point-set-topology

[obtuse meteor]

It seems f^{-1} must be continuous.

This is a consequence of invariance of domain

(An important theorem I have a habit of forgetting)

[heady skiff]

i don't really understand how p not being in X implies that every point of the closure of A in X is less than p, can somebody help me pls

[queen prism]

I don't want to work through the details but basically everything in A is less than p -> everything in the closure of A is less than or equal to p -> everything in the closure of A is less than p since p is not in X and hence cannot be contained in the closure of A

[gritty widget]

i am a little confused by this example because i dont see how the map given can be a section of the tangent bundle of S1, because if z in S1 then iz should also be in S1, not in the fiber above z?

it looks like ps(z) = p(iz) = iz, not z so ps is not the identity and hence s is not a section.. what am i not getting here lol

[opaque scroll]

This is a bit of an abuse of notation. You identify the fiber above (x1, x2) with the tangent line through (x1, x2) which you again identify with the orthogonal complement to (x1, x2) in R^2

[limber wren]

A is contained in (-infinity, p] which is closed, so the closure of A is also contained in (-infinity, p]

[sullen bear]

this problem is really tripping me up. All Ive been able to deduce is that there exists some epsilon neighborhood of 0 whose preimage is an upon neighborhood of p.

My current idea is that as epsilon tends to 0, the open neighborhood of p will get smaller until it doesn't include any other point q.

but im afraid im thinking too Eucliean-y

[queen prism]

o I was doing that one earlier

[ebon galleon]

any inverse image of an open neighborhood of 0 would give you an open neighborhood of p.
Fix p and q distinct in X, so you know there's f as stated in your assumtions; f^{-1}(0) = {p}. What can you say about f(q)?

[sullen bear]

This is the part that’s confusing me. How can I say anything about q?

I don’t see any problem with there being an open set containing just p and q

[ebon galleon]

Well, why don't we focus first on their image in R, since we know R is Hausdorff, we know that distinct points have disjoint neighborhoods

We know that f^{-1}(0) = {p}. Could we also have q in f^{-1}(0)? That is, could f(q) = 0?

[sullen bear]

Well by the problem statement no?

Oh so once the neighborhood of 0 gets sufficiently small, q can no longer be in it?

Rather the image of q

[ebon galleon]

Yeah so f(q) is not 0. So since f(p) = 0, we can find disjoint open neighborhoods U of f(p) and V of f(q), right? This is happening in R

[sullen bear]

And then their pre images are disjoint

Hence Hausdorff

Ty goat

[ebon galleon]

[glossy talon]

Looking at last sentence isnt the second part technically wrong?

sets in C are cylinder sets

if instead C was topology generated by cylinder sets wouldnt this be correct?

because an example of a set in C_1 would be \prod_i U_i which can't be written as union of cylinder sets

It goes on to say C_1 \subset \sigma(C) [\sigma(C) is notation for smallest sigma algebra containing C], which is correct given they meant that every set in C_1 if a countable union sets of the topology generated by C.

If what im saying is true then I don't know how S is seperable applies here

[glossy talon]

I guess the word countable as opposed to arbitrary?

[sinful cloak]

yeah I think it is just a subbasis.

[glossy talon]

Yeah I cross referenced with another proof

Pretty sure seperable is just so that things stay countable

[sinful cloak]

Wait, was the actual question something different?

[glossy talon]

No I asked another question

[sinful cloak]

yes

that's it

[glossy talon]

But I think seperable is there because otherwise the equality wouldnt be true

I dont honestly want to think of counter examples

[sinful cloak]

It's probably not that crazy

What if we use an uncountable set with discrete topology.

The product topology will be discrete again.

yeah and for example the diagonal is not included in the right side

I mean ${(x,x)|x \in M}$ with M the uncountable set we product with itself.

[gentle osprey]

M8732

[sinful cloak]

This set would be measuable with the borel sigma algebra but not with the product sigmaalgebra.

[glossy talon]

wdym the diagonal is not included in the right side?

[sinful cloak]

^

Okay i will try be more precise.

Let's set $M = \mathcal{P}(\mathbb{N})$.

[gentle osprey]

M8732

[sinful cloak]

We use the discrete topology for this. You know the discrete topology?

[glossy talon]

yes

[sinful cloak]

All sets are open

Let's think about the topology of M x M.

Do you agree, it is still discrete?

[glossy talon]

yes

[sinful cloak]

so all sets will be measuable

since the borel sigma algebra is genrated by all sets.

[glossy talon]

yes

[sinful cloak]

Now let's look on the other side.

[glossy talon]

ok

i see

[sinful cloak]

B(M) is still everything

but when you look at the generated sets in the product

you can take only finite countable combinations

you get for example all single points

[glossy talon]

you mean countable?

[sinful cloak]

yeah

sorry

[glossy talon]

yea i see now

[sinful cloak]

But to get the diagonal set

you need to union uncountably many finite sets.

[glossy talon]

how often do you consider measure space not seperable?

[sinful cloak]

[glossy talon]

aren't they pathological anyways?

[sinful cloak]

Pretty rarely but we are mathematicians and not physicists.

[glossy talon]

its done in physics?

[sinful cloak]

nah but physicsts tend to forget things that don't happen in practice.

"we measured it, so it is this way."

[glossy talon]

i mean im sure it will find its way later in my life

depending if I do weird probability things later

[sinful cloak]

I did some probabilitsy things that can be considered pretty weird

basically I ended up putting a countability condition in many places that is equivalent with separability. So I exclude the cases that wouldn't work.

I think in non separable spaces measure theory is difficult.

[glossy talon]

i wonder if its interesting at all

it probably shows up if you are doing weird geometry?

[sinful cloak]

Good question. I haven't foudn a use but if you look hard enough there probably is.

[sinful cloak]

Hardly

Like any finite dimensional space is separable. Many infinite dimensional spaces are separable too.

[glossy talon]

yeah

the first thing that came to mind is manifolds

but those are seperable

but maybe schemes aren't

[sinful cloak]

Schemes are a bit out of my expertise.

[glossy talon]

the definition of seperable for schemes is weird though

[sinful cloak]

if there is a use in algebraic geometry then idk about it.

[glossy talon]

same

but i think there might need to be a different definition of seperable too

[lime sable]

"separated morphism" in scheme theory is basically hausdorff, because both of them say that the diagonal is closed, albeit the first is the diagonal in the fiber product while the second is the diagonal in the cartesian product of topological spaces

i looked it up and "separable" in AG may refer to separable algebras, which are related to etale covers (AG version of covering spaces) and grothendieck's galois theory of course

[grave solstice]

the only irreducible closed sets in Q_p are singletons {x} right

[coarse night]

what's Q_p?

[grave solstice]

p-adic numbers, completion of Q wrt the p-adic metric

[opaque scroll]

Yes, Q_p is Hausdorff

[lusty trench]

Let $S_n$ act on $\mathbb R^n$ by permuting coordinates. Let $\sigma : \mathbb R^{n^2} \to \mathbb R^n / S_n$ be the map that sends a matrix to its eigenvalues and let $\pi : \mathbb R^n \to \mathbb R^n / S_n$ be the obvious quotient map. What's the pullback of the cospan formed by $\sigma$ and $\pi$?

[gentle osprey]

Eduardo León

[lusty trench]

Mmm, now that I think about it, do we have $\mathbb R^n / S_n \cong \mathbb R^n$, after identifying the equivalence class of $\lambda = (\lambda_1, \dots, \lambda_n)$ with $(e_1(\lambda), \dots, e_n(\lambda))$, where $e_1, \dots, e_n$ are the elementary symmetric functions?

[gentle osprey]

Eduardo León

[opaque scroll]

In general the pullback of A -f-> C <-g- B is the subset of AxB for which f(a) = g(b).

So in this case this will be the set of all pairs (M, v) where M is a matrix and v is the vector of eigenvalues of M in some order.

[lusty trench]

Yeah, but I was hoping for a characterization that would make it easier to compute, say, the homotopy and cohomology groups of the pullback.

Oh, wait.

[opaque scroll]

Isn't it contractable? Like

h((M, v), t) = (tM, tv)

[lusty trench]

Oh, dumb me.

Right.

Thanks!

And here I was like “what kind of complicated variety could that be?”

[strong swift]

Consider the quotient map $p:\mathbb{Q} \rightarrow \mathbb{Q} / \mathbb{Z}$. Show that the map $p\times id$ is not a quotient map, where $id$ is the identity map on $\mathbb{Q}$ .

[gentle osprey]

Eduardo291299

[strong swift]

I'm kind of stuck with this I don't know how to proceed, I suppose I need to find some non open in $\mathbb{Q} / \mathbb{Z} \times \mathbb{Q}$ whose inverse image is open in $\mathbb{Q} ^2$

[gentle osprey]

Eduardo291299

[fading vale]

This is an entirely different notion of separability. In functional separability means you contain a countable dense subset. In scheme theory it is a notion of hausdorffness

[tawdry widget]

Are you sure it isn’t? Because I think
$(p \times 1) (\cup U \times V)=\cup (p \times 1) (U \times V)= \cup p(U) \times V$ so it’s an open map?

[gentle osprey]

Cogwheels of the mind

[narrow cairn]

is there a natural way to define a topology from a measure?

theres a way to define it from an order and from a metric

was wondering if there was a way to do it with a measure

[ebon galleon]

p is a quotient map, not necessarily an open map

I don't think it's open

[lusty trench]

Are you expecting something coarser than the topology generated by the basis of measurable subsets?

[tawdry widget]

But this time isn’t p open? Any open U in Q, p^-1p(U) is open since U is union of open intervals W of length , say <1/2, p^-1p(U) is union of p^-1p(W), and p^-1p(W) is countable union of shifts of W, therefore open…

So p times id is open and surjective

[ebon galleon]

p open is "U open implies pU open". I think taking U = (-1,1) or something works as a counter example

[tawdry widget]

It’s not a counterexample, p^-1p(U) for your U is exactly Q

[ebon galleon]

Is it though? p^{-1}p (-1,1) would be (-1,1) U Z. This is not open

OKay

We are thinking of two differnt things here I think

[tawdry widget]

Wait this quotient Q/Z

[umbral panther]

Generally a point is measurable, so this gets you the discrete topology. Maybe you can do better by asking for measurable sets with positive measure?

[ebon galleon]

Is this Q/Z quotient group or Q/Z by contracting Z to a point

[tawdry widget]

Glue all points Z as one point, or the quotient group Q/Z

Yeah

[ebon galleon]

This distinction matters

[tawdry widget]

I think we thought different quotient

[ebon galleon]

@strong swift which type of quotient is this?

[ebon galleon]

Sorry lol, I've mostly worked with just contracting a subspace to a point. I forgot for a sec the other type exists

[tawdry widget]

Yeah if your case it’s not open

[surreal burrow]

the more i look at this the more confused i get

someone please help 😭

bro nvm i'm trolling

[narrow cairn]

is this true?:
let X and Y both be totally ordered spaces endowed with their respective order topologies, and assume that Y has the least upper bound property. let f: [a, b] -> [c, d] be strictly increasing, and assume the restriction of f to [a, n) is continuous for all n < b. finally, let f(b) = d = sup f( [a, b) ). is f continuous?

oh and also assume f(a) = c

[narrow cairn]

okay proved this elsewhere

then i can still use this! since the domains of q_n are locally finite on [0, 1), i can use the gluing lemma to get a map q on [0, 1), and i just need to assume that all the paths are strictly increasing, which still works by induction since q strictly increasing

[glossy talon]

is there a corresponding definition then?

oh wait

the underlying topology could be seperable

still would be confusing

and im not sure of examples of spaces of interest that arent seperable

[ebon galleon]

l^infinity (bounded sequence space) is not separable.

[fading vale]

the analytic notion of separabillity in AG is not a particularly interesting condition because of the existence of open points whose closure is any irreducible closed subscheme

[glossy talon]

well im talking for purposes of needing to rethink measure theoretic things

also i never once thought of measures on schemes

[fading vale]

the closest thing is motivic integration but this is very sophisticated

[late iron]

Let p: E \to B be a vector bundle of rank r

Does anyone know like any reference or why for the fact that, an isomorphism of the fiber at b \in B can be extended to an isomorphism of the entire bundle when r is odd?

And why does a counter example exist for the case of TS^2?

[umbral panther]

First show that the answer is invariant to a path in the space of isomorphisms. Thus it is sufficient to show it for one isomorphism in each connected component. In the identity component you use the identity isomorphism. If r is odd, then in the other component choose the -1 isomorphism.

I don’t know what the general case is, but in the case of a 2d vector bundle if you have an isomorphism that at one point has both a positive eigenvalue and a negative one, then at every point it has one of each and thus decomposes into a sum of two eigenspaces. But not all 2d vector bundles do that

[late iron]

Thank you so much!

Could you elaborate a little bit on what you meant for the case of odd rank bundles?

[late iron]

So are you saying that if there’s an isomorphism at the fiber (for TS^2 for instance) that has 1 negative and 1 positive eigenvalue, then TS^2 could be written as the direct sum of 2 line bundles over S^2? Which is not possible.

[civic verge]

guys a question the infinite ball in the extended plane of complexes has to be at the origin or has to be in the set?

[coarse night]

??

[scarlet turtle]

??

[civic verge]

I have a doubt when I want to know if the set is open or closed in the extended complex plane, when I make the disc, but then I don't know how to argue it.

[tawdry widget]

The topological structure is defined this way: it has a basis consisting of either open disks on C, or complement of closed bounded subsets on C (the latter is called open neighborhoods of infinity)

(Under stereographic projection from North Pole, an open neighborhood of infinity simple is the image of an open disks containing the North Pole )

[tidal lynx]

why are we so high

[tribal palm]

wow my uni library is throwing away some of their old topology books

among which bourbaki’s general topology and even veblen’s analysis situs

i took bourbaki for obvious reasons but let veblen be

though i should’ve probably grabbed veblen for antiquarian purposes

https://en.m.wikipedia.org/wiki/Analysis_Situs_(book)

[broken nacelle]

bruh

that's ridiculous wtf

why would they do that??

what kind of library throws away books???

[tribal palm]

i think they have multiple copies of these specific ones

[broken nacelle]

oh

but still!

[tribal palm]

probably most that have a lot of them and is still getting in new ones

[broken nacelle]

it's still unjustified in my books unless they literally have no spaces to store them

[narrow cairn]

wait, is (1, omega + 1) in the long line homeo to (1, inf) in R? its like a 'countably long line', which also describes R, but i cant think what omega itself maps to. (trying to show the long line is locally euclidian)

[narrow cairn]

wait, i dont think it is

but then, how do you find a locally euclidian neighborhood of omega?

[umbral panther]

Yes. Omega is {1/n}. Or if you prefer {1-1/n}. Omega+1 adds in {1}

Every countable ordinal embeds in R. That was Cantor’s line of thought. He was studying sets of failure of convergence of Fourier series and kept designing worse sets

[distant lichen]

"Kept designing worse sets"

Rather apt way to phrase it

[narrow cairn]

wait, but where do you send omega + eps?

like i get how that works for a bicontinuous map that takes (1, omega] to [0, 1) but neither of those are open

[narrow cairn]

wait nevermind i am a stupid idiot person that can barely breathe

lol

[formal tide]

The notes I'm reading say
"A function f between topological spaces is continuous iff for all $x \in X$, $f(\mathcal N_x)$ is a neighborhood base for $Y$ . "

I'm not sure what they mean, as I only understannd neighborhhod base for a point

[gentle osprey]

Shiranai

[formal tide]

here $\mathcal N_x$ is the neighborhood filter

[gentle osprey]

Shiranai

[ebon galleon]

Should be "f is continuous at a point x if and only if f(N_x) is a neighborhood base at f(x)", so then f is continuous iff this holds for all x in X

[formal tide]

I see, thanks!

[narrow cairn]

if you extend the long ray by adding a point (ω₁, 0), is it still path connected?

[red yoke]

No

But it is connected

I wonder if you can do stuff like long paths

[narrow cairn]

how would you prove that

[ebon galleon]

ω_μ path-connectedness

Certainly someone has looked into something like that

[red yoke]

How would you define a long path

[red yoke]

Sequence in [0, 1] converging to 1

[ebon galleon]

Cts function from the long line + a freely added endpoint or something?

[red yoke]

Seems bad

Not transitive

What if we do piecewise long line

So it can be long at multiple points in the middle

Then α×[0,1) is cf(α)-connected

Or sth

[gritty widget]

im having a hard time understanding why a map of spaces f:X->Y induces a map on cohomology in the other direction

[sinful cloak]

The chain complex of singular cohomology can be thought of like assigning any possible simplex in the space an integer. When you element of the chain complex for space Y and have a map f:X->Y, you can assign integers to X, by pretransferting any simplex of X using f and then asking the element of the chain complex in Y what integer it assigns to.

it's kind of like always when dualizing.

If you have a linear map X -> Y between vector spaces, this induces a map Y* -> X*.

[radiant epoch]

Anyone have any thoughts on Armstrong's topology book?

I am floored by some of the exercises

[formal tide]

I've been unable to prove i => ii or i => iii from the pic. Any ideas on how to do this? It seems to me that it can't be done

If I'm not wrong, the identity map from N + inf (order topology) to N + inf (order topology plus {inf} open) seems to not satisfy ii but satisfies i

[safe torrent]

Is this the right place for knot theory?

anime is a topic here huh

[umbral panther]

Yes, knot theory is a type of anime

[safe torrent]

makes sense

[tawdry widget]

What is the definition of prefilter? I only know filter.

[formal tide]

oh right

prefilter = filterbase

honestly just replace prefilter with filter and it's still pretty much the same thing

Also I asked this here too https://math.stackexchange.com/questions/4772756/if-a-function-preserves-all-filter-limit-points-it-is-continuous and found the mistake in my counterexample

[ebon galleon]

Never heard of convergence of prefilters. I can't imagine there is much use for considering that over just filters

[formal tide]

yeah it's a slight generalization that seems kinda pointless imo

[ebon galleon]

well okay, so (ii) and (iii) are equivalent by just passing to the associated filters, since it's true for filters (by the notes you sent on MSE, prefilter converges iff the associated filter) . Never heard of limit points of (pre)filters, but I would guess it's a similar argument

because generalizing it to prefilters has no use, even for much more general structures than topological spaces (like convergence spaces even)

[tawdry widget]

I am not sure I get it. x is a limit point of a filter base F. Does it mean that any open neighborhood U of x, there exists an element B in F such that B is contained in U?

Is that how limit points of prefilter is defined ?

[formal tide]

x is limit point of a filter base F iff every neighborhood of x intersects every element of F

[tawdry widget]

I see

[ebon galleon]

I think the usage "limit point of (pre)filter" is not particularly standard.

[formal tide]

it makes sense tbh but weird at first

[tawdry widget]

And converging to x what does it mean?

[formal tide]

that the filter it generates refines the filter neighborhood of x

[tawdry widget]

I see

[formal tide]

Interesting point, just checked, Bourbaki defines it like this, which is equivalent

[ebon galleon]

well okay. The same argument as above should work though: check it for actual filters and just pass from prefilters to filters (since again prefilters are worthless here)

Anyways, calling those points "limit points" is dumb as well. If the filter converges to x, then x should be a limit point. Bad.

[tawdry widget]

If you are sure limit is defined by this. Idk I can find a counterexample.

[formal tide]

wut but it is

[tawdry widget]

X={0,1} with indiscrete topology, Y=R field of real numbers with the usual topology, f just the inclusion map, F={{0,1}}
0 is limit point of F, and F converges to 0
But we only have f(0) is a limit point of f(F), we don’t have f(F) converges to f(0)
So i)->ii) fails

[ebon galleon]

Sorry, I meant that should be like an iff definition imo

[formal tide]

huh nice, I think you're right!

wait...

[ebon galleon]

That's not what i) => ii) claims.

i think?

[tawdry widget]

I am dumb yeah… rethinking…

[formal tide]

The prefilter F={0} has as a limit point 1, but f(F) does not have 1 as a limit point

[ebon galleon]

Yeah, I don't think the condition of i) holds for all (pre)filters

[tawdry widget]

Oh wait

I think in my example it does satisfy i)

And x=0, F={{0,1}} is a counterexample of ii)

I better check it rigorously

[ebon galleon]

huh?

This is a for every statement

ii and iii are equivalent, I know that part is correct. This is not a continuous map if that's what you're thinking

[tawdry widget]

I agree ii) <-> iii)

And f indeed is not continuous

I am double checking f:X->Y I gave satisfies i)

[ebon galleon]

.

[tawdry widget]

My bad

Yeah…

So how do we prove i)->ii)…

[ebon galleon]

yeah it's an awkward condition. Not entirely sure..

[formal tide]

idk

[ebon galleon]

Assume (i). By Prop 5.8, x in clY iff x is a limit point of {Y} as a prefilter. So
f(cl(Y)) = f(limit points {Y}) subseteq (limit points {fY}) = cl(fY).
So (i) implies continuity

since (i) states that the image under f of {limit points of F in X} is a subset of {limit points of fF in Y}

[tawdry widget]

What book it is from?

[ebon galleon]

The notes Shiranai sent on the MSE post shared

[tawdry widget]

Oh

[formal tide]

I remember I tried doing something along those lines but I failed, can't remember why tho 😦

[ebon galleon]

Conversely, assume (ii). Suppose x is a limit point of F in X. Then by Prop 5.7, there is a refinement of F, let's say G, such that G converges to x. By (ii) then, fG converges to f(x). But image under f preserves refinement (order by inclusion), so since fF is refined by fG, another application of Prop 5.7 gives that f(x) is a limit point of fF in Y. So (i) holds

So they are equivalent.

New definition of continuity unlocked

[formal tide]

proceeds to never use it again

but yeah nice proof, thanks!

[ebon galleon]

very true

But yeah I think(?) these should work, maybe with a bit more elaboration

[tawdry widget]

Omg you are GOAT

[formal tide]

also it seems the fact that the notes deal with prefilters instead of filters is because of this

the image of a filter is not a filter

[ebon galleon]

Yeah but that's very easily remedied by taking the upward closure

[formal tide]

yeah that's exactly the filter generated by a prefilter

[gentle osprey]

Ryx (Home for flowers)

[ebon galleon]

Or "pushforward" of filters I've seen it called

[heady skiff]

ok so i'm trying to prove that any closed balls in R^n are homeomorphic right

so i got the translation part down

now my dumbass is struggling with, let's say, expanding the ball

i know it obviously has to do with multiplying by a constant

at first i thought it was the norm of the point or some shit

wait let me try the norm of the center

nvm

i also tried the maximum of the coords

[gritty widget]

how much do you have to scale a ball of radius 1 to get a ball of radius 2? ball of radius 46/9? ball of radius r?

[heady skiff]

yeah this is basic geometry huh

uhhhh this is embarassing but idk... my guess is just the ratio of the radii

nvm

that doesn't work

[gritty widget]

why wouldn't it?

[ebon galleon]

If you can always translate them, make your life easier by assuming they're centered at 0

[heady skiff]

wait LOL

sorry i'm so stupid i forgot high school geometry

i thought it was (x - h)^2 + (y - k)^2 \leq r where r is the radius oops

okay that makes a lot more sense then

wait nvm

wait

okay, so i'm considering the circle centered at the origin with radius 1, and the circle centered at (2, 3) with radius 2. so i've come up with (x, y) \mapsto 2(2 + x, 3 + y), but if i plug in (.5, .6) then I get (5, 7.2), which is not contained in the circle that i'm mapping to.. so my map is probably off then

or i' mhigh

[heady skiff]

ohh okay

[gritty widget]

who cares about the circle formula. what happens when you multiply the coordinates of a ball of radius 1 by some scalar, r, say? what is the radius of the resulting ball?

[heady skiff]

it becomes a circle with radius r, doesn't it?

am i high?

cuz (1, 0) \mapsto (r, 0)

[gritty widget]

https://tenor.com/view/killer-fish-killer-fish-fish-pog-pog-gif-22185503

[heady skiff]

i mean does it not

i'm multiplying points inside the circle of radius 1 centered at the origin by 2

oh wait

i'm ritarded

[gritty widget]

ban any%

[heady skiff]

i have to scale up the point first and then translate it

wait that's interesting huh

wait let me think about that

translating then dilating doesn't work

but dilating then translating does

[ebon galleon]

translate to 0; scale appropriately; translate to whatever other center. gg wp ez normed vector space moment

[heady skiff]

that's crazy

wait wtf

why is that the case

that the order of operations matters

[ebon galleon]

hmm i wonder why order of operations matter let me think back to 5th grade

[heady skiff]

damn bruh y'all have no chill 😭

[real reef]

wp

[heady skiff]

nah but fr if it's so trivial, can you explain why translating the ball to the center of the bigger ball then dilating doesn't giving the big ball

[ebon galleon]

If r is the radius of the ball we want to translate to (let's assume the base ball has radius 1 and is centered at 0) and x is the center of the new ball, then it's just that t |--> (x + rt) [dilate and then translate] and t |--> r(x+t) [translate and then dilate] are generally not equal

[heady skiff]

oh shit ur right i didn't even think about them as functions lmaoo

thanks boss

[ebon galleon]

think my guy

[obtuse meteor]

Bruh don’t say that it’s a slur

[coarse night]

retired

[trail charm]

other than brower fixed pt thm, what are other uses of this thm 2.13?

[ornate berry]

That's the long exact sequence. It's very helpful for computing the homology of various spaces, as the immediate corollary demonstrates. It's a fundamental tool in homology in general, not only in simplicial homology.

[trail charm]

do u have any examples? other than the corollary (brower fixed pt thm)

i get why relative homology is useful, and we can consider this as just a special case of the long exact sequence of relative homology

[hidden crag]

2.14 and similar computations

Think of it as homology being well behaved wrt to quotients

Which is something that severely fails for homotopy

You’ll meet this a lot in computations

[unreal stratus]

Idk how you could compute without the long exact sequences

[distant lichen]

You can view excisive functors as taking homotopy pushouts to pullbacks

But this is really just an obfuscation of the LES

Or an obfuscation of fiber sequences in a topos

[unreal stratus]

Yeah

I didn't meant quite so literally like i couldln't compute at all, but I meant to emphasise how important the LES is to computing

[tidal cedar]

My alg top class did singular first and proved a fuckton of stuff simply using excision and spheres and the like

[unreal stratus]

Yeah, I guess as for why it's often useful like

The hypotheses always hold for CW complexes

If that's more your question

[trail charm]

yeah all of yalls stuff makes sense

thanks yall

[coarse night]

we did the same but using axioms only

[tidal cedar]

Like Eilenberg Steenrod axioms? That's what I mean

[coarse night]

yes the axioms. You said you did with singular tho

[trail charm]

i don't really see what's special about the subgroup. if U covers X, wouldnt any chain of Cn(X) also satisfy the condition?

like each σ_i of a chain of Cn(X) already has an image in a set in the cover U

[tidal cedar]

Sorry I should be more specific

We proved that singular obeys ES

then only worked from the ES axioms to prove computations

[coarse night]

Lol I see

axiomatic view sucks tho, gives no geometry or motivation.

[tidal cedar]

I don't think it necessarily does

But I can see that

We then did cellular/simplicial

[trail charm]

unless it means something like each sigma_i is contained in a single U_i

which i suppose is a subgroup since the image of any sigma_i isnt necessarily contained in a single U_i

[coarse night]

no, the definition says each individual simplex is contained in one of U_i. You can have simplices in X that lie in more than one U_i.

[trail charm]

ah yeah ok

just didnt read closely enough

oops

[coarse night]

spoiler: you can break simplex down to satisfy that each part lies in one of the U_i

[trail charm]

yeah i was just thinking about that lol

is that this whole barycentric shenanigans that im about to read about

[coarse night]

we never did simplicial lol. Cellular we again did axiomatically (almost)

we only proved one step using singular

Our instructor wanted to show that we only need the axioms to get everything

[tidal cedar]

Ahh yeah

That's kinda what we did

[coarse night]

cool stuff

[tidal cedar]

We just like wanted to show singular actually obeys said axioms

[coarse night]

who was your instructor lol

[tidal cedar]

Simplicial is singular w/ trivial attaching maps as well, which is kinda nice

Mine?

A stable homotopy theorist

[coarse night]

(you don't have to answer)

ok invariants don't match so we had different instructors

mine was an algebraic geometer

[trail charm]

my alg top directed reading prof is a diff geometer

originally i was gonna do a diff top reading w him and then realized i like alg top more so here we are

i dont regret it i love diagram chasing

[coarse night]

soon there will be no diagrams to chase

[trail charm]

NOOOOOOOOO

fine i'll go to hom alg then

where i can chase pretty butterfly lemmas

[floral bear]

are knots in R^3 ambient isotopic to the unknot if composed with the inclusion map :R^3->R^4?

ie, can all knots in 3d space be untangled in 4d?

[distant lichen]

Yes

We can pass the knot over itself in the extra dimension

[feral copper]

By codimension reasons, you can always unknot any embedding of S^1 into 4-space

[distant lichen]

This is the intuition

https://math.stackexchange.com/a/1426599

[feral copper]

There's just 'too much space'

[distant lichen]

This SE answer has a formal argument

[floral bear]

ah yes

[feral copper]

But you can still look at embeddings of S^2 in 4-space then

This is interesting

[distant lichen]

Yes

There are some smoothness / embedding things to consider with that definition

If you consider higher knots to be piecewise-linear embeddings then S^n -> R^(n+2) can knot and higher dimensional ambient spaces won't

If you consider smooth embeddings though you can knot spheres in far higher dimensional spaces

https://www.jstor.org/stable/1970561?origin=crossref

[feral copper]

(Alert, digression warning)
What's interesting is that if you look at the space of immersions of circles in 3-space with only transverse double points, this space decomposes into subspaces by counting the number of such double points. Zero double points is embeddings, i.e. knots.
Now, you know you can go from any knot to the unknot by changing crossings, i.e. adding and then removing one of those double points (crossing the codimension one stratum). You may need to cross this stratum several times; the minimal number of crossings needed is the unknotting number.
Fun fact: if you allow two double points at once (i.e. crossing the codim 2 stratum), you can always unknot every knot by only crossing this stratum once!
(end of digression alert, and mandatory @tiny ridge ping in case they wanna have some fun as I did when I discovered about this fact this Monday xD)

[distant lichen]

I believe this is the motivation for Vasilliev type invariants, no?

[feral copper]

Finite type invariants you mean?

Yes it is 🙂

But I had no idea people did this for a living, I discovered them and was fascinated

[distant lichen]

Yes this is a synonym

I like this sort of higher homotopical approach to this area

[feral copper]

I suspect this would break for 2-knots in 4-space Maybe you'll need to use a higher codimension stratum to unknot anything in one move only

Would you happen to know about this maybe?

[distant lichen]

I do not, sorry

[feral copper]

Np, don't apologize!

[distant lichen]

Have you seen the connections of this to Chern-Simons theory

[feral copper]

I tend to avoid CS at any cost

[distant lichen]

The Wilson loops of perturbative Chern-Simons are a universal Vasilliev invariant

[feral copper]

I honestly have zero clue what any of those words mean

[distant lichen]

https://arxiv.org/abs/math/0211062

This is an introduction but there is a serious amount of physics background

So don't worry about not getting it immediately

[feral copper]

Thank you for the kind reference!

[distant lichen]

Also may I present this quite absurd diagram of Hisham Sati & Urs Schriebers

[feral copper]

I'll give that a read this weekend, I was out of bedside stuff to read 🙂

[umbral panther]

What would “one move” even mean for S2 in S4? Immersions with intersection are generic. The codim 1 strata have a curve of self intersection

[feral copper]

Huh you're right
But thinking of broken surface diagrams, it would amount to switching who's above who in the fold lines I guess?

[umbral panther]

Ismar Volic studied codim 2 embeddings from a general position point of view that wasn’t supposed to work in codim 2 and recovered finite type invariants. I thought he studied S2 in S4 and got something, but I can’t find it

[distant lichen]

What does "wasn't supposed to work" here entail

[feral copper]

You mean this?
https://arxiv.org/pdf/1310.7224.pdf

[distant lichen]

Is this due to exotic structures on R^4 or similar?

[feral copper]

No I was just randomly thinking and typing on my keyboard. And no, I mean the standard 4-sphere / 4-space

[umbral panther]

Yeah, that covers higher dimensions

[umbral panther]

The Goodwillie calculus. This produces a sequence of functors called the n-th degree approximations. The first approximation for embeddings of M in N is immersions. The second approximation is the Haefliger functor. The Whitney theorem says that the map from embeddings to immersions is bijective on components of M is half dimensional. The Haefliger theorem says his functor is bijective on components of M is 2/3 dimensional. Really they say that the map from embeddings to the other functor is a k-equivalence where k ~ n-2m for Whitney and k ~ 2n-3m for Haefliger. So it got better by the codimension n-m. Careful calculation shows it actually got better by n-m-2. There are infinitely many of these approximations and if the codim is at least 3 each is better than the previous and so they capture the entire homotopy type of the space of embeddings. But in the knot case, they don’t get better.

[distant lichen]

Ahh interesting

I'm dimly familiar with Goodwillie calculus but not enough to really strongly follow this

What functor are we approximating exactly?

https://mathoverflow.net/questions/97377/codimension-two-embeddings-in-goodwillie-weiss-manifold-calculus-and-the-diffic

[umbral panther]

I think he got regular 3d finite type invariants telling you something any S2 in S4. But not about different knottings. Instead they show up in the higher homotopy groups of the space of unknots

[distant lichen]

Tom Goodwillie replied that he would attempt to come up with an answer, 11 years ago

Haha

[distant lichen]

Is the functor here the bifunctor taking M, N to diffeomorphism classes of embeddings?

[umbral panther]

You can even work with the trifunctor Emb(M, NxV) and use three different flavors of calculus to differentiate in the three variables

I was talking about fixing N and varying M. It’s a version of sheaf theory, no spectra magically appearing

[distant lichen]

If I'm reading this MO answer correctly the failure you're speaking of is the "non-analyticity" of Emb(N, M)

no spectra magically appearing
sad.

[umbral panther]

Right, an analytic functor is one in which each polynomial is a better approximation than the previous

[distant lichen]

Wonderful

[tribal palm]

i’m panicking, in a flurry of confidence some two weeks ago i randomly sent an email to my topology prof saying i have loads of questions and i basically just invited myself to his office, but now i suddenly can’t think of any questions, and it’s tomorrow

i didn’t really have topology in particular in mind, rather i think i wanted to ask him about doing maths in general… he’s the first prof i’ve had who is very active in research (and has been for almost 40
years (!))

[ebon galleon]

start cooking up some questions then!

[queen prism]

my measure theory prof is also panicking

because he’s giving a talk on GR in a few hours and he accidentally nuked all 56 of his slides

[tribal palm]

eep

[distant lichen]

Ouch

At that point it may just be time for honesty

Tell them what happened and postpone the talk

[queen prism]

he miiight just barely be ready in time

he deleted everything this morning and he has to be ready by 20:30

[tribal palm]

this is why you set up a backup-system

[rough bloom]

Could someone help me with the following?

Let $p,n\in\mathbb{N}^*$ with $p\leq n$ and denote $\mathcal{M}{n,p}$ as the vector space of $n\times p$ matrices with real entries. Denote $\Hat{\mathcal{M}}{n,p}$ as the subset of $\mathcal{M}{n,p}$ with rank $p$. Show that $\Hat{\mathcal{M}}{n,p}$ is open in $\mathcal{M}_{n,p}$.

[gentle osprey]

SeanLayz

[gritty widget]

given an n by p matrix A, consider the p by p matrix A^T A

or consider what it means for a matrix to have maximal rank in terms of its minors

[gritty widget]

because i already said this in the other channel, lemme give a bit more of a hint: what would invertibility of A^T A imply about A?

[rough bloom]

A would have a left inverse?

[gritty widget]

phrase it in terms of the rank of A

it is true that A will have a left inverse but you can take it a bit further than that

[rough bloom]

It’s equal to (A^TA)^-1 A^T?

[tawdry widget]

Something non-zero. In the small neighborhood of it it’s still non-zero thing…

Like determinant, it’s continuous

[gritty widget]

what does having a left inverse say about rank?

[rough bloom]

Basically pre image of Determinant of A^TA is non zero is open

[gritty widget]

we're not at that part of the argument yet

we're still doing the linear algebra

[tawdry widget]

Oh

[rough bloom]

Full rank

Equal to number of columns

[gritty widget]

and how many columns does our matrix have?

[rough bloom]

p

[gritty widget]

nice

[rough bloom]

Lol

But p is a singleton!

[gritty widget]

so an n by p matrix A has rank p if and only if A^T A is invertible (well, we only proved one direction. you can do the other)

now as you say we can use the determinant to see that these form an open set

[rough bloom]

Wait

Right

Let me tex this up and lmk if it’s right

$\Hat{\mathcal{M}}{n,p}={A\in \mathcal{M}{n,p} \vert \det{A^TA}\neq 0}$
Since the determinant is continuous from $\mathcal{M}{n,p}\to \mathbb{R}$ and $\mathbb{R}\setminus{0}$ is open, then its pre-image is open in $\mathcal{M}{n,p}$, which is exactly $\Hat{\mathcal{M}}_{n,p}$

[gentle osprey]

SeanLayz

$\Hat{\mathcal{M}}_{n,p}=\{A\in \mathcal{M}_{n,p} \vert \det{A^TA}\neq 0\}$
Since the determinant is continuous from $\mathcal{M}_{n,p}\to \mathbb{R}$ and $\mathbb{R}\setminus\{0\}$ is open, then its pre-image is open in $\mathcal{M}_{n,p}$, which is exactly $\Hat\{\mathcal{M}}_{n,p}$
```Compilation error:```! Extra }, or forgotten $.
l.58 ...,p}$, which is exactly $\Hat\{\mathcal{M}}
                                                  _{n,p}$
I've deleted a group-closing symbol because it seems to be
spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.

Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2020/texmf-var/fonts/map/pdftex/updmap/pdftex.map}]```

[gritty widget]

the determinant is not defined on M_{n, p} (when n > p). you are treating it as a function on M_{p, p} when taking the determinant of A^T A

other than this, yes this is good

[rough bloom]

!!!!

Thank you I got it

[heady skiff]

why is it that A \cap Z is both open and closed in Z?

can we just view Z as a subspace of X?

then we get that A \cap Z is open by the definition of the subspace topology

[gritty widget]

A is open and closed in X, so, by the definition of the subspace topology on Z, A \cap Z is open and closed in Z

[heady skiff]

that would also show it's closed under the subspace topology

[heady skiff]

can somebody explain the notation $S^n$ in $\mathbb{R}^{n + 1}$? it's the n-dimensional sphere whose points lie in $\mathbb{R}^{n + 1}$, but why don't we just call it $S^{n +1}$ then?

[gentle osprey]

okeyokay

[heady skiff]

because if it's points lie in $\mathbb{R}^{n + 1}$ does that not make it $n +1$ dimensional?

[gentle osprey]

okeyokay

[fading vale]

No

the n-sphere is an n-dimensional manifold

[heady skiff]

i don't know what a manifold is

[queen prism]

when we talk about a sphere, we mean just the surface of it, not counting the inside

if you look at just the surface, it looks pretty curved

but zoom in enough on any point of the sphere and it starts to look just like flat 2D space

that is what we mean when we say that the sphere in R^3 is a 2-dimensional surface
and that’s why we call it S^2, because if you zoom in enough it looks like R^2

if you took a plane in 3D space that would also be considered a 2-dimensional surface

[heady skiff]

ahh ok that actually makes a lot more sense - so the definition has to do more with manifolds or whatever?

[fading vale]

Yes

it takes n+1 coordinates to describe a sphere globally yes but it can be written locally using n coordinates

(remove any pole and use stereographic projection)

[heady skiff]

ah i see, i guess that makes sense

thanks guys

ok sorry i know i've asked this question before, but i still cannot for the life of me understand why G sends points on the cube to points on the sphere. I can understand F, because you're essentially making sure that one of the coordinates max is equal to 1 (dividing by it) so that it gets sent to the cube, but i don't understand why dividing by the norm gives you some point on the sphere

[queen prism]

what does it mean to be a point on the unit sphere

[pulsar lagoon]

Is S1\{(1,0)} open in S1 where we give S1 the subspace topology from R2 standard

[gritty widget]

yes

points are closed in hausdorff spaces

[pulsar lagoon]

thats true

thank you

[gritty widget]

i think the property of having closed singletons is called T_1 or something but hausdorff implies this and all spaces are hausdorff so who cares

[pulsar lagoon]

true, hausdorff is based

[coral pawn]

Can someone help with this?

The path is given by H(-,x_0) where H is the homotopy between f and g

But idk how to prove the equality on morphisms of fundamental groups

I feel like this is a "follow your nose" type problem but I haven't been able to figure it out

[fickle elm]

Take a homotopy H between f and g. $H:X\times I\rightarrow Y$ where H(-,0)=f and H(-,1)=g. Then $\alpha=H(x_0,-):I\rightarrow Y$ gives a path in Y from $y_0$ to $y_1$. This gives rise to an isomorphism $\hat{\alpha}$ between $\pi_1(Y,y_0)$ and $\pi_1(Y,y_1)$. And you need to argue $g_=\hat{\alpha}\circ f_$ by looking at the path this map defines and the key here is $\alpha$ taking from the homotopy H.

[coral pawn]

Oh so you actually do have to just write it out

[gentle osprey]

Dong_Valentino

[tribal palm]

pretty weird choice of terminology: two topologies are equal iff one is both finer and coarser than the other

[fickle elm]

Just like for the sets. Two sets A and B are equal if and only if A is a subset of B and B is a subset of A.

[tribal palm]

yeah they are just sets after all, i’m not talking about the logic, just the choice of words

[ebon galleon]

Right, that's T_1 or Fréchet (can also be stated in terms of neighborhoods but closed points is the better way)

[grave maple]

Is there a reference that discusses the category of topological spaces and partial continuous functions?

[ebon galleon]

Like partially defined continuous functions (only defined on some subset of the domain)? I have not heard of these being studied before

[distant lichen]

Partial functions can be given by spans where the left leg is a monomorphism

I think you may also be able to view Top_part as the EM category of the maybe monad

[ebon galleon]

Partial functions can also be modeled with pointed sets and basepoint preserving functions

So I would suggest pointed spaces instead

[distant lichen]

This is the Maybe construction when applied to the category of sets

[ebon galleon]

I know

[distant lichen]

I'm not aware of a treatment of this in the literature for topological spaces

[ebon galleon]

Yeah that's part of why I'd suggest pointed spaces, since that is something people have written about

[distant lichen]

I would also wonder if it turns out to be a useful notion at all

[distant lichen]

What is your motivation for this, if I may ask?

Thinking about this a little more I'd have to say I much prefer the span method

[grave maple]

My motivation: partial continuous actions on spaces.

[distant lichen]

Ahhh

As in group / other actions, or dynamics?

[grave maple]

Yes. Like partial group actions.

*partial topological group actions

[distant lichen]

Ok yes then I'd suggest looking at the subcategory of Span(Top) with left legs monic

More and more recently I've been falling in love with spans, they're quite neat little gadgets

[grave maple]

I have seen some books mention pointed spaces for dealing with partiality. I don't quite understand the span stuff.

[distant lichen]

The span is interpreted as

The left leg is the subspace inclusion, of the subspace where the function is defined

The right leg is the function from that subspace

We form composition via pullback

[grave maple]

Is this like (E,M)-factorizations?

[distant lichen]

I don't recognize (E, M) factorization, sorry

[grave maple]

Well, every function f : X -> Y can be treated as a composition X -> img f -> Y where the first part is epic and the second monic.

So you're saying I should treat a partial function f : X -> Y as a composition dom f -> X -> Y where the first part is monic.

[distant lichen]

So an example span from a space X to Y, and therefore also a morphism in a span category, would be something like

X <--- S ---> Y, where S is a subset of X and the map is an inclusion

[ebon galleon]

Would this interpretation have the partial functions as the objects?

[distant lichen]

No, the objects are still spaces

Span(C) has objects of C, and morphisms given by spans.

[ebon galleon]

Oh

So a morphism X --> Y would be a span as you sent?

[distant lichen]

Yes

[ebon galleon]

X <-- S --> Y

interesting

Okay yeah I see how that's better than pointed spaces for this

[grave maple]

How is it better?

[ebon galleon]

I was thinking Top^{span category}

[distant lichen]

For pointed spaces I am a little uncomfortable trying to reason about elements and continuity thereof

[ebon galleon]

So I was confused lol

[distant lichen]

Compared to simply packaging the subset data like this

Though as mentioned I am a little infatuated with spans

[ebon galleon]

So if you take pointed spaces to model this, you still need to worry about continuity on a global scale (i.e., the "partial" function still has to be continuous outside of the domain we are interested in). This is a better model since it would only require continuity on the subspace (and is perhaps a more direct interpretation of partial functions)

[grave maple]

I will get my hands dirty with both approaches and see what difficulties arise.

[ebon galleon]

Is this a common construction? I have not actually seen much done with categories of spans.

[distant lichen]

Relatively yes

There are a number of examples in algebraic geometry at the very least

[grave maple]

Not sure I follow the global argument.

[ebon galleon]

good to know, AG is on my reading list

[grave maple]

Continuity proofs with partial functions are basically the same as the with regular functions except you have to trace out your filters with the domain of definition.

[distant lichen]

For the pointed space approach all spaces need a separate disjoint point as the basepoint

For example S^1 with a basepoint is not a model for partial functions

We want something that looks like

$X \rightarrow Y \coprod {*}$

[gentle osprey]

CatsCradłe

[distant lichen]

Where mapping to * is considered undefined

I believe?

Actually I don't know if there's a more general definition of partial continuous function here

You mentioned filters and I could imagine allowing some sort of "sparseness" that the conts. injection of a subspace, as in the span definition, would not capture?

[ebon galleon]

I'll explain what I had in mind and what's wrong with it.
If we consider a function f : (X,x) --> (Y,y) as a basepoint preserving function, if we are really interested in the domain S subset X (let's say we do this by mapping X - S to the base point y, which we will treat as a "vaccuum" of some sort), if we want this to be continuous we need to worry about what happens on X - S still. In particular, if {y} is closed in Y, we could only model this over open subspaces (since f^{-1} (y) = X - S would be closed). But if you use CatsCradle's method, you only need S --> Y to be continuous; this problem with continuous around the basepoint y does not show up

[distant lichen]

It might be bad to admit but my opinion of the span method being nicer was mostly on general "category philosophy" of not considering elements and then I actually thought about this problem

[grave maple]

I think I understand. Apparently the theory of partial continuous functions defined on open (or closed) subspaces is nicer than in the general case.

[ebon galleon]

That makes sense

And I guess you can get around this by freely adding a "global" point to your space: if your original space is X, add an extra point p and define a new topology whose open sets are open sets in X union {p}

so given that i guess there's merit to both approached (or ofc any existing approaches in the literature, if they exist)

[distant lichen]

And once more we have recovered the Maybe monad

Though I think an Eilenberg Moore category is a far less direct method

I'm also far less comfortable simply asserting that it works here

[grave maple]

There is a computational subtley though with the Maybe monad/pointed space approach - it makes partiality a computable property.

[distant lichen]

Whereas the span definition can be quite easily checked

[ebon galleon]

What wouldn't work?

[grave maple]

So in the theory of partial continuous actions on top. spaces, if G acts partially on X, then it is usually assume that the domain of g in G, considered as a function X -> X, is open.

[distant lichen]

I'm not saying it wouldn't work I'm saying that there is a more nontrivial amount of stuff to check for the EM category approach

[tribal palm]

the lecture notes in my course gives a lemma for comparing topologies generated by fancy B and fancy B prime using bases: T is coarser than T prime iff for for each basis element B in fancy B and each point x in B there is a basis element B prime in fancy B prime with x in B prime subset B

then there’s a note below saying it is not necessary that fancy B subsets fancy B prime

but isn’t that whole messy sequence of quantifiers over basis elements and points in them just equivalent to the much more elegant: fancy B subset powerset fancy B prime?

[distant lichen]

I can't really parse this text formatting, sorry

[tribal palm]

one sec

[tawdry widget]

This is an example:

R^2 with two metric

Two norms I mean

max{|x|, |y|} and sqrt(x^2+y^2)

No one is subset of the another

One base containing open squares, another one containing open disks

[ebon galleon]

[tawdry widget]

No one contains another one but satisfies the condition in your text

[distant lichen]

Thank you I was flipping through munkres to find this definition typeset

[tribal palm]

the lecture notes in my course gives a lemma for comparing topologies generated by $\mathscr B$ and $\mathscr B'$ using bases: $\mathscr T$ is coarser than $\mathscr T'$ iff for for each basis element $B \in \mathscr B$ and each point $x\in B$ there is a basis element $B' \in \mathscr B'$ with $x \in B' \subset B$

[ebon galleon]

Missing a prime somewhere with the Ts

[tawdry widget]

Yeah beginning of fifth line

[gentle osprey]

Jens

[tribal palm]

but isn’t that just equiv to $\mathscr B \subset \mathscr P(\mathscr B’)$ ?

[gentle osprey]

Jens

[distant lichen]

It is not

[ebon galleon]

But yes, as Cogwheels mentioned, we are more interested in cases where two, a priori, completely different bases give comparable or the same topology

[tawdry widget]

Counterexample

[ebon galleon]

Required One to be a subset of the other is much more restrictive

[tribal palm]

oh, right

[ebon galleon]

(in fact, then it's obvious that one of the generated topologies is coarser than the other)

[limber wren]

I don't think that inclusion would even make sense, B and B' contain subsets of X, whereas P(B') contains sets of subsets of X

[tribal palm]

ok this is what i meant: $\mathscr B \subset \bigcup \mathscr P(\mathscr B')$

[ebon galleon]

Identify sets with their singletons

[gentle osprey]

Jens

[ebon galleon]

That's just B'

[tribal palm]

god i should probably just go home and sleep

[ebon galleon]

The union of all subsets of a set X is the set X itself

[ebon galleon]

Go slep

[limber wren]

https://tenor.com/view/cat-sleep-tired-cute-fall-gif-17673985

[tribal palm]

i think i am just using the notation wrong here, i mean to write on the right side of the relation the set of all unions in $\mathscr B’$

[gentle osprey]

Jens

[tribal palm]

would it still be wrong?

[limber wren]

I would think that the cleanest way to state it would be t(B) is courser than t(B') if t(B) is contained in t(B')

[tribal palm]

well yeah that’s the def

[limber wren]

like everything open in t(B) is also open in t(B'), meaning there are less open sets

[ebon galleon]

Okay, that works. Since then that's stating "if B is in T a topology, then T(B) is coarser than T"

And then you can take T to be T(B'), the topology generated by B'

But still, having a simpler condition to check like what is given in the text can be useful. Writing things in terms of large unions is not exactly the nicest thing

[tribal palm]

i like large unions

is there not a neat, compact standard notation for [ \left{ \bigcup \mathscr C \mid \mathscr C \subset \mathscr B \right} ]

[gentle osprey]

Jens

[tribal palm]

?

oh wait that’d just be T(B) wouldn’t it

ok i’ve condensed my confusion into the following, are the following two statements equivalent?: (i) for every $B \in \mathscr B$ and for all $x\in B$ there is some $B'\in \mathscr B'$ with $x\in B' \subset B$ and (ii) for every $B \in \mathscr B$ there is some $\mathscr C \subset \mathscr B'$ with $\cup \mathscr C = B$ ?

[gentle osprey]

Jens

[tribal palm]

i would think yes

[tidal cedar]

point set

[tribal palm]

what

[soft zephyr]

I think this is one of the first few things proven in topology Munkres. Though ur notation is a bit off cuz ur taking union of collections of sets at the end when u probably mean just union of all elements in the collection

But idk I’ve just never seen that notation

[gritty widget]

Can I prove the openness of a set U by showing that for every point x in U, there's a closed ball centered about x and contained in U? I know that usually, people use open balls for this, but is it possible to interchange the two and still have a definitively open set?

I typed out a short proof in LaTeX, but I'm not entirely sure if it's valid.

[tribal palm]

i write $\cup \mathscr C$ for $\cup_{C\in\mathscr C} C$

[gentle osprey]

Jens

[queen prism]

\bigcup

[tribal palm]

picked it up from some lecturer i think

not all that unstandard notation is it?

[gritty widget]

$\textbf{Proof:}

A subset $U \subseteq \mathbb{R}^{n}$ is open if every point $x \in U$ is the center of an open ball contained in U.

For all $\vec{p} \in U$, $\exists B_{\epsilon}(\vec{p})$ centered about a point x, such that $B_{\epsilon}(\vec{p}) \subseteq \bar{B_{r}}(\vec{p})$, where $\bar{B_{r}}(\vec{p})$ has the same center and $\epsilon \leq r$.

Therefore, x must be an interior point of U, and U must be open.$

[gentle osprey]

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

[gritty widget]

Oop-

Sorry.

[queen prism]

cursed and I’d never use it but also arguably not wrong

because you might write something like
[ \bigcup \mathcal{A} = \bigcup \left{ A_i : i \in I \right} ]

[gentle osprey]

Bladewood

[tribal palm]

and?

[soft zephyr]

Idk I don’t like it either cuz when ur actually considering unions of collections it will get confusing

[queen prism]

every closed ball contains a concentric open ball and every open ball contains a concentric closed ball, at least in R^n
so sure

[tribal palm]

in any metric space

[gritty widget]

Ah, alr! Are there any glaring errors in my proof?

I'm somewhat new to proof writing, since I'm a freshman.

[queen prism]

idk I’m not sure
maybe your balls look like B(x, 10), B(x, 9), ..., B(x, 1), and then when you get to the last one there’s no way to continue
but I don’t know if that kind of metric space is possible

[queen prism]

well a better way to say what I’m thinking of might be [ \bigcup \mathcal{A} = \bigcup \left{ A : A \in \mathcal{A} \right} ]

[gentle osprey]

Bladewood

[queen prism]

and then you can think of bigcup as a function taking families to sets

which it is

but I prefer writing \bigcup_{A in mathcal A} A

[uneven bronze]

In these notes (page 8, bottom), they give the following definition:

https://people.math.rochester.edu/faculty/lubkin/topology.pdf

[gentle osprey]

sunside

sunside

[limber wren]

the idea of a net of functions converging is a generalization of a sequence of functions converging, which would normally be like "f_n -> f pointwise if for all x, f_n(x) -> f(x) as n -> infinity", and the same idea for uniform convergence. You wouldn't talk about f_n(x) -> f(x) as n -> N or anything like that, it's the same idea

or even just think of the definition of a sequence x_n converging to x

you define it in terms of n->infinity, not in terms of n -> a where a is a finite number

[uneven bronze]

hmm, ok, but I have come across a net of functions f_a which converge to f as a->0. Does the above definition apply to this case?

[ebon galleon]

are sequences not sufficient for pseudometric spaces?

I mean it should be first countable, no?

[limber wren]

I think you're right that the above definition is a generalization of the idea of n -> infinity

otherwise I guess you'd have to say "f_n converges to f uniformly as n->a if for all e > 0 there exists N such that N < a and for all N < n < a d(f_n(x), f(x)) < e for all x in X"

I think, is that how you'd define it?

lol I've only seen the idea of n>N for nets

[uneven bronze]

in my case, I have a net of functions where the set D in the above definition is (0,a], where a is a finite real number. Then it doesn't make any sense to me to only speak of n>N.

[limber wren]

ahh yeah, my best guess is that the definition for f_n -> f as n -> a would be slightly different, I don't see how the definition above could apply to both cases

lol by best guess though, I could totally be wrong

[ebon galleon]

That would converge to f_a then (potentially other functions too? Since it's only a pseudometric, not a metric)

Since you could always take N(epsilon) to be a

I assume pseudometric is like a metric except you can have distinct points with distance 0? (at least that's what I've seen for pseudometrics). But this should essentially mean that you can replace and pointwise value of f_a with a point with distance 0 away from it?

[uneven bronze]

that was sloppy of me, my a in (0,a] should not be confused with the index in f_a

[limber wren]

I assumed so, np

got what you meant

[ebon galleon]

Well okay so I think there are two things you could do for something like this

If you have a net D, and you want to consider the limit as your points approach a value d in D, you could

1. Consider the directed set of everything less than or equal to d, for which it should converge to f_d (and some other functions which are "indistinguishable" from d by the pseudometric)
2. Consider the directed set of everything strictly less than d, which would be more interesting

So like here, to use a again as your limit over the net (0,a], it would be much more interesting (at least I would imagine?) to consider what the restricted net on (0,a) converges to

In particular, it sounds like it would be most interesting to ask whether f_a is the limit of the strictly lesser functions in the net. Like we could regard the limit of a sequence as the "infinity'th" value of the sequence, this should be a similar idea I think

[umbral panther]

I assume a pseudo metric space is a space with many pseudo metrics. There are topological vector spaces that naturally have this structure. If you have countably many, you can combine them into a single metric, but this is unnatural and also loses the homogeneity of a norm. But there might be uncountably many pseudo metrics. Then the space is not first countable and probably sequences do not suffice

[ebon galleon]

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

It should be first countable by taking balls of radius 1/n, as with metric spaces. But it will be metric space iff it's Hausdorff (or more generally I guess T_0)

But if you have a specific net you're interested in then that makes sense I guess

[umbral panther]

Do people really use these? If points are indistinguishable, why do you have them? Maybe you want to consider this topology, but not alone. I’d think you’d consider Minkowski space primarily with its metric topology and consider the pseudo metric as an auxiliary structure

[ebon galleon]

Compared to some of the other shit people use, this is relatively nice lmao

You can pass to the metric quotient (identify indistinguishable points) but you lose some structure (altho I remember my metric space Prof implied you should just pass to metric space quotient regardless)

I guess it depends on the specific scenario as to which is best

[tribal palm]

i went and barely had the time to ask any questions before he started talking about all sorts of topics for almost two hours lol

it was lovely

though it’s just now some 10 hours later i realized how big of a name he is

john rognes

he was the editor of acta mathematica in a four year period

[ebon galleon]

(don't doxx yourself)

[umbral panther]

He’s on the AT discord

[tribal palm]

whaat?

john is?

[unreal stratus]

Rognes bruh

Yeah he is one of the most famous topologists around lol

[tidal cedar]

^

[hidden crag]

His book on spectral sequences is good

[distant lichen]

The TMF one?

I was typing a joke about how one could say he wrote the book on that

[hidden crag]

TMF?

[tidal lynx]

topological modular forms

[coral pawn]

Let $\beta$ be a loop around $x_0$, what will the (endpoint preserving) homotopy between $f\circ\beta$ and $\alpha(g\circ \beta)\overline{\alpha}$ be in terms of $H, f, g,$ and $\alpha$? I can draw a picture for it but I can't write one down explicitly

[gentle osprey]

Finitely Many Bananas

[coral pawn]

i think i got it

[heady skiff]

are there any criteria for limit points existing besides explicitly producing a limit point? lol tryna think of one rn and coming up short

man

[limber wren]

first thought is that any sequence in a compact metric space must have a convergent subsequence

seems like a good place to start

[heady skiff]

oh my prof said i can't use any properties of compact spaces besides the closed interval of a real line being compact lol

so i'm assuming this a constructive proof maybe?

[limber wren]

lol well that's an annoying constraint

[heady skiff]

indeed

hmmm what could possibly be a limit point of A...

too bad it's not an open interval

[ebon galleon]

bruh why

[heady skiff]

idk

[limber wren]

maybe the sup of the set?

[ebon galleon]

"Let's build up machinery only to not let students use it"

[heady skiff]

i mean tbf the section did come before "properties of compact spaces"

[ebon galleon]

No. [0,1] U {2}

[limber wren]

whats the infinite subset of a close interval in that case?

oooh

nevermind

I got it

[ebon galleon]

Yeh it's in some interval but not [0,1] obv

[limber wren]

yup

[heady skiff]

ok dumb thought: A is contained in the union of a finite amount of open sets, is it true that A must be equal to a subset of those open sets then?

[heady skiff]

oh

nvm

[ebon galleon]

Well okay you can try contradiction I guess. Suppose A is infinite but has no limit points. Arrive at a contradiction. You could do it in a more topological manner by reasoning about what this says on the subspace topology of A

[limber wren]

lol hold on you're allowed to use the fact that the interval is compact, but not any facts about compact sets?

[ebon galleon]

But idk if that would be accepted by your prof

[limber wren]

how does that make sense lol!

[ebon galleon]

I imagine the open set definition as well?

You need something at least

[heady skiff]

the only things given in this section were the heine borel theorem

[queen prism]

what’s a compact set to you

[heady skiff]

a topological space X such that every open cover of X has a finite subcover

[ebon galleon]

Okay, suppose that A is a subset of a topological space X, and that a in A is not a limit point of A. What can you say about the point a in the subspace topology on A? (you should get an iff characterization even; a point is an isolated point of A iff (----) holds in the subspace topology on A)

[heady skiff]

hmm let me think

oh wait so if A has no limit points, then A cannot be an open interval right

or rather a union of open intervals

[limber wren]

or hey just take a sequence of points in A, and do the typical thing of dividing [0, 1] into two halves (assuming A is in [0, 1] wlog), picking a point in the sequence in a side with infinitely many points of the sequence, and repeat that (using induction) to produce a cauchy sequence of points in A, which has a limit in [0, 1] because [0, 1] is closed

[ebon galleon]

Okay but there's a simpler (and better because it abstracts to infinite subspaces of arbitrary compact topological spaces as well) argument with what this condition means on subspace topology by looking at ||the sets {a} for a in A||

[heady skiff]

i guess i'll return to this problem when my brain is functional tmrw

thanks tho

[heady skiff]

true

[hot locust]

okay so is this basically a topology where only the singletons are excluded, am I interpreting it right?

no wait, that doesn't seem right

so basically, the compliment of ever open set is a singleton, so it's sort of like a subspace of a co-finite topology?

am I correct?

no wait, that doesn't seem right either

it's just that for some of the open sets, their compliment is a singleton, such that the compliment set for every singleton of every element in X is present in the space.

so not necessarily co-finite

[novel acorn]

No because {x} can be clopen

[opaque scroll]

If singeltons are closed, then cofinite sets are open. But you can have more open sets, so all you can say is that the topology is some refinement of the cofinite topology.

[hidden crag]

I meant these

https://www.mn.uio.no/math/personer/vit/rognes/papers/notes.050612.pdf

https://www.uio.no/studier/emner/matnat/math/MAT9580/v21/dokumenter/spseq.pdf

[unreal stratus]

Oh cool where did you find those

[opaque scroll]

Presumably on his website

https://www.mn.uio.no/math/personer/vit/rognes/

[unreal stratus]

Dw I was just kidding cause I was the one who showed timo those resources iirc

[hot locust]

wait so why isn't (a+c)/2 in [c, d]

[ebon galleon]

Because a < c since c in (a,b).

So (a+c)/2 < c

[hot locust]

okay, got that but wait why isn't it a>c since it's a subset?

why is c in (a, b)?

[ebon galleon]

You would know a >= c. But you could also use that and trichotomy to get a contradiction

[ebon galleon]

By assumption. If [c,d] is open, there must be an open interval (a,b) around each point x with x in (a,b) subset [c,d]

[hot locust]

ohh okay, that makes sense. So it's not necessary that c is in (a, b) but we are assuming that say it is wlog and arriving at a contradiction

[ebon galleon]

kryojyn art

[limber wren]

love paint.exe

[hot locust]

thanks

[ebon galleon]

c is definitely in some (a,b); the specific assumption that we make for contradiction is that we can choose an (a,b) that's a subset of [c,d] as well. But yeah trichotomy is a better proof that (a+c)/2 or whatever anyways

[hot locust]

gotcha

thanks

[limber wren]

you can also see it algebraically, since c - (a + c) / 2 = 1/2 (c - a) > 0 if c > a

[hidden crag]

The first one yeah

The bigger one I found myself I think

[unreal stratus]

Hot

[hot locust]

Singletons are closed over R with the usual topology but open with discrete topology over R?

[queen prism]

open and closed in the discrete topology

[hot locust]

also closed because their compliments are open as well?

but won't that just make them...closed? But they are there in the discrete topology so open.

so open and closed huh

am I wrong

[kindred cairn]

they are clopen

[ebon galleon]

Clopen sets my beloved

[queen prism]

there is nothing wrong in topology with saying that a set is both closed and open

[ebon galleon]

Well there is because that means disconnected if it's a proper subset and disconnected sets don't exist

[queen prism]

empty set

[ebon galleon]

Not proper

[queen prism]

you don’t see \emptyset as a proper subset?
or are you talking about connectedness

[ebon galleon]

I don't think empty set is usually considered proper, is it?

[coarse night]

empty = empty ∪ empty

union of 2 "proper" disjoint open subsets

[unreal stratus]

sets aren't connected or disconnected

spaces are

😼

[coarse night]

Set embeds in Top

[unreal stratus]

yes

[coarse night]

fully and faithfully

[unreal stratus]

Using discrete topology? sure nice lol

[coarse night]

yes

[unreal stratus]

that's the free functor right

[unreal stratus]

It is yeah

[ebon galleon]

Or indiscrete

[unreal stratus]

A proper subset of X is just any subset of X besides X

[ebon galleon]

[coarse night]

wym by free topological space on X? it's adjoint to forgetful

[unreal stratus]

left adjoint

is what i mean

[ebon galleon]

It's free in the sense of left adj to forgetful yeah

[coarse night]

if by free you mean that then yes

[unreal stratus]

Then there is also the cofree topological space lol

[ebon galleon]

Indiscrete then is "cofree" as right adjoint

[coarse night]

yes that's the other adjoint

[ebon galleon]

Idk where I got that from then

[queen prism]

nontrivial vs proper?

[coarse night]

everything is trivial

[ebon galleon]

That might've been it

[safe galleon]

Guys I have a problem proving something

The only problem this is in frensh

[ebon galleon]

This looks like #linear-algebra no? Unless I'm missing something about topology since it's in French

[safe galleon]

Yeah it's the ideals of a grp of matrices

[safe galleon]

I wanted to ask about the " espace vectoriel normé" if you can look up the translation

Look up ||x||inf=max(Xk)

I forgot it's rule

The Norme infinity of x I suppose

[ebon galleon]

Okay I am now realizing that I don't know enough French math to properly help with this 💀

[safe galleon]

Yeah language barrier is a bitch

Look up Bolzano-Weistrass thoerem of normes maybe u will find what I'm looking for I hope

[narrow cairn]

"espace vectoriel norme" sounds like "normed vector space" but idk any french

[unreal stratus]

Yes

[limber wren]

which part? (this is more linear algebra though, there's nothing topology related in this)

[ebon galleon]

Yeah I don't see where it says that they're normed, but anyways it's a finite dimensional vector space, so it's homeomorphic (through a linear map) to R^{n×n}. So whatever you can conclude about R^m for arbitrary m will hold for M_n(R)

[safe galleon]

No I already told you that was a miscliked

<@&286206848099549185> I was talking about this ine

One*

This part

La normes des séries numériques dans un K-algebre normé

Idk the terminology

[feral copper]

#linear-algebra

Also, you'll probably find more help by not pasting a non-English homework rotated 90° in poor quality, just saying

Your exercise is just to show that R[X] with the sup norm is not Cauchy. It considers the sequence formed by partial sums of the exponential function, and makes you prove it's Cauchy but does not converge in R[X] (that is, the limit is not a polynomial function). I don't see much problems here, so I must ask: what have you tried so far?

(alternatively, R[X] is not a closed subspace of C^0; still not much topology at play, rather linear algebra and computations using analysis...)

[safe galleon]

I already said forgive my goofy ah cmera bro T-T

also why would i post english exercices when im a frensh student

you feel me

and the rotation it is properly positioned on my screen idk about urs

and yes this is topology not linear alg why everyone tells me its lenear when its not T-T

normed vector spaces are part of the topology

isnt it?

[hidden crag]

Because it’s an english server

[narrow cairn]

well kinda, theyre a linalg concept but you can use a norm to define a topology on any normed vector space

[ebon galleon]

Even still, I would say for finite dimensions it falls under lin alg, infinite dimensions under functional analysis

[safe galleon]

you guys' maths is weirdly classified

@ebon galleon are you familliar with grps andd fields?

come to #groups-rings-fields i need some questions anwsered if you dont mind

its past midnight my time i really need to sum up this

[ebon galleon]

I would normally oblige but I am at a family dinner today

[safe galleon]

Sorry to disrupt proceed

[eager vigil]

Does this perhaps somehow connect to the identity theorem in complex analysis? Can one somehow realize holomorphic functions as covering maps?

[sullen bear]

I think this belongs here: How do I prove that the linking number of two disjoint knots is always an integer? We defined the linking number as (1/2)(#LeftHandCrossings - #RightHandCrossings)

I have that the linking number is an integer iff there is an even number of inter-knot crossings

but idk how to proceed

[heady skiff]

i'm not really sure what you mean about what i can say about the point a in the subspace topology on A

it's a closed set ig? i'm not sure

also why are we considering only limit points inside A?

i'm trying to think about what we can deduce from A having no limit points in [a, b]

atm i'm not so sure if this would imply that A is closed because there's no limit points to begin with

[ebon galleon]

If a set has no limit points then it is closed. One way you can define the closure of A is clA = A U {limit points of A}

[heady skiff]

ye i was just thinking about that ig

[ebon galleon]

So {a} is closed. But the fact that a is not a limit point of A tells you more

[ebon galleon]

{a} closed is just true in R (true in all metric spaces since they are Hausdorff)

[ebon galleon]

Think about what this means about open neighborhoods of a in R, and how they relate to A

[heady skiff]

ok wait maybe i'm on to something?

wait

yeah

that would imply that every open neighborhood of {a} intersecting A would have finitely many terms

so maybe we could cover A with all of these neighborhoods and contradict the fact that A is infinite, idk

[ebon galleon]

Something stronger than this even. A point x is a limit point of A if and only if, for every neighborhood V of x, the intersection V n (A - {x}) is non empty.
What do you get if you take the negation of this?

If you haven't seen this characterization/definition of limit points (since this is quite topologically minded) maybe this isn't the most helpful approach, it was just the first thing that came to my mind

[heady skiff]

well my definition of limit points is what you just wrote down

[heady skiff]

isn't it just that if a is not a limit point then the intersection is either empty or contains at most finitely many terms distinct from a

idk, could we do something stupid like consider a neighborhood of a of radius large enough to cover it?

[ebon galleon]

Close

So if x is not a limit point, then by negation we should get there exists a neighborhood V of x such that this intersection with A is empty (not just finite)

[heady skiff]

oh right

[ebon galleon]

So if a in A is not a limit point, there's an open nbrhd with V n (A-{a}) empty. So what's V n A?

[heady skiff]

is it not just a?

[ebon galleon]

It is {a} yeah

Now V is an open set in R

Whose intersection with A is {a}

What do we know about a in the subspace topology then

[heady skiff]

oh that it's already open

which is our contradiction

cuz we must have O(a) \cap A open by the definition of the subspace topology

right?

[ebon galleon]

We get {a} open yeah, but that's not our contradiction yet