point-set-topology

[prime elbow]

to separate all naturals in R by disjoint sets i think (n,n+1/2) work,right?

[red yoke]

There are no naturals in (n, n+1/2)

[white oxide]

Careful,

[prime elbow]

yeah my mistake

[white oxide]

Yeah, what the fox said (🎶)

But yes, that works

[prime elbow]

(n-1/2, n+1/2), n in N

[red yoke]

Can you turn this into a cover of 1/(Z+)

[prime elbow]

this also work right?

[red yoke]

What is it explicitly

[prime elbow]

what if we take (1/n - r(n) , 1/n + r(n) ) where r(n) = max{ (1/n - 1/(n+1))/2, (1/(n-1) - 1/n)/2 } ?

[red yoke]

min?

[prime elbow]

it has distance less than <1/2 and there exist only one natural number so there cannot be two distinct natural number in (n, n+1/2)

[red yoke]

That's not a cover of N though

[prime elbow]

yes

[red yoke]

Yes that works

[prime elbow]

(n, n+1/2), n in N?

[red yoke]

That doesn't contain any naturals

n is not in (n, n+1/2)

[prime elbow]

why (1, 1 + (1/2) )?

[prime elbow]

(n , n + (1/2) )

[red yoke]

1 is not in (1, 1.5)

It's an open interval, not closed

[prime elbow]

yes i am too bad

my mistake again

[prime elbow]

( 1/(n + 1/2 ), 1/(n-1/2) )?

[red yoke]

Yup

[prime elbow]

Thank you ❤️

[gilded steeple]

can anyone please tell me what do characteristics function have to do with this proof? Here the author's using the discrete topology for the set ${0,1}$ and the Tychonoff topology for the generalized Cantor set

[gentle osprey]

lazur__

[tender halo]

members of the space D^m (where D is the two point set) are the characteristic functions of subsets of m

[gilded steeple]

damn, I completely missed that

thank you

[prime elbow]

{ x^2 + y^2 = 4, (x,y) in R× R } is compact because it is closed and bounded

[naive trench]

Yes, Heine-Borel theorem

[real granite]

Suppose some metric $d$ induces this topology.

We know open balls are open, so $\forall x\in X,\ \forall r>0,\ B_r(x)\subseteq X$ is open.

In particular, $B_{\rho} (1)={1}$ is open, where $\rho<d(0,1)$. However, this is not an element of the topology and so cannot be open.

Therefore no such $d$ exists.

[gentle osprey]

Douglas

[real granite]

Is that a correct proof?

[prime elbow]

It seems correct

And if you are familiar with Hausdorff property then you can also show that every metric space has Hausdorff property but the given space has not so it cannot be metrizable

[alpine nest]

Yeah, that would be a general proof of nonmetrizability of the nondiscrete topology, but your proof for this specific case works fine.

[real granite]

Yes I have seen Hausdorff spaces but this is before they are mentioned in this set of notes

[alpine nest]

Also the longer you can get away with not knowing what the separation axioms are, the happier you'll be

[real granite]

I think in my case only T_2 is mentioned

[alpine nest]

^Statements dreamed up by the utterly deranged

[real granite]

Which I assume is the one physicists care about

Or I guess just like... applied mathematicians in general

[alpine nest]

T_2 is pretty much the most important of them because the difference between being T_2 and not being T_2 changes the most properties in the most surprising ways

T_3,T_4,etc. get progressively nicer but it's not as much of a qualitative difference as between having T_2 and not having it

[real granite]

Then again, I do remember reading that in some areas of physics/computing, e.g. quantum computing, there are perhaps unexpected applications of algebra, e.g. finite fields, whereas one might assume physicists only work with infinite fields

So maybe there are areas where T_2 is not assumed

[alpine nest]

I think Zariski topology is the most famous/widely used ones where you don't have T_2

I think you don't even have T_1 in that one

OK, I think you do have T_1

At least sometimes

[real granite]

Is there a general result about a graph of a function being open/closed in the standard topology?

I would assume the graphs of continuous functions are closed because of the sequential definition of continuity, i.e. $f(x)$ is continuous at $a$ if $\lim_{x\to a} f(x)=f(\lim_{n\to\infty} x_n)=f(a)$, where $x_n\to a$.

[gentle osprey]

Douglas

Outsider

[alpine nest]

In particular, for metric spaces it's easy to show pretty much in the way you outlined.

[real granite]

Lovely stuff

[alpine nest]

The converse (closed graph implies continuity) doesn't hold in general, even for metric spaces, although it does in some very important special cases.

https://en.wikipedia.org/wiki/Closed_graph_theorem#Closed_graph_theorem_in_point-set_topology

[tender halo]

it is true for graphs into hausdorff spaces

[alpine nest]

Yep

[tender halo]

pretty easy proof too

for a f: X -> Y consider the mapping f x idY: X x Y -> Y x Y, it is continuous, the preimage of \Delta_Y (which is a closed set because Y is Hausdorff) is the graph of F and is closed

[alpine nest]

Neat! I didn't know that proof.

The one I recall involves some faffing about with disjoint neighborhoods

Although how immediate is it that the mapping you defined is continuous?

[real granite]

What actually is there to prove?

Like, if you suppose $U_x$ is a neighbourhood of $x$, then $x\in U_x$ where $U_x$ is open, so from a metric perspective, it can be written as a union of open balls. Is the point then to show that the open balls are open and therefore elements of the topology (and hence the neighbourhood itself is)?

To show the converse, is the idea that we need to show all open neighbourhoods (i.e. arbitrary elements of topology that contain $x$) can be written as unions of open balls?

I ask because this seems like something that is true at a very basic level, so I wonder what assumptions we would make intuitively we are expected to prove.

[gentle osprey]

Douglas

[tender halo]

it is pretty obvious that cartesian product of mappings is continuous because preimages of basic sets are basic

[alpine nest]

The idea is that you started out by defining "neighborhood" using the notion of metric, and on the other hand you have a general topological notion of "neighborhood", so superficially you have two ways to understand the concept of "neighborhood".

But they are in fact equivalent.

As you observe, it's not difficult to show this equivalence, it's just making sure that it does in fact hold.

[alpine nest]

yeah, I realized that as soon as I'd typed my remark

[tender halo]

for another proof in that vein, consider the diagonal of f: X -> Y and idX, it is a map X -> XxY and is in fact an embedding because idX is a homeomorphism

so X is homeomorphic to a graph of f

with the projection onto X being the inverse function

[real granite]

But in order to show that (neighbourhoods wrt metric)==>(neighbourhood wrt topology), aren't we sort of assuming that open means "is an element of the topology"?

Because by the metric definition, open just means you can put open balls inside the set at every point. So ok, we've shown that open balls are open (wrt metrics), but then how do you show that it coincides with the topological definition ("is an element of the topology") other than to assert it as a brute fact.

[alpine nest]

We are, yes. When you have a metric space, you have a definition of open sets that uses the metric. Those open sets in fact form a topology, so now you have a general topological space (in which the topology arose from the original metric).

So yes, a set is an element of the induced topology if and only if it's open according to the metric definition.

And that's not something you need to check because that's the definition of a topology induced by a metric.

(the thing that you do need to check is that the sets defined as "open" in metric language do in fact form a topology, i.e. are a family closed under arbitrary unions and finite intersections, but I assume that's already been done by this point)

[real granite]

Yeah that would be straightforward to show, going from metric to topological. But then what about the converse? We can assume metrizability (otherwise there is no point in this exercise) and show that every element of the topology can be written as the union of open balls (which we know will be open)?

[alpine nest]

You don't have to assume anything, the induced topology is by definition metrizable because it was induced by the metric. In general I think you're overthinking it

[alpine nest]

So yeah, this is true because that's how the induced topology was defined.

It's not anything deep

[real granite]

Well yeah, I'm inclined to agree. It's just that when proving two statements are equivalent, sometimes it will be not at all obvious, but here they are "within touching distance", so I'm left thinking "...well, what is their to prove, it will just end up being a little circular"

[alpine nest]

You don't have to assume metrizability because you're specifically talking about a topology induced by a metric

Yeah, especially if "neighborhood of x" was defined as "open set containing x" then I don't see much value in that remark you screenshotted.

Since the notion of "open set" means the same in the metric language and in the topological language when the topology was induced by the metric; but that's a definitionally true fact with nothing to prove.

[prime elbow]

I want to prove that metrizable space is normal space.

I proved for closed balls, let A and B are disjoint closed ball.

Let A = B[x, r] and B = B[y,s]then if we take t = (d(x,y) - r -s)/2 such that U = B(x,r+t) and V = B(y, s+t) are disjoint open balls contains A and B respectively.

Is it correct?

[full merlin]

This is nonsense right?

[alpine nest]

It's the output of an LLM so odds are high

[full merlin]

Question: Is the correct definition for two functors F, G: C to Ch_{>=0}(Ab) to be chain-homotopic the following: "there exist natural transformations a: F -> G, b: G -> F such that ab is homotopic to id and ba is homotopic to id"? Am I way off or missing anything? Also, when exactly are two natural transformations chain-homotopic? Do we only require that for all objects, the maps when evaluating the transformations at that object are chain-homotopic, or do we also require that the chain-homotopy between the maps is somehow natural? Or is that automatic?

[prime elbow]

If f: X -> Y where X and Y are Topological space and f is one-one and onto function then if f is open mapping then it will closed mapping?

[tender halo]

its a homeomorphism so yes

[plucky veldt]

its not necessarily a homeomorphism - you can have a one-to-one and onto function between an indiscrete space and a discrete space that's an open mapping but not a homeomorphism

[tender halo]

oh right

sorry i read f: X -> Y and it is automatically continuous in my mind

also open maps are continuous in my mind too

still the answer is yes because just take the complement

[plucky veldt]

a bijection maps complements to complements, so if it maps open sets to open sets it should map complements of open sets to complements of open sets, shouldn't it?

[thorny siren]

yes I think the same

[pseudo beacon]

yeah, since the complement of image will be image of complement

[prime elbow]

Identity mapping work here, right?

[prime elbow]

Is there any weaker condition than bijection which implies image of complements is complement of image?

[plucky veldt]

yes if the spaces have the same underlying set or are of equal cardinality

[alpine nest]

Actually that's equivalent to bijectivity, so no.

[plucky veldt]

yes but for the general case, notice that if you have disjoint closed sets A and B then every point in A has an open ball about itself that is disjoint from B

[prime elbow]

Yes

Let x in A and U_x is an open ball that contains x and disjoint from B then union of all U_x, x in A is an open ball which contains A and disjoint from B, right?

Similarly for B

[plucky veldt]

yes but you need to do this symmetrically

the union of all Ux is not necessarily an open ball but an open set, to be pedantic

[prime elbow]

So let U is union of all U_x and V is union of all V_y

[prime elbow]

Yes

[plucky veldt]

those open sets can still have an intersection though

[prime elbow]

U_x has no intersection from B so U_x has no intersection from V_y, right ?

[plucky veldt]

no

[prime elbow]

Why?

V_y contained in B

[plucky veldt]

Vy is larger than B, necessarily

[prime elbow]

Oh because it is closed?

[plucky veldt]

sorry no, V the union of Vy is larger than B

because it contains necessarily all of the points of B and then some

although no, now that I think about it it can be equal, in a discrete space

damn I'm constantly thinking in Euclidean spaces

[prime elbow]

So how every point in A has an open ball about itself that is disjoint from B

[plucky veldt]

but look at A = [0,1] and B = [2,3], (0,2) is an open ball around 1 disjoint from B and (1,3) is an open ball around 2 disjoint from A but they have an intersection

you need a little trick to make them disjoint

[prime elbow]

So it is not true?

[plucky veldt]

no, it is true

it's just that you need a little more to guarantee that those open balls are disjoint than just saying "Ux is disjoint from B and Vy is disjoint from A"

[prime elbow]

How?

[plucky veldt]

notice that (-1/2, 3/2) and (3/2, 7/2) are disjoint open sets that contain A and B in my example

[plucky veldt]

oh sorry, if you have different sets of the same cardinality you can't have an identity mapping obviously, but you can have bijection

[prime elbow]

And if I have the same underlying set ?

[plucky veldt]

well then an identity mapping is necessarily a bijection, and if you're mapping from an indiscrete space to a discrete space you have an open and closed mapping, but not a homeomorphism or even a continuous mapping, if the set has at least 2 elements

[prime elbow]

No I meant in general is there any weaker condition than bijection such that if f:X->Y is mapping then image of complement is complement of image

[alpine nest]

This is equivalent to bijectivity

[prime elbow]

How?

[alpine nest]

Exercise for the reader

[plucky veldt]

if X -> X is not an injection it's not a surjection, and vice versa

actually no, I'm thinking of finite sets

[alpine nest]

Yeah, for infinite sets that needn't be the case.

[prime elbow]

Okay 🥲

[alpine nest]

But if f: X -> Y has the property that for every subset A of X the complement of its image is the image of its complement, then f is both injective and surjective, i.e. it's bijective.

And I don't think it's a hard proof

[plucky veldt]

if a complement of {x} always maps to the complement of f({x}), then f(x) =/= f(y) for x =/= y

so you have an injection

[prime elbow]

Let f(x) = f(y) and x≠y.
Then X\x ≠ X\y so f(X\x) ≠ f(X\y) it implies that f(X)\f(x) ≠ f(X)\f(y) so f(x) ≠ f(y).

Is it correct?

[alpine nest]

Yes, that's the injectivity argument.

[prime elbow]

Yes

[alpine nest]

This one is more concise and the one I was thinking of, but both are correct.

[plucky veldt]

and if f(X\A) = Y\f(A) then f(A) + f(X\A) = Y

so you have a surjection

[alpine nest]

Yep, although I'm not loving the use of + to mean set union

[plucky veldt]

didn't realize unicode has ∪

[prime elbow]

Actually here I made a mistake it should be Y/f(x) ≠ Y/f(y)

[tender halo]

need to install a latex interpreter into your brain so it can recognize \cup and \cap in plaintext

[alpine nest]

This but unironically

Well, I wouldn't recommend actually installing anything into your brain until they're done torturing the monkeys.

[tender halo]

• is symmetric difference

[alpine nest]

But I unironically have no problem with people using some latex syntax in a plaintext that isn't being compiled.

So long as you don't get too silly with it

[prime elbow]

Let f(X\ ∅ ) = Y\ f(∅), right? So it implies f(X) = Y.

Is it correct?

[alpine nest]

yes

[alpine nest]

See, it wasn't so bad

[prime elbow]

Yes, thank you ❤️

[alpine nest]

I tend to use \bigtriangleup for symmetric difference

[tender halo]

on paper yeah

in chat + is my choice

[alpine nest]

Also because you do sometimes consider algebraic addition of sets

A+B being the set of sums a+b where a ranges over A and b ranges over B

[tender halo]

not real math

yoo im active

time to log off

[alpine nest]

#linear-algebra in shambles

[plucky veldt]

actually we've just done the implication from mapping complements to bijectivity, not the other way

but you can get there

[fringe thorn]

[prime elbow]

Let X is compact space and Y is Hausdorff space and f : X -> Y be continuous bijection. I need to prove that f is homeomorphic mapping.

So I need to prove that f maps closed set to closed set

Now let A be closed subset of X then it will be compact and its image also compact and since Y is Hausdorff so compact image is closed.

Is it correct?

[tender halo]

yes

[prime elbow]

Yes

Any hint to show every metrizable space is normal space.

[plucky veldt]

well I gave you hints, all you need to do is make the final step

what do you still have a problem with?

[prime elbow]

i am not sure about every point x has open ball which has center at x disjoint from B.

i know if x is not sphere point then it has but if x is sphere point then B(x,r+t), work?

[prime elbow]

please consider this

[real granite]

Is this well defined?

If $(X, d_1)$ is a metric space, it is $d_1 : X\times X \to \bR$. So, first issue, $x=(x_1, x_2)\in X\times X$, and second issue, if $y_1\in Y$, then $d_1(x_1, y_1)$ is mapping an element of $X\times Y$.

[gentle osprey]

Douglas

[plucky veldt]

y1 is presumably the first coordinate of y, which is an element of X

since y is an element of X × Y

[alpine nest]

Yeah, this is incorrect, they mixed up the indices.

x and y should both be elements of the product space, so x_2 should be an element of Y and y_1 should be an element of X

[plucky veldt]

actually, it should say that $x=(x_1, x_2)\in X\times Y$ and $y=(y_1, y_2)\in X\times Y$

[gentle osprey]

mzdunek

[alpine nest]

Yeah

[plucky veldt]

since B is closed, X\B is open, so and since A is disjoint from B it's contained in X\B, so every point in A has an open ball about it that is contained in X\B, aka disjoint from B

that just comes from the definition of a metric space

since open balls around each point are a basis for the space

now the only question is - how do you select such balls around points in A and B that they don't overlap if they are centered in different sets?

[prime elbow]

Yes now I got that point

[plucky veldt]

right, so what else do we need to do

[prime elbow]

So Ux is disjoint from B and Vy is disjoint from A

So take union ?

I think its not work

[plucky veldt]

no it doesn't yet work, because you have no guarantee that Ux is disjoint from Vy

[prime elbow]

Yes

[plucky veldt]

but you can find such balls

notice that if there's a ball of size ε around x that is disjoint from B, that means the distance from x to any point in B is at least ε

[prime elbow]

Yes

[plucky veldt]

whatever here's the solution - if you take an open ball of size ε/2 around each point, then Ux and Vy will always be disjoint, since a distance from x to y is δ ≥ ε, so Vy has size at most δ/2 and δ ≥ δ/2 + ε/2, so they don't overlap

[rancid umbra]

does anyone know of a space which is compact but not normal?

[tender halo]

any T1 space is embedded as a dense subspace into its Wallman compactification

[alpine nest]

I mean, {0,1} with the antidiscrete topology is compact but not normal

[tender halo]

if the starting space is not normal, the Wallman compactification will not be Hausdorff

so you can generate as many nontrivial examples with that as you want

[rancid umbra]

what about if point sets are required to be closed

oh i see

[rancid umbra]

okay thank you this is sufficient

i am trying to follow along with a solution online which shows that covering space of compact spaces whose covering maps are finitely sheeted are compact

and the person is quoting the shrinking lemma

which requires the space to be normal

[tender halo]

well generally compact spaces are hausdorff and therefore normal

[rancid umbra]

so you would need to impose the condition that the base space is hausdorff in order to guarantee that

but there are non-hausdorff compact spaces

[tender halo]

i call them quasicompact because im cool

[rancid umbra]

how do you deal with them in them here tho lol

[plucky veldt]

this exercise from the book topology without tears

(i) seems false - a cofinite topology is always T1, but a T1 space is not always cofinite

now that said, a T1 space always contains a cofinite topology, so the intersection of all T1 spaces on a set should be cofinite, seems to me

so there's an error here if I'm correct

[thorny siren]

yeah i think this is correct

this working

[prime elbow]

Are you doing topology without tears?

[plucky veldt]

yes

[tender halo]

what is the error?

i dont see it

[thorny siren]

meaning let X=R be and T finite topology on R and (X,T) a T1 space then this is working

[tender halo]

(cuz i think the exercise is true)

[plucky veldt]

it says (X, Τ) is the finite closed topology if and only if it's a T1 space

[tender halo]

no

there is a second part

right there on the page

[prime elbow]

What's your opinion on that?

[plucky veldt]

oh, I misunderstood

I thought these two were claims to be checked separately

so nevermind

[plucky veldt]

I quite like it, it was very accessible to me as a complete noob in topology, especially the initial chapters

[prime elbow]

I am doing it

[rancid umbra]

why does the author use V_i contained in U_i intersect (E - C)?
isn't it sufficient to take V_i = U_i intersect (E - C)?

trying to show that a finitely sheeted covering map into a compact space is closed, therefore perfect and proper

[red yoke]

Yea it's the same

[rancid umbra]

alright cool

is it also true that covering spaces of locally compact spaces are locally compact?

the proof that i have doesn't use the Hausdorff property at all, so i am thinking i might be wrong, since the exercise said to prove that covering spaces of LCH spaces are also LCH

[red yoke]

Which "locally compact"

[rancid umbra]

a space X is locally compact x if there is a compact set containing a neighborhood of x

by neighborhood, i just mean an open set containing x

Here is my attempted proof. $\newline$

It suffices to show that $E$ is locally compact. Fix $e_0 \in E$ and set $b_0 = p(e_0)$. There is an neighborhood $U_0$ of $b_0$ evenly covered by $p$ and some partition ${V_{\alpha}}$ of $p^{-1}(U_0)$ into slices with $e_0$ in the slice $V_{\alpha_0}$. $\newline$

Since $B$ is locally compact at $b_0$, there is a compact set $C_0$ containing a neighborhood $U$ of $b_0$. Since we also have $U\cap U_0\subset U\subset C_0$ and closed subspaces of compact spaces are closed, then $$\overline{U\cap U_0} = \overline{U}\cap \overline{U_0}\subset \overline{U}\subset C_0$$ is compact. $\newline$

By similar reasoning as in the case when $B$ was regular,
$$\overline{p^{-1}(U\cap U_0)\cap V_{\alpha_0}} = p^{-1}(\overline{U\cap U_0})\cap V_{\alpha_0}\supset p^{-1}(U\cap V_0)\cap V_{\alpha_0}.$$

Since $p\circ \iota_{V_{\alpha_0}}: V_{\alpha_0}\to U_0$ is a homeomorphism, then
$(p\circ\iota_{V_{\alpha_0}})^{-1}(\overline{U\cap U_0}) = p^{-1}(\overline{U\cap U_0})\cap V_{\alpha_0}$ is compact. It follows that $e_0$ is contained in a compact set containing a neighborhood of $e_0$, thus $E$ is locally compact at $e_0$. $\newline$
Since $e_0\in E$ was arbitrary, then $E$ is locally compact.

[gentle osprey]

c squared

[red yoke]

Why is the closure of U in C

[rancid umbra]

U is contained in C

[red yoke]

But C may not be closed

[rancid umbra]

ah, the part where compact subsets of Hausdorff spaces are closed

okay, then does this seem legit?

since we found where I was using it?

[red yoke]

I'm trying to come up with a counterexample

Idk whether this holds

[rancid umbra]

wait wut

it should be true that covering spaces of LCH spaces are LCH

[rancid umbra]

assume B is LCH here

[red yoke]

Oh for LCH it works

[rancid umbra]

yea, it would be interesting to see a c.e. for when the space isn't hausdorff

but i'd imagine it'd be kinda awful

covering spaces of the line with two origins?

[white oxide]

Local homeomorphisms ought to preserve all local properties, no?

They should afterall cover the space

[red yoke]

But is weak local compactness a local property

[white oxide]

Weak as in?

[red yoke]

^

No Hausdorff assumption

[white oxide]

:(

[rancid umbra]

1. and the definition that munkres uses coincide

[white oxide]

Although, given the selected compact set selected, any open cover restricts to the domain of the homeomorphism

and then you can select your finite subcover

[red yoke]

What homeomorphism

[white oxide]

this one

[red yoke]

What if there's no compact neighbourhood inside the domain of the homeo

[white oxide]

Yeah right, I vaguely remember there being a condition that K is contained in another open set

my bad

[rancid umbra]

all good

it was worth discussing, at least for me

[white oxide]

Actually, sanity check: covering maps are open since if U is open you can cover any point in the image of U with locally homeomorphic opens V_i.

Then intersect the pre-images of the V_i_j with U and map them homeomorphically to your base. The union of the V_i_j intersected with U then both cover f(U) and are also a subset of it - hence f(U) is open

But then we can just copy the proof of continuous images of compact sets being compact to open map pre-images of compact sets are open?

[red yoke]

N → {·} is a counterexample

[white oxide]

Hm, right. You only get the statement holding for the pre-images of the subcover, which in general fails to be finite

nor necessarily in the cover

[red yoke]

https://mathoverflow.net/questions/67181/lifting-local-compactness-to-a-covering-space/287202#287202

[eager vigil]

Hey, does anyone see how theorem 14.18 is being used first instead of using 14.19 twice? I don't see how it works. Am I missing something?

[prime elbow]

Proper finite subset of metric space X cannot be dense, right? Because the finite set is closed

[red yoke]

Yes

[prime elbow]

If X is metric space and only X is a dense subset of X, then what can we say about the topology?

[red yoke]

What does it mean for X\{p} to not be dense?

[prime elbow]

There exists an open set which is disjoint from X{p}

[red yoke]

Which means?

[prime elbow]

{p} is open?

[red yoke]

Yup

[prime elbow]

So means metric will be discrete

And it is true in non-metrizable topology

So it is true in arbitrary topological space

[red yoke]

Yeah it is

[prime elbow]

Okay, thank you

[gilded steeple]

hi there, could anyone extrapolate from this definition what a "system of sets" is? I checked the whole book and this is literally the first occurrence

[alpine nest]

It almost certainly means a family of sets, i.e. a set of sets (but people avoid tend to avoid saying "set of sets")

[gilded steeple]

alright thanks

[gritty widget]

hello

sorry if this is a stupid question

but I was trying to look at the felix hausdorff def of topological spaces

which uses neighbourhoods

so ik I need to first have a non-empty set X and a bijective function N which maps elements of X to collections of neighbourhoods of that element

and the four axioms that follow are

every neighbourhood N of a point x has to contain x

two neighbourhoods N and M of point x have to have an intersection that is a neighbourhood of point x

every superset of every neighbourhood of point x is a neighbourhood of point x

and ofc for every neighbourhood N of point x, there exists a subset M which is a neighbourhood of point x such that N is a neighbourhood for every point in M

so but then how would you differentiate this and a hausdorff space

bcz doesnt the last axiom make it so that no matter what, at least two neighbourhoods of diff points in X have to be disjoint

[red yoke]

It doesn't

[gritty widget]

how so

[red yoke]

Well how does it

[gritty widget]

ok lets say theres two points x and y in X

and I say theres a sequence of neighbourhoods P_n for x and C_n for y

such that P_n is a superset of P_n+1

same for C_n

now lets say P_n and C_n have some non-empty intersection M_n

as the sequences progress

ideally M_n should get smaller, no?

and if we take a limit for both sequences, M should tend towards the empty set

no?

[red yoke]

Ideally

But how do you guarantee that happens

[gritty widget]

im not v good at proofs but ig

oh wait

it wont happen unless M is countable

it can happen if M is countable

i think

or maybe im being dumb

[red yoke]

• Why does Pn and Cn strictly decreasing imply their intersection decreases? They could be decreasing by removing points outside the intersection.
• Even if Mn is strictly decreasing, why must it eventually be the empty set? This is only guaranteed if Mn is finite.

[gritty widget]

isnt it also guaranteed if Mn has the same cardinality as the naturals?

[red yoke]

• What if Mn is the naturals and each step you remove an odd number? There are still numbers remaining

[gritty widget]

hm

fair

[red yoke]

• Suppose any number in Mn does indeed get removed. How will your argument proceed?

[gritty widget]

i think im getting confused bcz im thinking of it geometrically

which is not a good way to think about topological spaces

ig

[red yoke]

You can think geometrically, but you have to be very careful

[tender halo]

it is an ok way to think if you know what part of the intuition to leave in and which part to discard

https://tenor.com/view/cirno-math-touhou-perfect-math-learning-math-gif-18399995

[alpine nest]

Also we could set X = {0,1} and define the neighborhoods such that the only neighborhood of 0 is {0,1}, and the only neighborhood of 1 is also {0,1}

[gritty widget]

yeah im not rlly good at that

[alpine nest]

This is not a Hausdorff space, and all the neighborhoods are finite even

[gritty widget]

but then wont this break the 4th axiom

that hausdorff defines

[alpine nest]

It won't

[gritty widget]

how so

[tender halo]

honestly the biggest caveat when thinking geometrically about topological spaces is thinking that all spaces are first-countable

[gritty widget]

oh wait

it doesnt say it has to be a proper subset

damn

[alpine nest]

Indeed

[red yoke]

Yea that too

[gritty widget]

ok yeah then the sequences Cn and Pn can be the same set repeated over n over

which would mean Mn never decreases

damn

ok i got it

i had one more q if its ok

how would you define the topology of R

[red yoke]

Generated by open intervals

[gritty widget]

yeah i was thinking that

one second

this person defines it as this

[red yoke]

This is a topology on R, but not the "standard" topology on R

[gritty widget]

oh

[alpine nest]

Induced by the Euclidean metric.

[alpine nest]

That's a topology on R

[gritty widget]

i meant to not include a metric

[alpine nest]

But the topology on R is the Euclidean one

[gritty widget]

i was thinking how you would show (a, b) is open w this def

[red yoke]

It's not in this topology

[gritty widget]

yea

that was my issue

thats all

[gritty widget]

ofc

i did do construction of R w/ completion of Q

so

that was my thinking only but i wanted to just know how i would write the std topology of R w/o defining the metric

ty @alpine nest @red yoke

[red yoke]

More generally there is also a notion of an order topology

Generated by open intervals on any totally ordered set

[gritty widget]

defined by a partial order

ah

yea that works too

[alpine nest]

As darnymuckhi says, the best way would be to define it as the topology generated by intervals of the form (a,b)

[gritty widget]

yep i was thinking of doing that

tysm

just

one small q

if its ok

would all metric spaces be hausdorff spaces

[alpine nest]

Yes

[gritty widget]

ok great

[alpine nest]

In fact metric spaces are perfectly normal Hausdorff spaces

[gritty widget]

idk what a normal space is

[alpine nest]

https://en.wikipedia.org/wiki/Separation_axiom

[gritty widget]

hmm

idrk how this stuff helps me with GR

but it looks interesting

ty

i was doing topology n diff geo to learn GR later

[alpine nest]

I don't know what GR is, but also Hausdorff (T2) is pretty much the most important of the separation axioms

[gritty widget]

general relativity

[red yoke]

All PCH manifolds are metrizable

[gritty widget]

but the GR books are not v rigorous to be fair

one book calls a manifold equivalent to a non-empty set and just says we endow it w a metric

[red yoke]

I guess the only topology you really need to care about is topology on R^n, since manifolds are just R^n glued together

[gritty widget]

yea

bcz at least one open neighbourhood of points m in M need to be homeomorphic to R^n

right?

[red yoke]

At least some open neighbourhood

[gritty widget]

yeah sorry

[white oxide]

No, but there need to be at least one such

The entire space is technically a neighborhood

[gritty widget]

yeah fixed it

at least one

[red yoke]

In which case you can pretty much rely on drawing pictures

[gritty widget]

lol

[novel plank]

Given a topological space X and a dense subset A, if we remove a point p from A such that A \ {p} is not empty, is it true that A \ {p} is still dense?

[tender halo]

no, take a discrete space

[novel plank]

What if the sapce is compact and T2?

[tender halo]

if A has isolated points, you cant remove them

[tender halo]

no, take a discrete space (but another one this time)

[novel plank]

but a discrete space is compact iff it's finite correct?

[alpine nest]

Take [0,1| union {2} ; the set of rational numbers in that set is dense, but if you remove 2 it ceases to be

[tender halo]

non isolated points you obviously can remove any finite amount

[tender halo]

yes

[alpine nest]

So basically you can't remove an isolated point from your dense set.

(also if a set is dense in a space, then it must contain all the isolated points of that space)

[tender halo]

if your space is t1 removing an isolated point will not make other points isolated so its safe to do all at once

[alpine nest]

If you remove a non-isolated point from your dense set then that should preserve the density.

At least if your space is Hausdorff, something weird might happen in the more pathological spaces

[novel plank]

so we don't need compactness at all then?

[tender halo]

i dont think that matters at all

[novel plank]

well, I was not expecting that haha thank you 🙏

[alpine nest]

I don't think so either, but I so rarely work with non-Hausdorff spaces I don't have a good intuition, so I'm hedging

But yeah, neither compactness nor haudsorffness matter.

You just need to ensure that the point you're removing from your dense set isn't an isolated point of the space

[tender halo]

yeah the difference is that you can remove a non isolated point and another point becomes isolated

which doesnt happen in T1 spaces

[real granite]

Let $(X, \tau)$ be a topological space and $A\subseteq X$. Then the subspace topology on $A$ is $\tau_A= { U\cap A | U\in\tau }$.

When revising this, I accidentally wrote down $\tau_A={U\in\tau | U\subseteq A}$.

Is this a valid alternative or is it wrong?

[gentle osprey]

Douglas

[red yoke]

Well they're definitely not equivalent

If A isn't open, it won't be a topology at all

If A is open, it does coincide with the subspace topology

[real granite]

Because if A is not open (not necessarily closed), then there might be no open U in A exist, and so this alternative would just give the subspace topology as the empty set?

[red yoke]

If A is not open, then the total space A is not in τA

[real granite]

which fails to satisfy the axioms

What about if you reversed the s.t., i.e. $\tau_A={U\subseteq A | U\in\tau}$

[gentle osprey]

Douglas

[red yoke]

Isn't that the same

[real granite]

I think that would run into the same issues

[red yoke]

If you know categorical language, the subspace topology on A ⊂ X is the final object in the category of morphisms Y → X with image in A

(otherwise ignore)

But basically the motivation is that a continuous map into A will be the same thing as a continuous map into X with image A

[real granite]

Yeah I was thinking that maybe {P|Q}≠{Q|P}, but the sort of thing I had in mind was slightly different. E.g. {n | √n in N} is not the same as {√n | n in N}, but here we are not simply reversing the order of the statements, the statements themselves are different.

Anywya

Makes sense

[thorny siren]

but for every U∈T

[thorny siren]

no, it is possıble if A closed and (X,T) a topological space T={X,empty set,U}we say X interior A=A and empty set interior A=empty set and yes {A,empty set} a topology on A and yea (A,TA) subspace of (X,T)

[red yoke]

I was referring to the second "alternative definition"

[thorny siren]

oh okay @red yoke sorry bro

[gritty widget]

Wouldn't this imply all subsets of A are in tau

[thorny siren]

why?

[gritty widget]

You're defining it that way

Literally look at it

[thorny siren]

yes

it is possıble but not everytime

[thorny siren]

this not work for every relative topology

[unreal gull]

how does the first (left) part of the union defining x not contain a limit point of the second part of the union defining X?

[opaque scroll]

Which point is it you're imagining?

Both sets are closed, so contain all their limit points, and they are disjoint.

[unreal gull]

well i thought as x goes to infinity the second part of the union would have a limit point in the first part of the union

[jaunty stratus]

Is there a reason why the arrows go from Y to X in the subspace and product topology yet they are reversed in the quotient and disjoint union topology?

It's about the universal property, btw

[ruby crown]

Can I ask about boundary operator in topological data analysis here?

[lucid geyser]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[red yoke]

I'm assuming it's the same boundary operator as in alg top

[ruby crown]

Why is ker(d_0) = Z^3 and ker(d_1)=Z in https://ibb.co/FWhJxDY ?

[merry geode]

I assume d_0 is a map to 0, then that part is obvious

The image explains why ker(d_1) is like this.

[ruby crown]

huh

Can you elaborate on that?

[merry geode]

ker(d_1)?

[ruby crown]

Yeah

[merry geode]

Do you see that e3 - e2 + e1 is in the kernel?

[ruby crown]

Yep!

Only pictorially though, wouldn't know how to reason about it algebraically

[merry geode]

Yeah, then to see that no more elements of kernel:
Consider Im(d_1) is isomorphic to C_1(X)/Ker(d_1).

[merry geode]

Ah, well, you simply compute d_1(e3 - e2 + e1)

The map is Z-linear, so you can apply definition of d_1

[red yoke]

This is basically solving linear equations

What is the kernel of the linear map ∂ in terms of the bases <vi> and <ei>?

[ruby crown]

uhm, how would you compute C_1(X)/Ker(d_1)?

[merry geode]

Hm wait

..that may not work, sorry

[ruby crown]

I'm lost

[red yoke]

C1 and C2 are vector spaces

(Z-modules actually since Z isn't a field)

[merry geode]

You can also solve this brute-force-y via

d_1(c_1 e_1 + c_2 e_2 + c_3 e_3) = 0.

[red yoke]

The boundary map d1 is given by the Z-linear map
d1(c1e1+c2e2+c3e3) = (-c1-c2)v0+(c1-c3)v1+(e2+e3)v2

Where {e1, e2, e3}, {v0, v1, v2} are the bases of the Z-modules

The kernel is the "nullspace" of this linear map

[merry geode]

Thinking about it, you can also try tensoring with Q, right?

[red yoke]

Yeah but it seems they're interested in Z here

[merry geode]

Then you can do proper LA, after that you pick out integers to restrict to Z-module

[ruby crown]

Oh yeah, so I just set two free variables to random vectors but wouldn't this show that the null space is spanned by two vectors?

[merry geode]

I don't think you get two free variables

[red yoke]

It should be spanned by e1-e2+e3

So it's isomorphic to Z

[ruby crown]

I don't really understand how the null space gets computed because v0, v1, ... are fixed

[red yoke]

The nullspace is the subspace of C1 that gets sent to zero

Zero as in 0 v0 + 0 v1 + 0 v2

Solving that gives you c1 = -c2 = c3

[ruby crown]

Huh, but can't there be a c such that the coefficients in front of v are non 0 such that the linear combination still sums to 0?

[red yoke]

No, because {v0, v1, v2} is a basis

[ruby crown]

oooo

[red yoke]

They are Z-linearly independent

[ruby crown]

Awesome, thanks!

[hollow geyser]

I had a real difficult time with this problem. I just want to know if it's perfect, or if it could be better.

(I should probably mention that g o f is locally constant)

And last lines should be {f(x)}, not {r}

[alpine nest]

Hm, there's a problem with the defintion of g, because you define g for a fixed x, so for different x you'll have different functions in the role of g.

Which means you can't consider the composition of g and f

Or if you do, then the only reasonable interpretation will make it a function that's constantly 0, because you've got g(f(x)), and you've defined g to be 0 if the argument is equal to f(x)

Ah, wait you fix some x, and then you define the function g_x, and consider the composition of g_x with f; that might be reasonable.

[hollow geyser]

Correct

[alpine nest]

Then yeah it seems to be fine, or at least nothing leaps out at me as a glaring mistake.

Hedging my bets because I'm not done waking up yet

I'd just alter the notation a bit to make the dependence of g on the fixed x more explicit

[quick bough]

has anyone read this paper on operads by any chance? https://arxiv.org/pdf/1906.12275

i have a couple of questions on this…

(in particular on how the composition product looks like if you restrict E to be Set, i’m not too familiar with computing coends)

[fair tusk]

Not sure I understand how containing attached 0-cells in a CW complex affects it - wouldn’t a line (1-cell) contain infinite 0-cells?

Also, when you only have one given 0-cell at one end of a line, its Euler characteristic is different from that of a line segment with a 0-cell at both ends

[unreal stratus]

Well being a cw complex is structure - you present it as a set of 0 cells to which you adjoin 1 cells, etc

You don't retroactively view it as containing this many 0 cells etc or yes you get nonsense like that

it also doesn't make sense to have a line [0,1] with only one zero cell, because you need to attach the one cell's boundary to the 0-skeleton

And if you just have one 0-cell that will give you a circle when you attach a 1-cell

[fair tusk]

Ah, so it’s always ascending - does that mean you can connect higher cells with two one cells (antipodal points on a sphere)?

[unreal stratus]

I'm not sure what you mean by that

I suggest you just review the definition as in hatcher or smth hm

[fair tusk]

Like if we start with two points and attach a surface between them

[unreal stratus]

Uhh as in D^2?

[fair tusk]

I didn’t really understand their disjoint union definition 😢

[fair tusk]

I was thinking more S^2 but I suppose that too

[fair tusk]

Oh my bad I meant 0-cells, not 1

[red yoke]

if it could be better

You mean more concise?

Let X be connected and f: X → Y be locally constant. For any S ⊂ Y and x ∈ f^-1(S), there is some neighbourhood U ⊃ x such that U ⊂ f^-1(f(x)) ⊂ f^-1(S), hence f^-1(S) is open. Thus if f(X) can be partitioned into two nonempty subsets S, T, then X is the disjoint union of nonempty open sets f^-1(S), f^-1(T).

[alpine nest]

Very elegant.

[red yoke]

Wait how do you show g·f is locally constant

[plucky veldt]

If every proper closed subspace of a (X, T) is compact, then (X, T) is compact, right? For an open covering Oi I can take an open set U, then its complement C is a proper closed subset ergo a compact subspace. Then the intersections Oi ∩ C are an open covering of C which have a finite subcovering Oj ∩ C, if I take Oj and U I have a finite subcovering of X.

[unreal stratus]

Yes that sounds good!

There is another way: consider the characterisation in terms of closed sets. So X is compact iff for every collection {C_i} of closed subsets whose finite intersections are non-empty, the whole intersection is non-empty. It is basically immediate from this as you can intersect everything with some fixed element of the collection

[plucky veldt]

makes sense

[plush folio]

In solution ii) here, what does it mean that the choice of f is unique by hypothesis?

[red yoke]

Maybe they just mean there's only one possible choice

[plush folio]

It's definitely the most obvious choice, but it's not immediately clear to me that it's the only choice

[red yoke]

I think you can show it is the case by looking where each element gets sent

πX · f = fX means the x-coordinate is given by fX

And similarly the y-coordinate is given by fY

[plucky veldt]

if you assume that f_X = π_X ∘ f, then the first coordinate of f(a) has to be f_X(a)

same with f_Y

π_X ∘ f gives you precisely the first coordinate of f

[plush folio]

Okay, makes sense 👍 So "unique by hypothesis" just means that it's trivial to show it is unique

[red yoke]

Or maybe they rephrased the question in their minds as

the unique map such that … is continuous

Otherwise "hypothesis" seems weird

[plucky veldt]

the hypothesis is that f is a function that composes with projections to form f_X and f_Y, which means it has to be unique

imo

[proven comet]

In my class we defined a Submanifold of dimension k of class m. I get that intuitively, the dimension k part speaks to the local isomorphity of the submanifold to Rk, what's the intuitive point of specifying a class for the diffeomorphism? How "smooth" the submanifold is?

[red yoke]

Does class m mean m times continuously differentiable

[drowsy iron]

Yes

[red yoke]

Yea it tells you how "smooth" the map is

All the data encoded in a class m manifold is essentially just the shape of the manifold (its topology) and which maps into/out of the manifold are C^m

[drowsy iron]

If I equip R with the opposite topology, where open intervals are "closed", and closed intervals are "open". Does that not mean that an "open" set can be compact? Because a cover of "open" sets, would be a cover of closed intervals, which trivially finitely cover closed intervals (which are "open")?

[alpine nest]

That won't work because it won't be a topology

[drowsy iron]

Oh yeah my bad lmao

[alpine nest]

You could start by requesting that intervals of the form [a,b] be open sets, but since topology must be closed under arbitrary unions, you can't avoid intervals of the form (a,b) also being open.

I think that is a topology that shows up in some (counter)examples

I.e. one generated by starting with closed intervals

[red yoke]

It's the discrete topology

[alpine nest]

What, already?

Oh yes, singletons will be open, never mind.

I mean, that means I was right that it is a topology that shows up

But not a particularly original one

But I could have sworn there was some kind of topology generated by some kind of intervals that's finer than Euclidean but not discrete.

Half-open intervals maybe?

[plucky veldt]

https://en.wikipedia.org/wiki/Lower_limit_topology

[plucky veldt]

I'm just asking because I have this exercise

seems to me that the condition of being a Hausdorff space is superfluous

[red yoke]

Yes

[plucky veldt]

and it's just confusing

not sure how to even prove it relying on the Hausdorff property

[ocean narwhal]

A lot of people only work with Hausdorff spaces. So: many times people may say "Hausdorff topological space" when it isn't strictly necessary.

[opaque scroll]

Hausdorff makes it an iff though, so that's something

I.e. X is compact iff every proper closed subset is

Actually, that's also true either way

[plucky veldt]

yes, a closed subspace of a compact space is always compact

[opaque scroll]

What I was thinking about was compact subspaces being closed

[ocean narwhal]

My point illustrated: with non-hausdorff spaces people sometimes assume that results just don't hold

[opaque zodiac]

yeah although a question you have to ask yourself is how much of your life you want to spend descending into generality hell

[surreal island]

Can anyone help me with this

exercise 1

[alpine nest]

That's pretty much why I tend to throw in the Hausdorff; it's not that I assume that results don't hold, but I know I shouldn't assume that they do, and I can't be bothered figuring it out.

I'm the lucky kind of person who doesn't have to consider non-Hausdorff topologies.

[merry geode]

Non-closed point

[rancid umbra]

if X is contractible and Y is path connected, then the set of homotopy classes of maps from X to Y, [X,Y], has a single element.

the fundamental group is the set of homotopy classes of based loops, so a subset of [I,Y], where I = [0,1]

why does the fundamental group even have any elements other than the homotopy class of a constant map? im having a fundamental misunderstanding of something

is it because in [I,Y], you can wiggle the end points around?

this has to be it

loops don't have to be equivalent to only loops since the end points are unconstrained

but [S^1, Y] is the fundamental group on Y, correct?

[merry geode]

Is fundamental group a subset of [I, Y]?

How is it?

[pastel linden]

[X,Y] is trivial if Y is contractible, not X

Sorry, [I, Y], not [X, Y]

[pastel linden]

Then your maps are all homotopic to a constant map

[rancid umbra]

both are

[umbral panther]

If X is contactable then [X,Y] is the components of Y, which is often a point

[pastel linden]

[I, Y] is the set of homotopy classes of paths. If Y is path connected, then this should be a singleton

Fundamental group is loops. That's obviously different from paths

[merry geode]

I don't see how [S^1, X] would be a subset of [I, X]

[pastel linden]

It's just [I, Y] with s(0)=s(1)

So you get induced maps by quotienting

[merry geode]

Is the confusion basically from that fundamental groupoid is composed of I -> Y?

[rancid umbra]

i think i am neglecting the significance of based loops

[pastel linden]

But yeah, S^1 is not contractible, so pi_1 = [S^1, Y] is nontrivial

pi_0 = [I, Y]

[merry geode]

pi_0 = [S^0, Y] also

[rancid umbra]

you don't need to fix a base point for pi_0?

[pastel linden]

If you fix a base point, then it's always just 1 for that connected component

We usually stipulate Y path connected for that kind of thing

[pastel linden]

S^0 is two points, so no

[merry geode]

Wait what

[pastel linden]

Yeah

[merry geode]

I am like.. why, how, uh

[pastel linden]

It's the equator of S^1, so two disconnected points

|x| = 1

[fading vale]

It's this if you talk about based maps

[merry geode]

Welp.

[fading vale]

It's this if the maps are not based

[merry geode]

Based maps?

Ah, like when you.. uh

[rancid umbra]

f(1) = y_0 for all ur paths

[merry geode]

How do you say this precisely

[fading vale]

Fix a point in the domain and codomain and demand that the first point is mapped to the second

[merry geode]

Ah, so that's where the basepoint comes into play

[rancid umbra]

okay, so [I,Y] for Y path connected is trivial since you can move the paths around and effectively tear apart any loops you create

since the endpoints of paths aren't bound together and there are no base points

[merry geode]

[I, Y] = [pt, Y], right

Since you can contract I

[pastel linden]

Yes

[rancid umbra]

even if you have based paths, [I,Y] is still trivial

[fading vale]

Yes, even if f(0) = f(1), a homotopy h from f to g does not need to satisfy h(0,t) = h(1,t)

So you can always just unwind the path

It doesn't have to be homotopic through loops, only through paths

[rancid umbra]

but when they are based loops, you loose the freedom to move paths around and split end points

[fading vale]

Ys

[rancid umbra]

so, how are [I,Y] and pi_1(Y,y_0) related, or maybe [S^1,Y] and pi_1(Y,y_0)

[fading vale]

The former is more related to pi_0

Ah

[merry geode]

When defining fundamental groupoid, do you say "double-based"

[rancid umbra]

okay, that i can buy

[merry geode]

How do you get an impression that [I, Y] is connected to pi_1? I forgor..

Maybe it's obvious

[fading vale]

The free loops and based loops are a little more complicated

There are exercises that tell you something about this in Hatcher

But basically the exercise is that homotopy classes of free loops in a path connected space corresponds to conjugacy classes in pi_1

[rancid umbra]

these two?

[languid patrol]

this mans has never slurped spaghetti

[fading vale]

Is that an ancient swedish cultural practice passed down from generation to generation of viking servant

[proven hollow]

I don't get how this is a branched covering of the sphere (this is taken from stillwell classical topology), I don't get why a and b are identified the way they are

the context is viewing w(z) = z^2 as a map on the reimann sphere

[umbral panther]

The Vikings were very active in Italy

[fading vale]

So true...

[fading vale]

Does it not make sense to you why this is a covering map outside of 0 and infinity?

Ignoring the bit about a and b for a moment

[proven hollow]

yeah that part makes sense

I just dont know how the diagram represents it

[fading vale]

The circles parallel to the equator correspond to a circle of radius r in C

Think about what the map does to that circle

The arcs a and b correspond to rays starting at the origin and going to infinity

Think about the image of those under the map

[proven hollow]

are the black annotated portions supposed to be gaps? I just dont get the picture

think I made some progress: slit is not the same thing as slice

[fading vale]

They're supposed to be longitudinal curves in the Riemann sphere (I think anyway)

With each pair differing by a 180 degree rotation

[deft crow]

Reading papers on the DInaburg definition of topological entropy; how does one show that :
$$\frac{1}{n}\log(\text{cov}(n,\varepsilon,f))$$
Is subadditive?

[gentle osprey]

Dealersgrip

[deft crow]

Found a paper with a mistake in it

[alpine nest]

Hm, I don't think it is; it's the log itself that's subadditive (with respect to n)

And then the well-known theorem tells you that the log divided by n will have a limit as n tends to infinity

[deft crow]

That's problematic

I'm reading a paper from Chicago university with a blatant error

https://math.uchicago.edu/~may/REU2014/REUPapers/Butt.pdf

Page 2-3, there seems to be an inversion of inequalities in the final stages of the proof, enabling the author to conclude erroneously

I may be wrong, but I'm pretty sure there's an error

[alpine nest]

There shouldn't be, the definition of entropy is fairly standard.

I'm not going to investigate the proof itself, but they seem to be showing that the log(cov(n,e,f)) is subadditive as a function of n.

Which is the correct path to take.

And then the limit of 1/n log(cov(n,e,f)) exists and is finite.

And that's h_e(f).

When you log this, you get the subadditivity of the log

[deft crow]

right, here the rightmost inequality is what is causing me issues

because earlier in the proof we show that:
$$|A|\geq\text{cov}(n,\varepsilon,f)$$
$$|B|\geq\text{cov}(m,\varepsilon,f)$$

sorry the other wayy round

[alpine nest]

Yeah, the inequality isn't correct as written, but the conclusion is correct.

[gentle osprey]

Dealersgrip

[alpine nest]

What they should have done is written $cov(m+n,\epsilon,f) \leq |C| \leq |A||B|$, and since $A$ and $B$ were arbitrary, then passing to their lower bounds we get the desired inequality.

[gentle osprey]

Outsider

[deft crow]

perfect, thanks I hadn't understood that

thanks a lot!

[alpine nest]

So you're correct that there's an error in the paper, but it's fixable

[quick bough]

is the wedge sum of co-h-spaces again a co-h-space?

[prime elbow]

They define X to be Lindelof space iff every open cover of X has a countable finite subcover.

I need to show that the second countable space is Lindelof space.

Since it is the second countable topology, there exists a basis which is countable.

So if X has open cover U_i then U_i is union of basis element which is countable union.
Then it is Lindelof space.

Maybe I need to do more work in my proof

[quick bough]

do you maybe mean a countable subcover?

your proof is not correct btw, each U_i is a union of countable many basis elements, but the initial open cover could have been possibly uncountable

so X is a union of uncountable many opens, which in turn are a union of countable many basis elements

the way to fix this is the following: consider J to be the set such that n \in J if and only if there exists an open set A of the cover you chose with B_n lying in A, where B_n is one of the basis elements

and the B_n are indexed over N since X is second countable

[prime elbow]

Yes

[prime elbow]

But eventually it will be a union of countable basis element

[craggy cedar]

The countable cover proposed here doesn't seem well defined

What open sets are you putting in it

[prime elbow]

Now I understand where I am wrong

[inland laurel]

I think a defined set can be open cover and a base

[prime elbow]

Sorry I don't understand

[inland laurel]

if X a topological space and and every subcovers of open covers is finite, meaning a base is finite and yea this topological space lindelof space and second countable because every subcovers of open covers is finite for a lındelof space and yea When we find all the bases, a bases or more than one bases is finite bro

[quick bough]

yeha, sorry, i said something wrong, you need to find a subcover, so you can’t just pick the basis elements because they may not lie in your basis

[inland laurel]

this is unimportant bro sorry

[quick bough]

then again, i think my suggestion leads to the solution

[craggy cedar]

"Every subcover of open covers is finite" is only true if the topology itself is finite, no?

That seems like an unrelated notion

[inland laurel]

yes ı mean "Every subcover of open covers is finite" for lindelof space ı said

[quick bough]

that’s not true

[inland laurel]

then this is false but why?

"They define X to be Lindelof space iff every open cover of X has a countable finite subcover. " he said

[quick bough]

i don’t rlly get what you are trying to say tbh lol

[inland laurel]

.

[quick bough]

the subcover needs to be countable, not finite

nothing abt this is finite

[craggy cedar]

Not sure if this has actually been resolved, so consider the sequence $B_{n_i}$ of basis elements contained in some element of the open cover

[gentle osprey]

خرشوف

[craggy cedar]

This is also false, unless, as stated, the topology is finite

[inland laurel]

maybe he want proof for finite topology

[quick bough]

i don’t think he does lol

[craggy cedar]

Lol what

[quick bough]

if the topology is finite, it’s trivial

since there are only finitely many opens

[craggy cedar]

A topological space with finitely many open sets is obviously both lindelof and second countable...

[quick bough]

yup

[inland laurel]

yes

[craggy cedar]

Anyways, this is how I'd start the proof

@prime elbow Where do you think you'd go from here

[inland laurel]

but my proof working for finite topology it is correct?

[quick bough]

it’s trivial if the topology is finite

[craggy cedar]

I mean yeah, but it's not a very useful proof

Given we want a proof for an arbitrary second countable space

[inland laurel]

thank you @quick bough and @craggy cedar

[worthy nebula]

hey

i'm having a hard time figuring out how the van kampen theorem for spaces implies the van kampen theorem for cw complexes, and if it really does

the version for spaces uses open covers of the space, the version for cw complexes uses subcomplexes

[unreal stratus]

I believe the point is that if you have a cover of X by subcomplexes A_i (i=1,2) then you can find open neighbourhooods U_i of A_i which deformation retract onto A_i

[worthy nebula]

oh, okay

a space and its deformation retract are homotopy equivalent, right?

[unreal stratus]

yes, essentially by definition

if i:A->B is the inclusion and r:B->A the deformation retraction then ir is homotopic to identity and ri is actually equal to identity

[worthy nebula]

yeah, true

thank you so much for the insight

[unreal stratus]

np

[craggy skiff]

Let $\alpha \in \mathbb{R}^*,, p:\mathbb{H} \to \mathbb{D} \setminus {0}, , p(z) = e^\frac{2 \pi i z}{|a|}$. I want to show that $p$ is a covering map but I dont't know how to make this. First, clear $p$ is continuous and surjective. I think I need to start with an $y \in \mathbb{D} \setminus {0}$ and take an open disk $D \subset \mathbb{D} \setminus {0}$ to be a neighbourhood of $y$. Now, because disk is open and connected, there exists a holomorphic branch of logarithm in $D$. It's ok this start? How I can continue? Thanks!

[gentle osprey]

Tibi

[pastel linden]

The branch cuts of log are rays from the origin because it comes from the branch cut of arg(z)

so you can form an open cover by removing {0} union positive or negative real axes from the disc. on each cover we can choose a holomorphic branch of log and the sheets are given by log|z| + iArg(z) + 2pi i n

[craggy skiff]

Ummm I still don't understand

Sorry 😦

@pastel linden

[prime elbow]

Sorry my mistake I wrote the wrong definition

[prime elbow]

The union of B_(n_i) is X, right?

[craggy cedar]

Yes, but can you prove that?

[merry geode]

How do you construct this sequence

What if I (with some mathematical trolling) picked B_{n_i} = B, a basis element inside some of the open cover?

Even if you require B_{n_i} to be distinct, you can resort to picking some subset.

[prime elbow]

Since $ X \subset \bigcup_{i} U_i$ and $U_i = \bigcup_{j = 1}^{\infty} B_{ij}$.

Then taking all $B_{ij}$ with respect to $i$ we get a sequence which union is $X$

[gentle osprey]

Notknow🙇

[craggy cedar]

If B_1 is contained in an element of the cover, set n_i=1

If not, check B_2

[merry geode]

Yeah this sequence does not necessarily cover entire X.

[craggy cedar]

This seems false

Let x in X, there's some cover element containing x

There's some basis element containing x contained in that cover element by definition

[merry geode]

Hm, so you exhaustively enumerate over the basis, picking all the ones contained in some elt of the cover? Fair.

[craggy cedar]

Do you have a different solution?

I'm sure there is one but I don't really see it

[prime elbow]

Can someone tell me the correct solution?I think my first solution seems correct but I don't know how to write

Because there may be many basis elements overlaps here

[craggy cedar]

There are overlaps but that doesn't really matter

[craggy cedar]

See here

[merry geode]

I do not have it different by much, but intuitively covering is one that every point is contained.
So for every point p, you pick a basis element B_p inside some U_p of the cover. (I guess this requires choice) Then adjust U_p so that it only depends on B_p.

Now, remove duplicates from {U_p}

[craggy cedar]

I'm not sure I see how the "adjustment" works

Also duplicates are removed by the cover being a set

[prime elbow]

It seems correct

[prime elbow]

Is it correct or not?

[merry geode]

If I get you correctly, one problem here is that you need a subcover, which means you need U_{n_a}'s. Covering by basis elements is not enough.

[prime elbow]

Yes

[craggy cedar]

Well yeah of course but you clearly just pick elements of the cover containing your basis elements

[merry geode]

Well, in another words: Gather such basis {B_i} picked through this process, obviously countable and covering X.
Pick U_i containing B_i.
Done!

[prime elbow]

But when get basis elements B_i that covers X then we can add some additional basis elements to complete U_i

[craggy cedar]

It seems like you kinda did what I did but with extra steps idk

[merry geode]

Dunno about extra steps part

Basically, I am saying your process was not clear for me.

[prime elbow]

Good it is now clear

Thank you

[craggy cedar]

Idk I feel like I just constructed the set {B in basis|B is in an element of the open cover} and indexed it

[merry geode]

Why would you add additional basis?

[craggy cedar]

But yeah clarity is a thing, my bad

[merry geode]

Ah, I see.

[prime elbow]

Let B_1 and B_2 cover X and B_1 in U_1 and B_2 in U_2 but B_2 not in U_1.

To complete U_1 and U_2 we can add some additional basis element then it will be U_1 and U_2

Sorry, I don't know how to explain this

[merry geode]

Ah, I mean, the subcover will be {U_1, U_2} in this case.

[prime elbow]

Yes

Just add some additional basis elements in the sequence if we make sequence

I think it is not necessary

[merry geode]

Yeah that was why I was confused, it is not necessary for finding U_1 and U_2.

But as you said, U_1, U_2 can be re-formed(?) by joining basis elements together.

[prime elbow]

Now I think I got it

[prime elbow]

To show every metrizable space is normal.

Let $A$ and $B$ are disjoint closed subsets of a metric space $(X,d)$.

Then we know that by Urysohn's lemma there exists $f:X\mapsto \bR$ which is continuous and $0\leq f\leq 1$ and $f=0$ on $A$ and $f=1$ on $B$.

Then we can let $U=f^{-1}[0,1/2)$which is open in X contained $A$ and $V=f^{-1}(3/4,1]$ which is open in X and contained $B$.

And $ U$ and $V$ is disjoint.

Is it correct?

[gentle osprey]

Notknow🙇

[red yoke]

If you already have Urysohn's lemma on metric spaces then yea

[prime elbow]

Actually they differently stated Urysohn's lemma that in metric space for any two disjoint closed set U and V there exists function f:X->R such that 0≤ f ≤ 1 f(x) = 0 on A and f(x) = 1 on B.

They define f(x) = d_A(x)/(d_A(x) + d_B(x) )

Is this only one choice or there exists more Urysohn function?

[red yoke]

Composing with any homeomorphism g: [0, 1] → [0, 1] with g(0) = 0, g(1) = 1 gives you another Urysohn function

[prime elbow]

And there are infinitely homomorphism g:[0,1] -> [0,1] such that g(0) =0 and g(1) = 1, g(x) = x^n, right?

[unreal stratus]

yes

Also just like

You can take any continuous function [1/4,3/4] -> [0,1] and then extend it to [0,1] with f(0)=0, f(1)=1

[prime elbow]

Means if we have continuous function f:[1/4, 3/4] -> [0,1] then we have function g:[0,1]->[0,1] such that g(x)= f(x) on [1/4,3/4] and g(0) = 0 and g(1) = 1 and g is continuous

[unreal stratus]

Yes

[quick bough]

is the wedge sum of two suspensions again a suspension? my guess is yes since the suspension functor is a left adjoint to the functor that takes a pointed space to its loop space and the wedge sum is a coproduct in the category of pointed spaces, so the suspension functor preserves that, so suspension(M) \vee suspension(N) = suspension(M \vee N)

[red yoke]

I think that's the reduced suspension in the adjunction

[gaunt linden]

What is "=" there? They are not homeomorphic, though intuitively I think they're homotopy equivalent.

[red yoke]

They should be homeomorphic with reduced suspension

Which is homotopy equivalent to the suspension if the space is nice (?)

[gaunt linden]

Sounds fair.

[unreal stratus]

Yes for both things darnymuckhi said lol

[quick bough]

yea, i mean the reduced suspension

[naive trench]

If $A_j$ is a set defined as the points $x\in\partial K$ (K a relative compact set of an open set) such that $u_j(z)\geq h(x)+\varepsilon$ How can I see that they are a family of closed sets?

[gentle osprey]

Alex

[naive trench]

u_j is upper continuous and h continuous

I can add that u_j is subharmonic and h is harmonic too but I dont think is needed for this

[naive trench]

Nvm I figured out

[earnest ibex]

Suppose $X$ is a 3-dimensional CW complex having no non-trivial element $\alpha \in \pi_1(X)$ satisfying $\alpha \cdot \alpha=0$. I want to show that $H_\left(X \times \mathbb{R} P^2\right) \cong H_(X) \otimes H_*\left(\mathbb{R} P^2\right)$. My idea was to use the universal coefficient theorem, but I'm not sure what the statement about the fundamental group will tell me about the homology of $X$ in general. $H_1(X)$ is the abelianization of $ \pi_1(X)$, but saying that $\pi_1(X)$ doesn't have 2-torsion doesn't necessarily mean that $H_1(X)$ doesn't have 2-torsion either, right?

[gentle osprey]

ImHackingXD

[real granite]

This is from the worked solution to the proof of (ii).

Isn't there a flaw in the reasoning? $U\in \tau'$ does not imply $U\in \tau$, which is what is being said when he writes "the same holds for $U\in\tau \subseteq \tau'$.

The way I proved $\tau\subseteq \tau' \implies (x_n\to_{\tau'} x \implies x_n\to_{\tau} x)$ is by doing contradiction. Assume $\tau'$ is finer than $\tau$ and that $x_n\to_{\tau'} x$ but $x_n \not\to_{\tau} x$. Then $\exists U_x\in\tau\ s.t.\ \forall N\in \bN, n>N \implies x_n \not\in U_x$. But, as $\tau'$ is finer than $\tau$, this neighbourhood is also open wrt $\tau'$, which contradicts convergence in $\tau'$ (as it must hold for all neighbourhoods).

[gentle osprey]

Douglas

[real granite]

Would someone please take a look at this?

Thanks

[shadow charm]

if tau' is finer than tau, then any reasoning you apply to opens of tau' will apply to opens of tau since an open of tau is automatically open in tau'

ie the implication you want is U in tau => U in tau' (which holds here)

[paper wedge]

consider Rx{1,-1}/~ where (x,1) ~ (x,-1) for all x!=0

is this trivially path connected but not arc-connected?

by arc connected i mean any two points can be the endpoints of a topological embedding

[opaque scroll]

Which points are you suggesting can't be endpoints?

[paper wedge]

0 and anything?

like

(0,1) and (0,-1)

[opaque scroll]

Right, yeah (0, 1) and (0,-1) seems like they're not arc-connected

[paper wedge]

yeah thats what i meant it's a problem i saw

and idk how to like

word this

thats why i emphasized "trivialyl"

"trivially"

[opaque scroll]

Well, I think there certainly is something to show, but imagine
f: [0, 1] is an arc connecting the two.

Then f attains a maximum, so by intermediate value theorem is not injective

[paper wedge]

i see

yeah i couldn't think of that i just thought there it is

thank you

[inland laurel]

hi ı working van kampen theorem and ı want gave easy a example and ı tried things and maybe ı being ridiculous.

this is correct a example?

[white oxide]

what is that topology 😭

[hollow geyser]

I'm stuck on my proof. I'm supposed to just use connectedness and intermediate value property, but I don't know how to proceed here without saying that [a, b] is compact (compactness will not come up until next chapter)

[real granite]

Yeah I get this, but just because U in T', doesn't mean U in T (cf. "given U in T', for n sufficiently large, one has x_n in U. In particular the same holds for U in T contained in T' ").

[unreal stratus]

||R has no non-trivial covers|| jk

[unreal stratus]

Hm i would think moore geometrically as you definitely shuldn't need compactness

lemme read more carefully

okay i mean i agree your solution is like the most geometric actually

@hollow geyser but one thing you can do is to use the same argument again on "the other side"

Uhh

wait hm

yeah idk using the fact cts functinos on [a,b] are bounded seems optimal

ah hm

[hollow geyser]

Yeah you're seeing my dilemma now

[unreal stratus]

yeah lol

ig all you need is cts on cloosed bounded interval is bounded which is standard but idk what exactly youo've donoe

like it follws from bolzano-weierstrass quickly

[hollow geyser]

I'll look at this

I don't see how that follows.

Ah the mysteries. But maybe that's why this problem is labeled as "hard"

But this book will usually say when a solution requires something so far out of the chapter. Makes me think there's some obvious thing I am missing

[unreal stratus]

i think i can imagine a nice topology way actually lol

i will think more

well i already mentioned an overkill alg top way lol

but there is a nicer way i can think but i'll lyk

[gaunt linden]

If you have Bolzano-Weierstrass, then assume f is unbounded on [a,b]. Pick x1 in [a,b] such that |f(x1)|>1, x2 in [a,b] such that |f(x2)|>2, and so forth. Take a convergent subsequence, and let c in [a,b] be its limit. But then f(x) cannot have a limit for x->c.

[hollow geyser]

Why can't it have a limit for c?

[gaunt linden]

Suppose such a limit L existed. Then for all n > |L|+1, we have |L-f(xn)| > 1, and (by definition of c) there are such xn arbitrarily close to c.

[hollow geyser]

Also, your sequence seems unbounded. Doesn't BW only apply to bounded sequences?

[gaunt linden]

I said to pick all xn from [a,b] -- that makes them bounded.

[hollow geyser]

Oh I see

[hollow geyser]

Okay I think I have an idea now. And I think I can even do it without BW. Let's see what I come up with

[red yoke]

||I think by IVT the part inside [a, b] should look like a long triangle right||

||You could use something like intersection of decreasing closed intervals to show there's a singularity||

[hollow geyser]

That requires proving that the image is bounded. Which is where I am stuck

[red yoke]

I mean, an infinitely tall triangle

Like 1/x² - 1

[hollow geyser]

oh I see

I'm doing something like that actually

[gaunt linden]

You could also say something like: Suppose f(a)=f(b)=0 and f is not bounded above on [a,b].
Then by the IVT, every y>0 is hit twice between a and b.
Let x_n be the smaller of the two values such that f(x_n)=n.
Prove that x_n is a bounded increasing sequence and therefore converges to some number c.
But (as before) the limit of f(x) for x->c cannot then exist.

[hollow geyser]

All good ideas. Mine's taking a bit of time to write, but I think it's what the book "wants"

[gaunt linden]

[Perhaps a bonus exercise: It is possible for a continuous function R->R to hit every number exactly three times].

[hollow geyser]

How? If $f :\bR\to\bR$ is continuous and surjective, then $\lim_{x\to\pm\infty} f(x)$ must certainly diverge to keep it unbounded, but where can a third divergence happen to guarantee three of every number? If divergence happens at any $c\in\bR$, then how can $f(c)$ be defined?

[gentle osprey]

SWR

[gaunt linden]

I'll wait a bit before I reveal that.

[hollow geyser]

I'll think it over as I finish my other problem

[coarse night]

|| think something similar looking to sin(x)+x works idk||

[gaunt linden]

Yes, but ||you need to scale either the sine or the x such that the local maxima are at the exact same heights as the local minima||.

[coarse night]

Ya i just drew it and gave an approximate form

[gaunt linden]

As written, your function is strictly increasing, so you need some scaling already there. :-)

[coarse night]

Lol sure

This is what i drew lmfaoo

[hollow geyser]

clever

[coarse night]

Never thought of the problem, nice one

[merry geode]

One of things that is mysterious to me is why turning this conceptual graph into a formula is difficult.

[hollow geyser]

okay how about four times now?

[coarse night]

Not my problem, there exists one

[merry geode]

Hmm, like intersecting 2 times is impossible, right

I bet similar happens with 4

Unless you admit incontinuity.

[next crystal]

Its easy to write out if we make it all straight lines in the picture

[merry geode]

Well, I am talking about Smooth

[next crystal]

Oh nvm then

[merry geode]

But I see, you can basically make it smooth I think, using partition of unity.

[merry geode]

Ah that's exactly this.

After learning Morse theory, I think I can see how this problem is Morse theoretic stuff.

[hollow geyser]

,w plot y=sin(x)+2x/sqrt(4+9pi^2)

[gentle osprey]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by spelwerdzrite

[hollow geyser]

ahhh

so close

[merry geode]

Hint: look at critical points

[hollow geyser]

I did, but my algebra pooped somewhere I guess

[gaunt linden]

I don't think there's a nice closed expression for the just right factor.

[hollow geyser]

idk where the error is

[merry geode]

Ah, sorry I thought you were going for 4 point thing.

3 point.. idk

[prime elbow]

Let ${ C_j: j\in J}$ be a family of closed compact subsets of a Topological space $(X,T)$. I want to show that $\bigcap_{j\in J} C_j$ is compact.

We know that a closed set of compact space is compact.

So, $\bigcap_{j\in J} C_j$ is a closed subset of $C_j$ so it will be compact.

Is it correct?

[gentle osprey]

Notknow🙇

[hollow geyser]

Yeah. Suspiciously simple

[gaunt linden]

The solutions of cos x = -B are not equally spaced with a distance of pi between each -- the ones on the upslope of the cosine curve are offset from the ones on the downslope.

[hollow geyser]

ah heck you right

[gaunt linden]

By symmetry you can just look for f(x2) = 0, but things still won't cancel out neatly enough to solve it to closed form.

[merry geode]

Better to connect lines, I guess

[gaunt linden]

It's neat to know the solution can be real analytic, though.

[merry geode]

You can kind of smoothe the lines using partition of unity, no?

[gaunt linden]

That can make it C^infty, but not analytic.

[merry geode]

Ah, brainfart. Sorry

[prime elbow]

Okay, thank you

[hollow geyser]

ugh

[red yoke]

It's also possible on a closed interval [a, b]

There is a continuous function on a closed interval such that all fibers have an even number of points

[worthy nebula]

If I have a homomorphism $\psi: G \to D$ with $N \subset \mathrm{ker} , \psi$, N being a normal subgroup of G, I can say I have an induced homomorphism $\varphi: G/N \to D$, right?