedge]

ehh, i enjoy confusion in general

but thx

[low stone]

Welcome.

Kbai, will go read bugundjis topology now.

[steep wedge]

bye

[low stone]

It seems to be a good book

[steep wedge]

yeh it is

just terminology is a bit old

and a few (one that i have seen yet) exercise are wrong(/ missing a condition)

but yeh

fun stuff

[low stone]

It's fun in the sense that it has a lot more intense dopamine boosts after solving questions through these kinds of stupidities

[steep wedge]

yep

thats why i like math, intense dopamine after difficult qn

[limber wyvern]

Hi, I have a question about naming:

Why do we call the elements of a topology like "open sets" instead of just elements?, my book does not specify this, and chatgpt tells me its just a way of naming, but for me it's strange to change the name in this way

[fast yew]

we use topology to generalize the notion of "openness" that we know from open intervals from R like (0,1). at this point we go general enough to be able to let any subsets be considered open depending on the topology you use

[limber wyvern]

Also, Im starting in topology, and I don't get the sense of space and sets in the same thing😅

[fast yew]

this part of teaching topology is a bit backwards chronologically because historically we used much stronger conditions on what constitutes an open set but we have stripped those away to allow a, frankly, crazy level of generality

[fast yew]

wdym

[limber wyvern]

I mean, I'm seeing that a topological space is a (x,t) where t is a topology

[fast yew]

yes

[limber wyvern]

And I imagine a space, like, a vector space

or like

in R2 I imagine the space as the typical x, y coordinates

[fast yew]

don't worry if you are confused because technical terms will be thrown around ad hoc often without the "why is it this way". it's this way because it works, and you gain intuition as you get further along the course

[limber wyvern]

I see

Yeah, it's feel hard for me to imagine a open set in a space

[fast yew]

try to stick to the definitions as much as you can because you will encounter a lot of pathological and counterintuitive examples along the way

[limber wyvern]

For R2 I imagine it as points in the plane, but I cannot imagine a open set of values, i just imagine points lol

[fast yew]

oh i see the issue

ok in R do you know open intervals

[limber wyvern]

yeah

[fast yew]

like (1,2) vs [1,2]

[limber wyvern]

yeah

[fast yew]

in R^2 the situation is only a bit more complicated. the open sets in R^2 are any sets where the boundary is not included

picture like a disk without its boundary or some blob

[limber wyvern]

I can imagine it like this

but it has no sense for some points

like

in the case X is a,b,c and we are in a discrete topology

the set a is included

but its just a element

not a range of elements

[fast yew]

yeah that you'll just have to accept tbh. if the topology says it's open, then it's open, there's no "intuitive" way to think about this

[limber wyvern]

where did u get this image?

[limber wyvern]

I see

[fast yew]

found this online
https://math.hws.edu/eck/metric-spaces/open-and-closed-sets.html

[limber wyvern]

Yeah, here is another doubt

so chatgpt tells me this:

That i cannot imagine it like a surrounding around the set because that only applies in metric spaces, not topologic spaces

[fast yew]

for example you can define a topology on Z called the cofinite topology, where the open sets are all sets whose complement is finite. you get really weird results, like the sequence (1,2,3,4,5,6,...) converges to every element in Z. there's no good way to think about this, it's just bashing definitions

[fast yew]

wdym? i might need more context than that

[limber wyvern]

give me a second

[limber wyvern]

So, basically, I asked chatgpt:

I imagine this points like having an arbitrary "boundary" around them which is open, but he told me that, this concept of boundary only applies in metric spaces not topological spaces

[fast yew]

i think the concept you are looking for is a neighborhood. in metric spaces you get access to open balls centered at your points, which you don't have access to in general topological spaces

[hollow pawn]

does this follow because one of the sets in the intersection is just the topology generated by the basis/subbasis itself?

[limber wyvern]

I see

[rancid umbra]

not sure what you mean

[limber wyvern]

Well, I see that I have to start imagining things and stick to the concepts

thanks for ur help @fast yew 🙂

[fast yew]

i'm also taking a topology course rn lol which is why i know the struggle

[quasi forum]

this gets you half of the argument, youd still need to show that in some sense the topology generated by A is the "smallest"

[hollow pawn]

i mean, every set containing A should contain the topology generated by A, and the topology generated by A itself is a set in this collection, so i assumed the intersection equals A

[quasi forum]

in that if T is the topology generated by A and T' is any other topology containing A, then T is a subset of T'

[rancid umbra]

ah, i misread, thanks for clarifying tho

[limber wyvern]

What i find curious is that, I'm seeing topology in my first year of university. I don't know if thats normal

[tranquil cosmos]

not really

depends what course it is though

[limber wyvern]

It's computational math

[hollow pawn]

that follows right because the topology generated by A is all unions of its subcollections and any topology is closed under unions of any subcollection?

[tranquil cosmos]

so a topological space has a set X of points

and then you can have subsets of X that contain some of the points, right

some of the subsets are open, some of them are closed, some of them are neither

T is just all the sets that are open

[quasi forum]

yeah i think youre right

it should be a one liner

but you still have to mention it

[tranquil cosmos]

in R^n, T is all the sets that have the open ball property

(i.e. for each point x, there's a little open ball around it)

but a lot of top spaces (X, T) are really, really weird

[hollow pawn]

alright ty 👍

[quasi forum]

np

[steep wedge]

One intuition of open sets I have which works pretty well is "one sided distinguishability"
So like, elements inside are distinguishable from sets disjoint from that open set,

This explains arbitrary union pretty well;

If an element is inside the union, it is inside atleast one open set, if a set is outside the union, it's outside all of them, so there is an open set which contains the element but is disjoint from the set we are comparing to, so we can use that open set to distinguish the element

This also explains finite intersection;

If an element is in the intersection, then it is in all of them, but if a set is outside the intersection, then I can break the set into finitely many parts, such that for every part there is a open set disjoint from it
(Like if you have A,B,C as open sets, and Z as the outside set Z-A, ZintersectA - B, ZintersectAinstersectB - C)
So if i distinguish the element from each part, then I am done, and since there are only finitely many parts, i can just distinguish the element from the first part, then the second part, then the third part and so on until I am done.

Now I'll give a few examples of topologies and how this intuition applies.

In an indiscrete topology, we arent allowed to distinguish anything at all, we just lack that capability

In a discrete topology
We can tell if a element is outside a set, for any element and set, quickly

In the sierpinski topology
{{},{1},{1,2}}
we do have a ability to tell 1 from 2, but we can't tell 2 from one, to make it more clear imagine it like you have a cliff, you have a point at the bottom and one at the top, if you are at the bottom, you can clearly see that the top is different, but if you are at the top, you might slip at any point, and fall to the bottom, this not being able to distinguish where you are and the bottom.

for the cofinite topology, given any finite set you can tell that a point is outside it, maybe you do this just by checking every point in the set, but if it is infinite, you cant, as at any point of time, the element might just be later in the set.

For an order topology, you can basically just notice the fact of a set that it is smaller than or equal to some value, or it is bigger then or equal to some value, so if one set is completely bigger then some element, you can distinguish it, and if some set is completely smaller the some element you can distinguish it.

closed sets in this intuition would just be a set which i can distinguish from

Clopen sets would be sets which I can both distinguish from, and can be distinguished from.

Interior would be set of points in a set that can be distinguished from all sets outside the set

Closure would be the minimal set containing a given set so that you can distinguish any point outside the set from it.
(Or the maximally set so that any point you can distinguish from a given set is also distinguishable from this set)

G delta sets whose elements you could distinguish from outside if you could do countable infinite number of things in finite time (like those popmath videos about supertasks)

F sigma sets would be sets you could distinguish from if you could do supertaks

[tranquil cosmos]

I kinda like that

points in an open set don't have an identity crisis, but points outside of open sets might

especially if they're on the boundary

[finite token]

This reminds me of like, thinking of open sets as predicates

[steep wedge]

How does that work?

[finite token]

Well they just like take in an element and output a truth value depending on whether that element is in the set or not

[steep wedge]

But according to that wouldn't you have all open sets would be clopen? Cause the negation of a predicate is a predicate

[gritty widget]

Open sets wouldn't just be arbitrary predicates

There would just have to be some predicates deemed open on a set X such that:

The predicate of "being in X" is open

The predicate which no element satisfies is open

If two predicates P and Q are open, the predicate "P(x) and Q(x)" is open

If arbitrarily many predicates P_a indexed by a in A are open, so is the predicate "P_a(x) for some a in A"

I personally think, when thinking of open sets as predicates, it is much better to refer to them as "local predicates", since "open predicate" sounds kinda odd imo

[steep wedge]

That seems like a weird way of rephrasing the topology axioms?

[gritty widget]

Indeed it is

[steep wedge]

I mean instead of saying "subset" you are saying "predicate"?

[gritty widget]

Pretty much yeah, you're just switching out subset with the predicate of belonging to that subset

[steep wedge]

Is there any change to the intuition?

[gritty widget]

Similar(or exactly the same, depending on your POV) to thinking of subsets of X as functions from X to {0, 1}

[steep wedge]

Yeh

[gritty widget]

I think it can sometimes help to think of open sets as local predicates, I think you get an inductive-esque definition of compactness like this, but idk any other examples

[steep wedge]

I havent done compactness yet.

[unreal stratus]

Where local is expressed in terms of open sets? Jk

[fast yew]

need help with question 3a), question 2 included for context

i think the only sequences that converge to 0 are the ones that converge to zero w.r.t. the standard topology on [-1, 1] because the neighborhoods of 0 look like regular intervals (a,b)

but idk how to properly show it

ping on reply pls

[steep wedge]

What exact issue are you getting in showing this? Just applying the definition should work?

[fast yew]

definition of convergence?

[steep wedge]

Yup

[fast yew]

oh right so it's just x_n -> 0 iff it's in every neighborhood of 0 which is (a,b) which means x_n < b for all b > 0 and x_n > a for all a < 0

okay bruh that was a brain fart lol thanks