point-set-topology

[opaque zodiac]

and where does my boi Borel come into it?

oh

no but how did Borel get into this

Mathematicians are only allowed to do one thing

[dim perch]

So I'm not too familiar with the actual history, but the "obvious" etymology of Borel equivariant cohomology is that it's the (nonequivariant) cohomology of the "Borel" construction

i dont actually know if borel is responsible for the borel construction

sounds wild to me

[molten dirge]

it was a real hootenanny shindig hullabaloo

[molten dirge]

Wait which Borel are we talking about

[opaque zodiac]

||this one||

[white oxide]

||Great borel algebras and also the best suitor for turning your homology into cohomology too||

[white oxide]

What are some examples of it being used in non-derived/spectral settings of AG?

[cedar pebble]

one that comes to mind is that there are analogues of all these complex oriented cohomology theories in the setting of motivic homotopy theory, and sometimes these are useful things to study

e.g. the complex K-theory spectrum KU gets replaced by the algebraic K-theory spectrum KGL, the complex cobordism spectrum MU gets replaced by the algebraic cobordism spectrum MGL, etc

[white oxide]

Okay, not really avoiding using the two as long as it ties back to "classically" motivated questions

Serre's intersection formula comes to mind

[cedar pebble]

well sure but that's not so much related to this spectral stuff as it is related to derived AG in a more basic sense

[white oxide]

I see. What motivated you to familiarize yourself with the topic?

[cedar pebble]

I mean I like homotopy theory and I like motivic stuff

I should say a very nice application of the algebraic cobordism spectrum is you can characterize it and construct it in terms of double point degenerations, and Levine-Pandharipande used this to prove a bunch of conjectures on degree 0 Donandson-Thomas theory for threefolds

so these fancy constructions do have concrete applications in AG

[white oxide]

Aren't there applications of the aforementioned stuff to motivic stuff?

[cedar pebble]

there are yeah

one route to studying motives is through motivic stable homotopy theory

in some sense motives are capturing everything at height 0 to use the chromatic analogy

[white oxide]

Dont the cohomology theories in AG care about "higher height data"?

[cedar pebble]

not the usual cohomology theories, but you can construct various cohomology theories that see things (mostly torsion phenomenon) of height >0

there is an interesting notion introduced recently by Vishnik of "torsion motives"

[merry geode]

Are you familiar with the spectrum stuff?

[cedar pebble]

(these are Chow motives whose identity morphism is killed by a natural number)

you can study these in a chromatic sort of way

[merry geode]

I thought you basically knew everything, including (kind of) side subjects

[cedar pebble]

I certainly do not know everything!

I used to be a lot more interested in stable homotopy theory stuff before I got more into AG/arithmetic

[merry geode]

Ahhhh

[cedar pebble]

I am kind of rusty on that stuff but I remember parts of it

[merry geode]

Well that's the impression I got (you knowing every math)

[cedar pebble]

like I know the basics but can't really keep up with actual working AT people who do this stuff

[hearty condor]

Imagine knowing everything about topology.

You'd become the second coming of J. Munkres, and the guy isn't even dead.

Jokes aside that's probably legit impossible to do.

Even for the most avid math goblin that lives off of cave moss and math articles on esoteric math subjects.

[white oxide]

Are these mostly of formal interest or do they arrive as more natural research problems?

[cedar pebble]

I think there are quite a few natural motivations for these things and they often have concrete applications

another application worth mentioning is that the current best computations of stable homotopy groups of spheres use the motivic stable homotopy category as a tool

[merry geode]

Motivic stable homotopy

[cedar pebble]

yes it's a fancier category but a natural one to consider for a lot of applications

[merry geode]

Feels like Motivic stuff is infecting everything

[white oxide]

"Why was 6 afraid of 7?"

[opaque scroll]

And here I am thinking it's because 7 was a registered sixoffender

[dim perch]

I wish I understood enough motivic homotopy to understand its burgeoning significance

ofc cellular C-motivic spectra are starting to become fundamental in stable homotopy because of the connection with synthetic spectra/the asseq

and R-motivic spectra betti realize to a large portion of the computational C2-equivariant results that have come out lately (dont think about odd primes youll get sad)

but I've sensed a lot of hype from my algebraic number theorist friends lately about non-A1 motivic homotopy and its eventual role in NT and AG, which seems quite interesting

[gentle osprey]

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

[rancid umbra]

whey did you wrap everything in $$?

Let $A$ be a closed subspace of a normal space $X$, and let $f: A \rightarrow Y$ be a continuous map. The adjunction space $Z$ of $X$ to $Y$ by $f$ is defined as the quotient space of the disjoint union of $X$ and $Y$ by the identification $x \in A = f(x) \in Y$ for all $x \in A$.

We want to prove that if $X$ and $Y$ are normal spaces, then $Z$ is also a normal space.

[gentle osprey]

c squared

[surreal island]

Yee

How can i prove it without theorem of Tietze ?

[white oxide]

Well, the disjoint union of X and Y is normal, and pre-images of disjoint closed sets are disjoint and closed.

What fails when you try to stitch the open sets separating both in the pre-image into open sets in Z?

[real granite]

When it says "$f_n\to f$ in $B(S)$", I assume "in $B(S)$" really means "with respect to the metric associated with $B(S)$"?

[gentle osprey]

Douglas

[real granite]

f is supposed to be an element of B(S) so I doubt "in" means "element of"

[wispy veldt]

yes

small remark: id avoid using the notation L^infty for this space, since it means a slightly different notion of bounded almost everywhere functions.

[real granite]

I assume the L thing is Lebesgue integrable?

[wispy veldt]

yeah

[real granite]

Yeah I haven't done measure theory and I don't think it matters for what I've done so far (or will be doing next year), so I will avoid using measure theoretic notation

[real granite]

follow up regarding the proof itself, when proving the converse it says "one has d(f_n, f)≤e<2e". How does this follow from the supposition that f_n converges uniformly to f?

I realise uniform convergence means that |f_n (x)-f(x)|<e for all x in S, but |f_n(x)-f(x)| might not actually attain the value sup|f_n(x)-f(x)|, e.g. if S is an open interval then there might not be a maximum value on S of the modulus, so we can't assume there is an a in S for which |f_n(a) -f(a)|=sup|f(x)-f_n(x)|<e

I hope that makes sense

[wispy veldt]

how does not attaining the sup change the proof tho? you still get

d(f,fn)

[real granite]

How else would it follow?

What I was thinking was that, naively, you might say "because $|f(x)-f_n(x)|<\epsilon,\ \forall x\in S$, then in particular we have $|f(x_0)-f_n (x_0)|<\epsilon$ for $x_0\in S\ s.t.\ |f(x_0)-f_n (x_0)|=\sup|f(x)-f_n (x)|$."

However, there is no requirement for such $x_0\in S$ to exist.

[gentle osprey]

Douglas

[wispy veldt]

all you need to use is $$|f(x)|<\varepsilon, \forall x\in S \implies sup_{x \in S} |f(x)| \leq \varepsilon$$

[gentle osprey]

James Banach

[wispy veldt]

regardless of whether f attains this sup

it does however need to exist ofcourse (again regardless of whether f attains it)

thats where B(S) comes into play

[real granite]

"it" being what?

[wispy veldt]

the supremum

[real granite]

Oh right yeah

[real granite]

I'd have to have a think about how to prove this but it makes sense intuitively (thinking about graphs)

[jaunty summit]

is there an example of a group with the following 4 properties

1. finitely presented
2. incomputable word problem
3. computable homology groups
4. infinite cohomological dimension

[umbral panther]

In the typical example it is hard to distinguish the group from the zero group, and thus the homology is all zero, in particular computable

You can force infinite dimension by multiplying by Z/p

[fierce lily]

Can someone help me check if my solution is correct?

[hexed steppe]

what is alpha

[fierce lily]

alpha refers to the character look like A in question

[lethal oxide]

if X a topology and A basis in X, mean A subset of X and every x∈A mean every x∈X because A subset of X and every point x on A x∈X

[real granite]

How are we getting the highlighted bit?

If there is no $d$-ball inside the $\rho$-ball, that just means there must be at least one point in $B_s ^d (x)$ that is not in $B_r ^{\rho}(x)$, but if this happens finite times and the final time is $x_m$, then you can still have $x_n$ being inside both balls for all sufficiently large $n>m$.

[gentle osprey]

Douglas

[rancid umbra]

i would write this out more carefuly. its difficult to understand what you are saying

[real granite]

Maybe I've expressed myself very poorly here, but hopefully you get the point I'm trying to make

[lethal oxide]

yes but i think he said X and S topology and A basis in X and S, let Y topology on A. X intersection S= Y

[rancid umbra]

the sequence of (sn,d) balls cannot be contained in the (r,rho) ball

[rancid umbra]

the "only finitely many" thing you're worried about can't happen

you have infinitely many radii sn going to zero

so you can always choose xn in the (sn,d) ball and not in the (r,rho) ball

[real granite]

Oh hang on... I think my concern is resolved by the fact that all B^d _{s_n} are not contained in the rho-ball, so x_1 can't be in the rho ball because of the d-ball with radius s_1,... and x_n can't be in the rho ball because of the d-ball with radius s_n, so you just end up with every x_n being outside of the rho ball

I was not thinking of the d ball as "contracting" as s_n tends to zero

But it makes sense now

[fierce lily]

I tried to prove that all union of basis elements A equals intersection of topology D.

so since A(the character look like A) is in intersection of all topology D, it implies that all union of elements A is contained in D

[rancid umbra]

call it script A or use latex to type this up so we don't have misunderstandings

[fierce lily]

and the other direction is to assume every x in D, it means x contained in every topology T. since there is a topology T1 generated by basis script A, then x is in one A, which means x is in all unions of A

[rancid umbra]

what is x?

[lethal oxide]

x a point on A and a point on X

[rancid umbra]

that is not what you need to show.

oh what

no, i was asking karlking.

[lethal oxide]

i think by definiton

[rancid umbra]

okay, x is in D

[lethal oxide]

if x not a point on A and x a point on X it is not logic

[fierce lily]

so is it correct to show all unions of A equals D?

[lethal oxide]

i think no

but ı not sure

[rancid umbra]

this part is correct. if $\mathcal{T}$ is any topology containing $\mathcal{A}$, then it will contain any open subset generated by $\mathcal{A}$.

for the opposite inclusion, if you look at what the intersection of all topologies containing $\mathcal{A}$ is: $$\bigcap_{\substack{\mathcal{T}\text{ a top. on $X$}\\mathcal{T}\supseteq\mathcal{A}}}\mathcal{T},$$and you know that the topology generated by $\mathcal{A}$ contains $\mathcal{A}$, so...

[lethal oxide]

D a topology isnt it?

[gentle osprey]

c squared

[lethal oxide]

if D a topology and A basis of D i think no it is not possible

[lethal oxide]

but if D a set and T a topology on D and A basis of T yes it is possible

[lethal oxide]

@fierce lily

[rancid umbra]

this actually leads to a one line proof

[real granite]

How to build intuition for fine/coarse?

I get that we're saying $x_n \to_{d_{\infty}} x \implies x_n \to_{d_1} x$ and so we say the topology generated by $d_{\infty}$ is finer, but I don't really have a good mental picture of what's going on.

I know that the discrete topology, generated by the discrete metric (under which only eventually-constant sequences converge), is the finest topology, so in my head finer=fewer sequences converge, but I'm not really sure how good this is

[gentle osprey]

Douglas

[hexed steppe]

is that really intuition or just

a rephrasing of the definition

[real granite]

That's my point

[hexed steppe]

if it feels intuitive to you then certainly it is fine.. as it is equivalent

[real granite]

Hence "I don't really have a good mental picture"

[hexed steppe]

not sure what kind of picture you are hoping for

[real granite]

Idk. How do other people understand this stuff intuitively?

[gritty widget]

I usually visualize finer topologies like a focused lense, with coarser topologies looking blurrier relative to them

[hexed steppe]

i mean you can think of the collection of open sets as like a measuring device

yeah

you can pretend that the topology allows to distinguish points up to open sets

like if i give you points x,y, and i tell you the list of open sets which contain x but not y, and vice versa

[gritty widget]

As an example:

[hexed steppe]

then depending on how fine the topology is, you may or may not be able to determine whether x=y

[gritty widget]

Consider a descending chain of topologies $T_1 \supset T_2 \supset \cdots \supset T_n$

If two points are topologically distinguishable in T_j, then they are in all T_i with i < j

However, points may be distinguishable in T_i, but not in T_j, with j > i

So, as you get closer to T_1, i.e. as you get finer and finer topologies, you can see the points more clearly

[gentle osprey]

eigenpuppet

[real granite]

So how does that align with the case of metrizable topologies?

I think my problem is that if we're saying $x_n \to_{d_{\infty}} x \implies x_n \to_{d_1} x$ means $d_\infty$ is finer, then that's equivalent to $B_r ^{d_\infty} (x) \subseteq B_r ^{d_1}(x)$, and so I think subset=finer, but when you look at the topologies as $\tau_1\subseteq \tau_{\infty}$, the finer topology is actually a superset

[gentle osprey]

Douglas

[hexed steppe]

i strongly doubt that implication is true

[real granite]

Which one?

[hexed steppe]

the only one

in your message

sprry

i meant equivalence

[gritty widget]

You can think of the ball inclusion as the sets being “tighter.” However, the topology itself is a superset, because it contains all of the old open sets, plus some new ones

[hexed steppe]

the ball inclusion is way stronger

than one topology being a subset of the other

[real granite]

Yes you're right I think

I'm forgetting "for sufficiently large n"

There is no guarantee that they will be subsets

Just because d_inf is finer

[hexed steppe]

what

there is no dependence on n in what you wrote

you said one ball around x is contained in the other

for a fixed radius

[real granite]

There is.

x_n converging to x under d_inf means that for all r>0, for sufficiently large n, x_n will be in the metric-d_inf ball of radius r

But what this means is that for small values of n, x_n can lie outside that ball

[hexed steppe]

yes that is the definition of convergence

but again

you are talking about comparing balls in different metrics

there is no reason why one would be contained in the other

[fierce lily]

so do you mean my forward proof is correct, but the opposite direction only needs to argue that intersection of topology is a topology generated by A because it contains A?

[hexed steppe]

in general

B^dinfty(x) subset B^d1(x) is not a statement which has any dependence on n

[real granite]

Yes... I realise that now. I'm agreeing with you that what I claimed was equivalent was not in fact equivalent

[hexed steppe]

ok

but im saying that

there is not really a way to salvage it as far as i can tell

which you seemed to be asserting

if i misunderstood then sorry

[real granite]

I'm not, I'm just clarifying where I've made mistakes in my reasoning

[hexed steppe]

i mean suppose xn was contained in B_r(x) for all n

[real granite]

The reason I said this was because I thought $x_n$ converges under $d_{\infty}$, and this implies convergence under $d_1$, so $x_n \in B_r ^{d_{\infty}} (x) \implies x_n \in B_r ^{d_1} (x)$ and so the former is a subset of the latter, but this is wrong

[gentle osprey]

Douglas

[hexed steppe]

yeah

the balls have many points

which are not xn’s

[real granite]

Ye

Is it normal to do metric spaces before not-neccesarily-metrizable topological spaces?

[hexed steppe]

however based on your notation i am guessing maybe you are thinking of the ell^1 and ell^infty distance on R^d

[hexed steppe]

yes

[rancid umbra]

yes

[hexed steppe]

balls in the ell^p metrics actually can be compared

but thats a very special situation

[real granite]

Idk what you mean by ell^

[hexed steppe]

why did you write infty and 1

as decorations

[slender glen]

Very much so, metric spaces are much nicer and more similar to other stuff you've seen

[real granite]

These were the examples the lecturer was using

[hexed steppe]

yeah these are called L^p distances (or ell^p when its a sequence space instead of a function space, imo a pointless convention)

for p = 1 and infty

anyway

note d_infty >= d_1

as i am sure you have seen already

[real granite]

yes

[hexed steppe]

so you really do have the inclusion of balls

[real granite]

but yh in general it doesnt work like that

[hexed steppe]

this is extremely strong

[real granite]

Do you have any good topology video recommendations? My intuition for this is pretty... lacking

And I think just taking notes from a textbook really isn't enough to build it

textbook/lecture notes

Point set/general topology

[hexed steppe]

i dont

[real granite]

@gritty widget what about you?

[plush folio]

This one is pretty decent: https://m.youtube.com/watch?v=vv3JNSPKeEU&list=PLd8NbPjkXPliJunBhtDNMuFsnZPeHpm-0

[fierce lily]

is {0} a basis element of R_k? R_k has the basis element of (a,b)-K. K={1/n| n is in Z}

[gritty widget]

I don’t really have any good topology resources, sorry

I learned the point-set I know by reading Kolmogorov & Fomin’s chapter on it and some papers my prof sent me

[rancid umbra]

why do some authors require that the sheets of a covering map be connected and some don't?

are those formulations equivalent?

[lethal oxide]

i think yes it is possıble

[umbral panther]

No, they aren’t equivalent. Consider the case where the base is just a point. Worse, the case where the base is 2 points

If the base is connected and has a chosen base point, general covering spaces are sets with an action of the fundamental group. Connected covering spaces are sets with a transitive action of the covering group. These are related to sub groups

[rancid umbra]

connected as in the sheets are connnected? or the covering space is connected?

[umbral panther]

I meant the covering space is connected. I’m not sure what sheets are, other than connected components of the covering space

[rancid umbra]

for example, lee defines covering maps like this:

but munkres defines them like this:

[umbral panther]

This definition fails unless U is connected. If the base is locally connected and you only look at connected opens, then the sheets will be connected

[rancid umbra]

which one?

[umbral panther]

Lee’s definition only makes sense for locally connected spaces

[rancid umbra]

anyways

[umbral panther]

Munkres’s definition is better. But Lee’s definition works for manifolds, the topic of his book, which are locally connected

[rancid umbra]

okay cool

thanks

[unreal stratus]

Oh yeah that's odd it mentions connected opens (to me)

[umbral panther]

I think it could be useful, but putting it in the definition is weird

[unreal stratus]

that's what i mean yeah

[quick bough]

im having a hard time trying to find a certain counterexample for the naturality of the splititng in the universal coefficient theorem, i want to find a chain map from a chain complex C to itself that doesnt produce a natural spliting, for this, consider the following in Z/2 coefficients

you should get a commutative diagram like this when applying naturality of the UCT

now, my question is: what is M_*

my guess is that its just M again, but im having a hard time trying to understand maps induced in homology

[merry geode]

Maps on cycles will induce maps on homology, so yeah.

Pedantically, it is not exactly M, considering it has different domains

[quick bough]

yeah, i see

well, what i’m saying is it can be represented by M

but yes, got it

[merry geode]

Indeed.

[fierce lily]

Can someone explain why it generates topology different from R_L?

[prime elbow]

Take [√2, 2), is it open in this topology?

|| Let this is open in given topology, then by definition of open set, for √2 there exist a basis element B such that √2 in B and B is a subset of [√2,2).

Let B = [a,b) where a and b are rational numbers.
Then a< √2<b imply that a<(a+√2)/2 < √2 lie in B.
But B is contained in [√2,b) so its contradiction.

Hence lower limit topology is strictly finer than given topology. ||

[prime elbow]

Let f : X-> Y be continuous. If Z is a subspace of Y containing the image set f(X), then the function g: X-> Z obtained by restricting the range of f is continuous.

Reason:
Let S be open set in Z since Z is a subspace topology of Y so S is the intersection of Z and U, where U is open set in Y.

Now g^(-1) ( S) is intersection g^(-1) (Z) and g^(-1) (U).
And g = f, right?
Since Z contains f(X) and f is continuous mapping.

Hence, g^(-1)(S) = f^(-1)(U). Thus g is continuous.

Is it correct?

[gentle girder]

try thinking of a set that is open in the lower limit topology that isn't open in this one

[quartz horizon]

I think so? This is essentially the universal property of the subspace topology

[prime elbow]

What does it mean by universal property?

[quartz horizon]

So a universal property, in short, tells you what a mathematical object does, in terms of how to use it

In this case, what the subspace topology on Z does is let you convert continuous functions into Y contained in Z, into just continuous functions into Z

It’s much like the set-theory subset

What a subset Z of Y does is let you convert functions into Y contained in Z, to just functions into Z

And vice-versa

In topology though it’s important that you put the correct topology on Z, in this case the subspace one

[prime elbow]

Good point

[quartz horizon]

Otherwise there’d no longer be a one-to-one correspondence between continuous functions into Z and continuous functions into Y that are contained in Z

That’s what your preimage check is for

This is nice to illustrate with a commutative diagram but

I’m assuming you don’t know what those are

[prime elbow]

Yes I am not sure about the Commutative diagram in topology

I learnt in abstract algebra

[quartz horizon]

I mean I could show you if you want

[prime elbow]

Yes

[quartz horizon]

Ok so like

We can start with the set theory one

[quartz horizon]

So suppose we're in set theory

Suppose Z is a subset of Y

[prime elbow]

Okay

[quartz horizon]

Then there's a (natural) correspondence between "functions into Y contained in Z" and "functions into Z"

Here I've drawn functions as arrows between the domain X and codomain

Note that $f$ and $\tilde f$ aren't equal - they don't have the same domain and codomain, for one

[gentle osprey]

Pseudonium

[quartz horizon]

Nonetheless, we can relate them

[prime elbow]

Yes

[quartz horizon]

Specifically, if Z is a subset of Y, there's an inclusion function from Z into Y, which we'll denote $\iota$

[gentle osprey]

Pseudonium

[prime elbow]

What do you mean by functions f into Y contained in Z but we assume Z is a subset of Y

[prime elbow]

Yes

[quartz horizon]

So what I mean is a function with codomain Y whose image is a subset of Z

[prime elbow]

Okay

[quartz horizon]

The outputs of the function are contained in Z

Does that make sense?

[prime elbow]

Yes

[quartz horizon]

E.g. if Y was the natural numbers, Z was the even numbers

[prime elbow]

Yes got it

[quartz horizon]

Then I'm referring to "functions whose output is always even"

[quartz horizon]

So we can relate f and tilde f with a commutative diagram

We can then phrase the universal property as follows

Given any f : X -> Y such that Im(f) is a subset of Z, there's a unique f tilde : X -> Z that makes this diagram commute

Conversely, given f tilde, you can recover f as f = iota o f tilde

[prime elbow]

f = tf' ?

[quartz horizon]

Yeah

$f = \iota \circ \tilde f$

[gentle osprey]

Pseudonium

[prime elbow]

Okay

[quartz horizon]

So in other words, $f(x) = (\iota \circ \tilde f)(x)$

[gentle osprey]

Pseudonium

[quartz horizon]

Which means $f(x) = \iota(\tilde f(x)) = \tilde f(x)$

[gentle osprey]

Pseudonium

[quartz horizon]

All we've done is change the codomain

[prime elbow]

Yes

So how does this Commutative diagram help us? To find f tilde ?

[quartz horizon]

What the commutative diagram does is relate f and f tilde

You'd still have to prove that the subset satisfies the universal property

[prime elbow]

This one

[quartz horizon]

Yes

[quartz horizon]

Which for set theory involves going into the weeds of what a function is

In terms of a function being a collection of ordered pairs (x, f(x))

[prime elbow]

Yes

[quartz horizon]

So that's the version for set theory

We can reuse our diagrams when talking about topology

[prime elbow]

With additional information?

Continuous function

[quartz horizon]

So, this is what the subspace topology does - continuous functions into Y whose image is contained in Z are (naturally) identified with continuous functions into Z

[quartz horizon]

And specifically, this is how they're related

So you should check that $\iota : Z \to Y$ is indeed continuous

[gentle osprey]

Pseudonium

[quartz horizon]

And that if $f$ is continuous, so is $\tilde f$

[gentle osprey]

Pseudonium

[prime elbow]

Yes

[quartz horizon]

Both of these involve going into the weeds of what the subspace topology is

[prime elbow]

So the diagram tells us how they are related to each other

[quartz horizon]

Yes

If you knew more category theory then the diagram gives you a way to prove it too

Cause the preimage is just a pullback

[prime elbow]

No

[prime elbow]

Means?

[quartz horizon]

Uh

It's probably too much to explain for someone who doesn't know category theory..

Basically

For topology you have to talk about preimages

[prime elbow]

No problem

[quartz horizon]

And there's a way to talk about preimages in category theory

As a universal property

[prime elbow]

Yes

[quartz horizon]

But it's a bit more complicated than the universal property for subsets

And I feel like just showing you won't really help

[prime elbow]

Okay

[merry geode]

X -- f --> Y
| |
f^-1(V) -> V

[quartz horizon]

Yeah that

[merry geode]

Whwn you have inclusion V -> Y, there is inclusion f^-1(V) -> X - basically subset

[quartz horizon]

Yeah

The proof in mind does have to use the universal property of the pullback though

Which involves cones and stuff

[prime elbow]

Inclusion implies f^(-1)(V) is subset of X , identity mapping?

[merry geode]

You can say, f^-1(V) is maximal among the sets which "satisfy" this diagram

[merry geode]

Yes.

[prime elbow]

Then what if there is element y in V such that there is no x such that f(x) = y ?

[merry geode]

Then such element y has no "contribution" to the preimage f^-1(V).

Do you know classical definition of preimage?

[prime elbow]

Yes I see my mistake

[merry geode]

Ah, I did not mean you made a mistake. Sorry if I phrased it like that

It's easy to be confused!

[prime elbow]

No no, actually thanks

[merry geode]

No problem!

[prime elbow]

I know it is the closure of {1/n | n in N } so it is closed but how does its complement look like ?

[white oxide]

Wdym

A contains all 1/n and 0. The complement everything else

[prime elbow]

Means?

[white oxide]

Think of a real number

Is it 0? Is it of the form 1/n? If you said no to both, then it lies in the complement of A.

Like, do you want to know how it decomposes as an union of disjoint open intervals?

[prime elbow]

Yes

[unreal stratus]

well it decomposes as like

union of (1/n, 1/(n-1)) for n >= 2

[prime elbow]

Sorry I don't understand this one

[unreal stratus]

as well as (-oo,0)

Like all i've left out there is the points 1/n and 0

[prime elbow]

And also (1, oo) ?

[unreal stratus]

actually what I said was wrong anyway

[unreal stratus]

Yeah

You can consider like, map x -> xsin(π/x) i guess lol

[prime elbow]

Got it, thank you

[unreal stratus]

and then the roots are 0 and 1/n for integer n

lol

[white oxide]

I mean, you hopefully know how the complement of 0 or a single 1/n looks like

[prime elbow]

Yes

[unreal stratus]

But that is a meme way to do it

[white oxide]

And how intersections of open intervals look like

From there you can justify it

[prime elbow]

Not sure

[white oxide]

Well, work it out from definition then. As a treat exercise to the reader

An open interval (a,b) is all elements bigger than a and smaller than b

[red yoke]

Can you draw the closed set

[prime elbow]

Finite intersections will be empty or open interval I am not sure about the infinite intersection of open intervals

[prime elbow]

I don't understand what you mean ? In the number line we can mention which are closed set

[red yoke]

Can you draw it visually?

Like plot the points in the closed set

On paper

[white oxide]

In general it's complicated. But in this case you can notice that any open interval stops getting smaller after finitely many steps

[prime elbow]

Yes because now I know it is union of (-∞,0), (1/n, 1/(n-1) ) and (1,∞) where n≥2

[white oxide]

Or like, draw it out on the number line

[merry geode]

Golden

[prime elbow]

According to this definition, is empty set contains in basis ?

I think no

Because it has one exercise which Topology has exactly one base?

I think indiscrete topology with non-empty set

[gaunt linden]

The empty set of points is the union of the empty set of basis sets.

[prime elbow]

Sorry, I don't understand

[gaunt linden]

The union of no sets is the empty set, so you don't need to list Ø explicitly in the basis for a topology.

But it's allowed to have superfluous elements in a basis (usually that cannot even be avoided) so there's no topological space that has exactly one basis.

[prime elbow]

But If I don't take empty set in basis then indiscrete topology has exactly one basis ?

[gaunt linden]

If you only count one of the two possible bases, then yes, there's exacly one of the one you count.

[prime elbow]

Okay, thank you

[unreal stratus]

This reminds me of a fun thing

How may open covers of the empty set are there

[red yoke]

Two

[quick bough]

i’m trying to understand the answer to this question in https://mathoverflow.net/questions/59938/examples-for-non-naturality-of-universal-coefficients-theorem:

1. what is the map given by f? i understand that you can collapse the 1-skeleton of RP^2 to obtain S^2, but then they include the bottom cell into the suspension of RP^2, what exactly does that mean?
2. i don’t understand the whole part where they show the self map on X induces identity in integral homology but not in Z/2-coefficients

i’m not too familiar with suspension, which might be the problem

[prime elbow]

It says that an open set is a union of the basis element but it is not clear that it implies converse

[alpine nest]

What converse?

That every union of basis elements is an open set?

[prime elbow]

Yes

[alpine nest]

Basis elements are open sets

And every union of open sets is open

[prime elbow]

Yes

But it is not clear from definition

[alpine nest]

That's literally part of the definition of open sets

[prime elbow]

It says each non-empty open set is union of basis elements but it does not say that union of basis elements is open.

Maybe I am too bad in English

[alpine nest]

And the definition of basis states that it's a collection of open sets

[prime elbow]

I am talking about this definition

[alpine nest]

Yes, and it says "A collection Sigma of open sets is the base..."

[red yoke]

Any union of open sets is an open set by the definition of topology

[prime elbow]

Yes

[alpine nest]

Yes, and the definition of basis presupposes that you already have a topology

And you pick the basis elements out of the open sets in that topology

[red yoke]

And all elements of the base are open

[prime elbow]

Got it

[prime elbow]

I missed that point I thought they assumed this not necessarily open

[red yoke]

If you've studied abstract algebra, you might've come across the notion of "generation"

[prime elbow]

Yes

[red yoke]

"this subset of vectors spans this space"
"this subset of elements generates the group under addition"

Etc

[prime elbow]

Yes

[red yoke]

A base is just that

A subset of the topology that generates the topology under unions

[prime elbow]

Oh that's a good point

[red yoke]

You might also encounter a "subbase", which is a subset of the topology that generates the topology under unions and finite intersections

[prime elbow]

Yes

[silver ridge]

How did we arrive at the axioms for a topological space? Looking at them, it seems hard to believe that they encode anything. Does anyone have any motivation?

[red yoke]

#advanced-lounge message

[silver ridge]

that's very clever

[alpine nest]

Yeah, I think the neighborhood idea is the best motivation.

[silver ridge]

My mind is kinda blown

[alpine nest]

Also note that there are lots of competing (equivalent) definitions of topological space

[silver ridge]

Mhm

[unreal stratus]

and only one that anyone uses

[slender glen]

there's at least two

defining a topology by its open sets, and defining a topology by its closed sets

[unreal stratus]

i joke though because presumably in pure poinset stuff you will use other things

i mean sure but those are kinda trivially equivalent

but yeah fair

[alpine nest]

The Kuratowski axioms

[silver ridge]

poland mentioned

[slender glen]

there are a bunch of equivalent definitions (and it's insightful to be able to jump between them), but I think you just have to get used what type of information a topology encodes, they're a pretty fundamental object.

[unreal stratus]

what i had in mind was more the inequivalent definitions

[alpine nest]

As a true patriot I use nothing else 🇵🇱

[unreal stratus]

as in like other models for spaces lol

but that is not helpful here

[wispy veldt]

only useful topology comes from open balls of a metric

[alpine nest]

This but unironically

[silver ridge]

algebraic geometers in shambles

[unreal stratus]

Sad

[red yoke]

Grothendieck topology

[unreal stratus]

Now try gluing things

[alpine nest]

Even the nonmetrizable ones are often constructed using systems of neighborhoods, so basically still balls

[slender glen]

based, reject non-metrizable topologies

[unreal stratus]

Oh I didn't even have sites in mind, but based

[slender glen]

the spectrum of a ring tho

[nimble pebble]

what even is a zariski

[unreal stratus]

there are two zariski topologies tbf

lol

[nimble pebble]

FINE what even are zariskis

[unreal stratus]

idk if that is a serious question or a meme question

[red yoke]

They sound like a brand of biscuits

[alpine nest]

I'm so happy I've never needed to know this

[unreal stratus]

Sad!

[silver ridge]

its a meme question dont bother answering. bloe is a known troll

[unreal stratus]

I have alg geo exam in 2 days lol so i do know.

[nimble pebble]

i culd pull a wrong answers only but i feel a bit nonengaging

[nimble pebble]

bet ure trembling in ur boots that ive finally decided to start chatting down south

[silver ridge]

you need to leave.

[unreal stratus]

lol

[alpine nest]

You can genuinely get quite a lot of mathematics done without straying far beyond the realm of metric topology.

[nimble pebble]

im allowed to be here cause i got some awesome topological excitations i want to talk about

[red yoke]

Topological excitations

[kindred cairn]

are you still a student ?

[unreal stratus]

Masters yes

[kindred cairn]

cool

impressive

I'm also a masters student but I don't think I have even 1/10th of your knowledge

[nimble pebble]

u can classify spin states in solids depending on which stable states u can land on by perturbations

so its kinda natural to think of them with topological info

[red yoke]

Is this be quantum material physics

[prime elbow]

i thought you are PhD student

[nimble pebble]

usually called condensed matter but yeams

spin fields are maps of the form S^2 -> S^2 so u get states that go by pi_2(S^2) = Z

same goes for all sorts of systems

[unreal stratus]

thanks

i'm sure u know a lot smh

[red yoke]

I should start my qm arc sometime

[kindred cairn]

I follow the "bare minimum" philosophy, so always trying to do the least possible in terms of effort and studying

[red yoke]

Be in tune with the Tao

[kindred cairn]

is he also like this ?

[red yoke]

Take the path of non-action

[slender glen]

what is the lagrangian for taoism

[red yoke]

The Tao that can be spoken is not the eternal Tao

[nimble pebble]

im reading off a book but the symmetry structure of certain materials give smth called their order parameter

superconductors have U(1), spin systems S^2 as above, liquid crystals with an axial symmetry RP^2 and He-3 S^2xSO(3) which is kinda neat

[slender glen]

lie groups my beloved

[nimble pebble]

higher homotopy groups matter!

[prime elbow]

any hint to find the closed ball which is open with respect to the metric topology

[opaque scroll]

Think about a one point space for example

[prime elbow]

One point space means X has only one point, then X is open and closed ball

[last gale]

Could anyone tell me what the difference between Hom F(Hn(C* ); F) and Hn(C*)'s dual is

[fathom steeple]

is this proof that if f is a continuous bijection from X to Y such that X is compact, Y hausdorff, then f is a homeormophism correct? it is easier than the sol i found online

Let V be closed in X. Then V is compact in X hence f(V) is compact in Y. Since Y is hausdroff f(V) is closed as well, qed.

[red yoke]

Yes

[unreal stratus]

They are the same (by definition)

i guess there is a small : there to indicate that that is the case

[last gale]

Oh v.s. denotes vector space

[unreal stratus]

Yeah

Ah i see the issue now lol

[last gale]

I thought it meant versus to ask me to contemplate the difference

[unreal stratus]

that is unfortunate

[last gale]

Lmao thank you so much

[unreal stratus]

lol

Np!

[umbral panther]

The difference is whether you take the dual before taking homology or after. If the ground ring is not a field, they might not be the same

[prime elbow]

If A is bounded subset of metric space X, does it mean for all x in X, a in A d(a,x) < M?

[slender glen]

no, it just means that for all a,b in A, d(a,b) < M for some M

[slender glen]

consider that this definition would make (0,1) an unbounded subset of R

[fierce lily]

if proving subspace topology. If there is an open set U in R * R such that U intersect Y={1/2} * (1/2, 1], then U should be {1/2} * (1/2, b) as b>1. Then we get an open interval U=( 1/2 * 1/2, 1/2 * b), is that argument correct?

[prime elbow]

Okay

[opaque scroll]

Yes, that works

[dim perch]

this

c.f. the UCT

[dawn geyser]

Hi 2 function f,g and f homotopic to g f,g:X---->Y continuous. If f homotopic to g and f,g:X---->Y continuous H:X x I---->Y continuous and H(x,0)=f(x)and H(x,1)=g(x). By definition. My question f homotopic to g and I just say H(x,0)=f(x)and H(x,1)=g(x) and I not explain as H(x1,0) H(x1,1) would it be true? Because I saw a book and this book just explain H(x,0)=f(x)and H(x,1)=g(x).

[kindred cairn]

@dawn geyser not sure I understand your question

[dawn geyser]

I mean A example just identify H(1,0)=f(1) and H(1,1)=g(1) but not identify another function mean H(2,0) H(2,1) etc. Can I say f homotopic to g?

[kindred cairn]

how do f and g take 1 as argument

f(1), g(1)

[dawn geyser]

X={1,2,3} T={X,{},{1},{2},{1,2}} Y={a,c} T|Q={Y,{},{a},{c}} and f,g:X---- >Y continuous and f(1)=a f(2)=c f(3)=a and g(1)=c g(2)=a g(3)=a now H:X x I---->Y I=[0,1] H(1,0)=a H(1,1)=c but I identify just H(1,0) and H(1,1) I not identify H(2,0) etc. My question can I say f homotopic to g now? @kindred cairn

[plush folio]

H is not even a function if you don't define what values it takes. Whether H is a homotopy depends on how you define it

[dawn geyser]

Mean H a homotopy on here and f homotopic to g

[dawn geyser]

.

[white oxide]

I could've sworn I saw the exact same question 10 times the last few days

Huh

[dawn geyser]

@white oxide I saw question on METU mathematics Olympiads before 2 day

[dawn geyser]

By Yıldıray Ozan

[white oxide]

Right, finite sets are an extremely contrived setting for homotopies though. Like usually you want a locally path connected hausdorff to do anything

[dawn geyser]

Yes

[gentle girder]

<@&268886789983436800> can yall do something about this

[muted arrow]

what

[gentle girder]

literally spamming this chat with the same example over and over

[muted arrow]

is this just like, the first time theyve actually gotten a response

weird that it's a different account

lemme check on the other one

[white oxide]

They are slightly different questions on closer inspection but like

[ebon galleon]

it has been 12 times, over 2 months, from 3 different accounts

[gentle girder]

#point-set-topology message here is one where they've gotten a response
#point-set-topology message here is another

[white oxide]

The same set and topology everytime?

[ebon galleon]

Yes.

[muted arrow]

Definitely weird but like

[gentle girder]

the list is really long

I will find more

[muted arrow]

if they've never gotten a satisfactory answer....

[white oxide]

T={X,{}, ...}. Interlinked.

[ebon galleon]

And each time, I or someone else has pointed out that the questions just simply doesn't make sense because things are not fully defined. There has been satisfactory answers.

[muted arrow]

sigh okay I'll verify this

I'll mute them for now

resume normal activities

[gentle girder]

thanks

[prime elbow]

In ultrametric space spheres are open.

Let S be sphere in ultrametric space such that d(a,x) = r, where a is fixed and x in S.

Now let y in S so d(a,y) = r. Let B(y,s< r) be open ball.

Let z in B(y, s) then d(y,z)< s < r.

We have, d(a,y) = r and d(y, z ) < r.

Now if d(a, z) < r it will contradict that d(a, y) is less than maximum of d(z, y) and d(a, z).

Similarly if d(a,z)> r then it contradict that d(a, z) is less than maximum of d(a, y) and d(z, y).

Hence d(a, z) = r so B(y, s) is contained in S.

Is it correct?

[prime elbow]

In discrete topology X, dense set is only X, right?

[rancid umbra]

yes, because the closure of any set is itself (as every set is both open and closed)

[prime elbow]

Because all subsets are closed

[rancid umbra]

yes

[prime elbow]

If X is a finite set and X is topological space then X is separable, right?

Because then I will let X which is countable dense set

[rancid umbra]

yes

any countable space is separable

because any space set is dense in itself

[prime elbow]

Yes

And indiscrete space is also separable

If (X, T) is topological space and T is finite then any hint what can we say about separable of X ?

[opaque scroll]

Think about how you did this for the indiscrete topology

[prime elbow]

If T_1 is finer than T_2 then every dense set of T_1 is dense in T_2, right?

[opaque scroll]

That's right

[prime elbow]

Let U_1,..., U_n be a proper open set in Topological space.

Then pick a_i in U_i for all i, then { a_1,...., a_n } is dense in a given space.

Is it correct?

[prime elbow]

Q is connected and Hausdorff space, right?

With respect to subspace of R

[red yoke]

Not connected

In fact it's totally disconnected: all connected components are points

[silver ridge]

Can someone confirm: ${\mathbb{N}}$ is closed because ${\mathbb{N}}^\mathsf{c} = \mathbb{N}$ in X, and $\mathbb{N}$ is a subset of $\mathbb{N}$ hence open?

[gentle osprey]

swifteeee

[red yoke]

Yes

(can you draw this topology)?

[silver ridge]

uhhhh

I don't even know where to start

[red yoke]

(it's homeomorphic to some subspace of R²)

[silver ridge]

I can kinda see this as N with a "point at infinity", {N}

whatever that means

I have no idea arki

[red yoke]

It's actually {0} ∪ {1/n : n ∈ Z+}

[silver ridge]

what's the topology

[red yoke]

Subspace of R

[silver ridge]

huh

what's the homeomorphism?

[red yoke]

n → 1/(n+1)
ω → 0

[silver ridge]

1 \in X is open in X. The homeomorphism maps it to 1/2. This set is open because we can find an epsilon ball that contains only 1/2 under the subspace top.

Case 2: complements of finite sets

does {1, 2, 3, {N}} count as a finite set in X?

i guess so?

[red yoke]

You can compare the basis for X and the basis for {0} ∪ {1/n : n ∈ Z+} inherited from R

If you've learnt about bases

[silver ridge]

I know that a basis is a set of open sets s.t. their unions generate all open sets

nothing beyond that

I know that a basis for R are all the e-balls, for example

[red yoke]

That gives a basis for any subspace of R as well

[silver ridge]

oh. then clearly {0} is closed because every epsilon ball intersects some 1/n

no e will guarantee that {0} is interior to {0}

[red yoke]

And what are the open sets containing 0

[silver ridge]

complements of finite subsets of N, in X

{{N}, some infinite subset of N}

[prime elbow]

Yes

[prime elbow]

Okay, thank you

[silver ridge]

applying the homeomorphism, the open sets containing 0 are {0, infinitely many 1/n}

[red yoke]

Cofinitely many 1/n

But can you show that from being a subspace of R (this is an analysis question)

[silver ridge]

cofinitely?

[red yoke]

All but finitely many

[silver ridge]

ah right

well, each {1/n} not in our set is open, so take finite union and then take a complement

wait

Consider the finite 1/n not in our set. finite subsets of R are closed, so each {1/n} is closed, and so is the finite union of {1/n} not in our set. Then we take complements

actually

I don't think we're guaranteed that open/closed sets map onto the topology of R from the subspace topology

[merry geode]

Oh 'cofinitely' is nice terminology to use! Thanks!

[silver ridge]

[0,3) is open under the subspace topology on [0,\inf), but is neither open nor closed uner the usual topology on R

[red yoke]

The definition of the subspace topology S ⊂ R is that if U ⊂ R is open then U∩S ⊂ S is open

[silver ridge]

oh I forgot about that characterisation

[red yoke]

Or you can think of S as a metric space

[silver ridge]

right, then [0,3) = (-1,3) \cap [0, \infty) for example

[red yoke]

Yup

[silver ridge]

thank you I appreciate the help

this subject is hard

[red yoke]

It may be hard at first but as you progress you may learn more clues that let you see this is obviously {0} ∪ {1/n : n ∈ ω}

This topology is also known as the one-point compactification of N, since you add one point at infinity and it becomes compact

As well as (the order topology of) the ordinal ω+1 = ω ∪ {ω}, since it expresses the fact that {n : n} converges to ω

[tender halo]

it is also called a (countable) Fort space

[silver ridge]

Is it worth 'properly' learning set theory for topology?

[red yoke]

Hm probably not necessary

[tender halo]

not real set theory anyway

[red yoke]

But the long line may appear as a common counterexample

Which involves the first uncountable ordinal

[silver ridge]

I think bredon has an appendix I'll take a look at

[sly geyser]

https://en.wikipedia.org/wiki/Furstenberg's_proof_of_the_infinitude_of_primes
this was pretty fun to read actually

[unreal stratus]

barely has anything to do with topology though lol

[tender halo]

well actually, the real point-set topology is half set theory. but what other mathematicians regard as topololgy does not use set theory beyond transfinite induction at most

[sly geyser]

I thought the topology proof of the infinitude of primes was this oh so super complicated thing but it was actually pretty darn simple

[tender halo]

and like Stone-Cech compactification

[prime elbow]

https://www.math3ma.com/blog/topological-magic-infinitely-many-primes

You can also look at it

[silver ridge]

The appendix in bredon takes 3 pages to prove existence of ordinals and has no exercises

[tender halo]

sounds about right

[silver ridge]

Is bredon specifying a basis or a subbasis here? Previously, he has only used "generated" when talking about subbases

[sly geyser]

from what I understand this is what would happen if you took an infinite amount of circles and then glued together all the (0,0) points together right?

actually wait....

I should review the definition first for this topology

it might be a bit different from that

subspace topology....

so it is just an infinite amount of circles glued together at one point

[red yoke]

A subbasis

[silver ridge]

thanks

[sly geyser]

actually wait

[red yoke]

Such that any open set containing the origin contains cofinitely many circles

ω+1 is a subspace of this

It is indeed the subspace topology w.r.t. R²

[sly geyser]

why cofinitely?

[silver ridge]

Sorry, one more thing. When we take finite intersections of the subbasis, does that incude taking the intersection of an element with itself?

[red yoke]

Yes, as well as the entire space

We take the intersections of finite subcollections of the subbasis

Including the collection with only one subbasis element

[silver ridge]

[red yoke]

(and the empty subcollection has intersection the entire space)

[silver ridge]

yep

[red yoke]

An epsilon ball at the origin contains all circles smaller than radius ε/2

It's different from "gluing infinitely many circles at a point", which wouldn't be a subspace of R²

[alpine nest]

1 is a finite number, that is true

[red yoke]

Indeed, {{q ∈ Q : q < 1}, {q ∈ Q : q ≥ 1}} has two elements

[prime elbow]

They defined A to be semi open if there exists open set U such that U contained in A and A contained in Closure of A.

I want an example such that a closed set is not necessarily semi open.

[tender halo]

I assume you mean closure of U here?

a point is not semi-open, anything nowhere dense is not semi-open

if a closed set is semi-open it is regular closed

so any closed set that is not regular closed is what you want

[prime elbow]

Yes

[rancid umbra]

the cantor set is closed but contains no open intervals so cannot be semi-open

[rancid umbra]

points sets in discrete spaces are semi-open since they are open

[tender halo]

[hearty condor]

[ocean narwhal]

I don't get this point. The Cantor set in itself is semiopen.

Singletons in R are the most obvious counterexample

[sly geyser]

if you mean the point where all the circles intersect that would contain parts of all the circles....

[red yoke]

Yea

And any neighbourhood of it has to fully contain cofinitely many of the circles

Since the neighborhood basis is given by epsilon balls

As opposed to the wedge sum of countably infinitely many circles, which has finer topology than Hawaiian earring

[sly geyser]

I don't get it, if you have an open ball around the intersection point, isn't every circle's points eventually gonna get close enough to be in whatever neighborhood/ open set you define around it?

sus.....

[opaque scroll]

Why this be sus?

[sly geyser]

he said it was finer...

but here it says it is homeomorphic

[opaque scroll]

What you showed is very much not a wedge of circles.

That's not compact, so can't be the compactification of anything.

[opaque scroll]

The wedge of circles and the topologist earing (or Hawaiian earring) are somewhat similar, in that both involve circles meeting at a point.

But the difference is that for the topologist earing the circles accumulate towards the intersection point, making the whole thing compact.

[sly geyser]

now that I think about it

oh

so that's what he meant by a cofinite amount of circles

🤦‍♂️

I misunderstood

[opaque scroll]

You can embed the wedge of circles in R^2 as

Union_n {(x, y) | (x-n)^2 + y^2 = n^2}

Notice here the circles get bigger and bigger, so don't accumulate

[sly geyser]

hmm interesting

[tacit basin]

In R^n with the euclidean topology, does every proper nonempty subset have a nonempty boundary set

[tender halo]

because R^n is connected and doesnt have clopen sets

[tacit basin]

So its true right

[tender halo]

yeah, empty boundary is the same thing as being clopen

[tacit basin]

I tried proving this and did the following instead of deriving a contradiction with respect to the connectedness of $R^n$. I think it's right but I'm not fully sure. Suppose there's some proper non-empty clopen set $S\subset\R^n$. We know closed implies $\partial S\subseteq S$, and the boundary is nonempty since $S$ is nonempty. Choose $x\in\partial S$, then $x\in S$ and by openness, there exists some $\varepsilon>0$ such that $B\varepsilon(x)\subseteq S$, but by being a boundary point, $B\varepsilon(x)\cap S^C\neq\varnothing$. This implies there exists $x \in S$ and $S^C$, a contradiction

[gentle osprey]

Sara

[tender halo]

and having non-trivial clopen subsets is the same thing as being disconnected

[tacit basin]

Oh ok, nvm my proof is circular ig

This works right? Suppose there's a proper-nonempty clopen set $S\subset \R^n$. Then $S^C$ is also clopen. $\R\subset S\cup S^C$, $S\cap\R^n\neq\varnothing$, $S^C\cap\R^n\neq\varnothing$, and $\R^n\cap S\cap S^C=\varnothing$, but this contradicts $\R^n$ being connected (since it's convex)

[gentle osprey]

Sara

[tender halo]

i mean im not sure whats your definition of connectedness is lol

to me connected just means "has no clopen sets"

[tender halo]

sure, your proof works then

[lethal oxide]

is there anyone to explain tychonoff's theorem in simple terms?

[tender halo]

which proof do you have in mind? there are several quite different ones

[lethal oxide]

ıdk ı read on book but i couldnt understand even a word

the easiest one would be better

[tender halo]

do you have a screenshot of the proof?

[tender halo]

which part is giving you pause?

[lethal oxide]

proof of theorem 2.5?

[tender halo]

i mean you need to ask questions

i cant divine the nature of the thing you dont understand by myself lol

[lethal oxide]

oh true wait

there was a text in wiki about product topology but there is nothing in here @tender halo

why do we need coordinate projections in here?

[tender halo]

so the idea of the proof is as follows: we are given a centered family of closed subsets, and we win if we find an element that lies inside the intersection of all sets of the family

that is what being compact is

we take a maximal centered family and project it onto each axis and take an element in the intersection which we know is there because each axis is compact

that element is the coordinate of x along i'th axis

now we have x and we prove that it lies inside the centered family we were given in the beginning

[lethal oxide]

thanks for everything @tender halo

i think im understanding it better now

[hollow geyser]

I know how to do this tediously with each metric. Is there some more clever way to prove this without having to do each metric separately?

(Context: Trying to prove that closed balls are path connected. I could do other paths if that's deemed easier)

But I was interested in knowing this way if possible

[wispy veldt]

triangle inequality

[hollow geyser]

That's my first guess, but my brain just isn't folding correctly. I keep thinking "triangle inequality, ok, but with which points?" Is your guess triangle inequality? Or do you already know for sure?

[wispy veldt]

it works

well

for R^n

[hollow geyser]

If you're certain that that's the approach, then I'll think it over more

[fierce lily]

If A is a subset of a topological space X and A' denote the set of all limit points of A, can i conclude that clousure of A equals A' instead of A union A'? I think A=A union A'

[wispy veldt]

okay actually i just spotted a issue

i dont think all metrics will even work

its counter intuitive but its true

[hollow geyser]

I've heard some people define a limit point as not being in the original set. So you may want to review what your definition of limit point is

[hollow geyser]

Oh you might be right. I hear 4-spheres get weird

[wispy veldt]

consider $d(x,y)= |x_{1}-y_{1}|^{p}+|x_{2}-y_{2}|^{p}$ for $0<p<1$

on R^2 here

[gentle osprey]

James Banach

[wispy veldt]

then the balls wont be convex in general

if it was a metric induced from a norm

then indeed all balls would be convex

due to the convexity of the metric itself

or more simply homogenuity of the norm

otherwise its not a given

[hollow geyser]

I was just asking about $d_1$, $d_2$, and $d_{\infty}$. I should be good there, yeah?

[gentle osprey]

SWR

[wispy veldt]

indeed certain sophisticated metric spaces have no convex open sets at all

[wispy veldt]

yeah

[hollow geyser]

But yeah, I wasn't even thinking of convex. I guess that's what I'm ultimately trying to prove huh

[wispy veldt]

indeed

convexity is ofcourse stronger

than path connectedness

[hollow geyser]

True. So "lol the ball is convex" would be pointless to say

[unreal stratus]

What does it mean to talk aboouot convex opens in a general metric space

[gaunt linden]

Does "convex" have a meaning in general metric spaces?

[unreal stratus]

Aha

[gaunt linden]

Ah, jinxed.

[unreal stratus]

I thought convex required some notion of straight line

[wispy veldt]

as long as you have a vector space you can speak about \$tA={tx: x\in A}$ and $A+B={x+y: x\in A, y\in B}$, and so a set C is convex when $$tC+(1-t)C\subset C, 0\leq t\leq 1$$

[gentle osprey]

James Banach

[gaunt linden]

I see.

[wispy veldt]

but in general metric spaces ofcourse that wont make sense, just in topological vector spaces

[gaunt linden]

Some searching found me a concept of https://en.wikipedia.org/wiki/Convex_metric_space>.
And as far as I can see, R with the metric d(x,y) = arctan(|x-y|) has no nonempty open sets that are "convex spaces" with the induced metric according to that definition ...

[unreal stratus]

Sure vector space always makes sense

[unreal stratus]

That's when you know it's a great definition

[hollow geyser]

What about all of R?

[gaunt linden]

Even all of R.

[hexed goblet]

hi math ppl, general question about topology here. i'm a rising senior physics major with a math minor (did all my calcs, ode, + linear algebra), and i'm really interested in topology. only thing is, i'm unable to fit in my uni's topology course + the real analysis pre-req. do y'all have any resources to help me out with learning a bit of topology? anything would be appreciated

[hollow geyser]

General reading is always a good resource

Online free courses if you want a syllabus

This is a very introductory book, but a good start to know the basics
https://www.amazon.com/Introduction-Metric-Topological-Spaces-Mathematics/dp/019956308X

Why are you looking to use topology for?

[eager vigil]

Hi, I'm going through prop 2.4 in Bredon, and I'm stuck on the (<=) part. I get what the author is saying, but I can't seem to figure out the map itself. How does one decode this diagram to find the map?

Here h_p is the change of basis homomorphism

Hm, nvm, I think I understand. If I'm not mistaken, this is simply a slightly handwavy application of the gluing lemma on a polygonization of the unit square? So to write out the function explicitly, one would have to define it on 6 different not-too-pretty polygons...

[proven condor]

Hey I've come across a question which seems really obvious but can't prove it

Am I suppose to ask help here or in another channel?

ok i get it now lol, help channel

but it's a good topology one

This one seems really intuitive, but I cannot prove it.
Let A in R^n be an open set, with n >= 2. Given a in R^n - A, the set A U {a} is open if, and only if, a is an isolated point in the boundary of A. Equivalently: there exists a ball B = B(a,r) such B-{a} is contained in A.

[prime elbow]

What is meant by a is an isolated point in the boundary of A?

[rancid umbra]

i think a fundamental example is a punctured disk; you can use this to guide your intuition @proven condor

if A U {a} is open, there is some ball B_r(a) contained in A U {a}. since a is not in A, B_r(a) - {a} is contained in A. since A is open, (try to finish this direction) ||this punctured disk doesn’t intersect the boundary of A, so a is an isolated point of the boundary of A||

the converse is a bit tricky

[prime elbow]

Means a is not a limit point of A?

[rancid umbra]

no, it is still a limit point because the boundary of a set is contained in the set of limits points of that same set, but it means that there is some open nbhd U around a such that U intersect the boundary of A is equal to {a}, or equivalently, U - {a} intersected with the boundary of A is empty

[prime elbow]

Oh boundary of A, sorry I am not familiar with this terminology

What is meant by boundary of A, closure of A ?

[rancid umbra]

closure of A minus interior of A

[prime elbow]

Oh got it

[rancid umbra]

a prototypical example is that the boundary of the closed unit disk is the circle

[prime elbow]

Tao use this terminology

[rancid umbra]

the set of all points in the plane with norm <= 1 is the closed unit disk.

the set of all points in the plane with norm < 1 is the open unit disk

the set of all points in the plane with norm = 1 is the unit circle

[prime elbow]

Thank you

So closure of closed unit disk is closed unit disk, right? One reason is because it is closed so its closure is the same as it is

Got it

The set of all points in the plane with norm = a called sphere, right?

[rancid umbra]

usually it is called a circle in R^2, a sphere in R^3, and a hypersphere in higher dimensions, but in general, the sphere in R^{n+1} is called an n-sphere, e.g., the circle is a 1-sphere, the usual sphere in R^3 is a 2-sphere, etc.

[prime elbow]

Got it, thank you ❤️

[civic verge]

name of book?

[eager vigil]

Topology and Geometry by Glen E. Bredon

[civic verge]

I was familiar with that book, I have it in physical, heh, too bad I can't even get to the heels of that book.

[civic verge]

I have that book in pdf, if you need it write me to private I will send it to you.

[civic verge]

Hi guys have you ever heard of the concept of ε-red?

[rancid umbra]

do you mean an epsilon net?

[civic verge]

ye

[real granite]

For the proof of (1), the point with not containing any points is that $\emptyset$ is open vacuously?

I.e. you can say $\forall x\in\emptyset, \exists r>0\ s.t.\ B_r(x)\subseteq \emptyset$ because $x\in\emptyset$ is never satisfied

[gentle osprey]

Douglas

[alpine nest]

Yep, every element of the empty set has a neighborhood that's entirely contained in the empty set.

And since I will never pass by an opportunity to repost this:

[real granite]

That is a worthy repost

[prime elbow]

The only continuous mapping from R to Q is constant mapping.

Because R is connected and Q is totally disconnected, right?

And Q is totally disconnected because let A be its subspace which has more than one element then let a and b in A and s irrational number such that a < r < b.

Now we have (-∞, r) intersection Q intersection S which is proper open and also closed set, right ?

[alpine nest]

Yep, and you don't even need to invoke the total disconnectedness of Q, just disconnectedness on its own is sufficient for the conclusion.

[prime elbow]

Sorry, I don't understand what you mean?

[alpine nest]

I mean "The only continuous mapping from R to Q is the constant mapping, because R is connected and Q is disconnected" is also a correct argument. Never mind, I'm wrong, you do need the total disconnectedness

[real granite]

Two metrics $d_1, d_2$ are equivalent iff every $B^1 _r$ (i.e. every open ball wrt to $d_1$) contains some $B^2 _s$, and every $B^2 _s$ contains some $B^1 _t$.

Is there a name for this?

[gentle osprey]

Douglas

[alpine nest]

You just used it, equivalence.

[quartz horizon]

I guess it means the corresponding metric spaces are homeomorphic

[real granite]

No I mean like the theorem

Like "equivalence-containment theorem" or smth

Cos the definition we've been taught is that convergence under d_1 implies convergence under d_2 and vice versa

And it is not immediately obvious that this is the same as the containment thing

So I don't really think of containment as a definition even though I suppose you might argue it is

[red yoke]

For any x and any r > 0, there are some s1, s2 > 0 such that B2(x, s2) ⊂ B1(x, r) and B1(x, s1) ⊂ B2(x, r)?

What happens if you allow "and some y s.t. B1(y, s1) ⊂ B2(x, r)"

Oh it fails

[real granite]

yes, I'm suppressing some of the notation for ease. The balls are centred on the same arbitrary point.

[alpine nest]

I don't think it's groundbreaking/influential enough to merit a standardized name.

It's just a criterion for equivalence of metrics

Because inded your definition involving convergence is equivalent (hah) to the one involving balls

And the fact that both definitions are equivalent is indeed a theorem, but I don't think it's a named theorem.

[white oxide]

To add to the above, try to memorize that anything of the form:
$\forall x \in \emptyset : P$
Is always true, since it being false would begin with "there exist an x in the empty set ...". The important bit is really just the quantifier used

[gentle osprey]

Kerr

[hexed goblet]

do you know of any good ones? also thanks for the book rec!
i'm looking to use topology for my future work in astrophysics/cosmology! i am also minoring in astronomy, and plan to pursue a phd in astrophysics, hopefully focusing on cosmological research. in particular, i find cosmic topology fascinating, as there is a surprising amount going on there, with different cosmic topologies potentially having different effects on the nature of universe. it's super cool, but also something i don't exactly understand very well when i try to read papers on it yk 💀

[hexed goblet]

that would be fire

[real granite]

I assume it's valid to say $$d(x,y)=\left| \frac{1}{x}-\frac{1}{y} \right| = \frac{|x-y|}{|xy|}$$ and therefore $$d(x_n, x)=\frac{|x_n-x|}{|x_n x|}$$ which tends to zero if and only if $d_2(x_n, x)$ tends to zero?

[gentle osprey]

Douglas

[alpine nest]

assuming you prove the if and only if part, then yes

[real granite]

Well isn't that trivial? Because $d_2(x_n, x)=|x_n-x|$?

[gentle osprey]

Douglas

[alpine nest]

depends what level of real analysis were treating as trivial, some would say yes, some would say no

[real granite]

Uh... what are the (potential) non-trivialities?

[alpine nest]

no, I just mean can you actually prove it if needed?

[real granite]

Well yes... if a=0 then a/b=0, and if a/b=0 then a=0
(I'm assuming b≠0)

We're just taking a=|x_n -x| and b=|x_n x|

Those equals shouldn't really be equals should they when talking about sequences

Yeah I think I kinda get your point, but it should be straightforward to do rigorously with some epsilon delta stuff

[alpine nest]

the limit of a_n/b_n being zero doesn't immediately imply that the limit of a_n is zero

I mean, it does in this particular case

but it's not a general fact

[real granite]

Well yes, you could have b_n -> infininty and a_n=const

But yeah

In this particular case it is immediate due to the constraints of the problem

[viral atlas]

Question from a friend not on Discord, would like suggestions, hints, or references: https://math.stackexchange.com/questions/4928933/open-neighborhood-of-a-geometric-knot-in-an-orientable-3-dimensional-manifold

[alpine nest]

now I'd really like to see a rigorous argument

[inland thistle]

I think this may sound like a weird question, but if there are two topologies, one with uncoutable basis, and one with countable basis, then can we say the topology with uncountable basis is strictly finer than the one with countable basis?

For this, each topology belongs to different topological spaces

for example, w1 has uncountable basis while R has countable basis

[red yoke]

Usually you can only compare two topologies on the same set

[inland thistle]

then I think if there is a homeomorphism between w1 and f(w1), then f(w1) also has uncountable basis? (not sure if this is true)

I see, I just give context first

so I wanna solve a question like, is there an imbedding from w1 to R?

I know it is impossible

but idk how to go about proving or arguing it

here w1 is omega-one (first uncountable ordinal)

[red yoke]

You could apply first-countability of R

[alpine nest]

also you can have an uncountable basis for the Euclidean topology on R

[inland thistle]

yeah true as well as countable basis

[inland thistle]

do you have any idea for this

my knowledge is limited, and I might wanna see the approach where we use continuity mostly

[alpine nest]

You can't compare fineness of topologies if they aren't on the same space.

[red yoke]

^

[alpine nest]

and on R you could have the Euclidean topology with a countable basis, and also the Euclidean topology with an uncountable basis

Neither of which is stricly finer than the other, because they are the same topology

[inland thistle]

that makes sense

also some general question, it is not necessarily true that if f: X -> Y is continuous, then for some basic element C in Y, f^-1(C) is also a basis element in X?

and how about it if f is a homeomorphism

my understanding is that for each topology, I think there is a basis generating it, is this wrong?

[red yoke]

The basis is not unique though

[inland thistle]

hmm I see

so are you saying we are not thinking about the corresponding basis between two spaces?

[tender halo]

let me rephrase your question so it makes sense - if we have two topologies T1 and T2 on X with weight of T1 smaller than weight of T2, is it true that T2 is finer than T1

the answer is obviously no

take two topologies on X with different weights and no isolated points (say, Sorgenfrey line and R) and adjoin an isolated point to one of them

it doesnt change the weight, but obviously the topology with the isolated point has to be the finer one (if one of them is in fact finer)

||note that im doing a sleight of hand here in that we in fact have two different sets now, so you will have to fix a bijection bewtween X and X + 1 and take the induced topology on X||

[civic verge]

If someone needs help and can be helped why not do it, you have to be an eogist not to do it, I think that here in this group there are only egoists, and I would start with you.

[white oxide]

thanks for the fresh copypasta material ig

[hollow geyser]

I've gotten nothing but wonderful help in here. Where are you getting this perception?

[real granite]

Instead of reasoning using an arbitrary point $x\in\interior A$, can you do as follows?

By definition, $\interior A=\bigcup_{x\in A} B_r(x)$. Any union of open balls forms an open set (axiomatically true given open balls are open and metric spaces are topological spaces), so $\bigcup_{x\in A} B_r(x) \subseteq \bigcup_{U\subseteq A} U$. Conversely, $\bigcup_{U\subseteq A} U$ must be open because the union of open sets is open. Any open set can be written as a union of open balls, so $\bigcup_{U\subseteq A} U \subseteq \bigcup_{x\in A} B_r(x)$. As each is a subset of the other, they are in fact equal.

[gentle osprey]

Douglas

[thin tide]

I have no idea where to ask this but is there a difference between “canonical” and “natural” isomorphisms/identifications or are they used interchangeably

I can’t seem to find a good answer

Or rather an agreed upon answer

[late moat]

seems right to me

[real granite]

supremely epic

[opaque scroll]

Both words are used with slight variation, somewhat informally.

But 'natural' also has a technical meaning (in category theory) which tries to capture the informal meaning somewhat.

[thin tide]

Ah ok

[gaunt linden]

It sounds like you were expecting bases for topological spaces to be a bit like bases for vector spaces. For vector spaces, if we have two spaces V and W each with a basis, and a bijective correspondence between those bases, then the spaces are "essentially the same space".
This is not the case for topological spaces and bases for topologies. A basis for a topology is a lot "sketchier" than a basis for a vector space -- one topological space can have bases of different sizes. It's almost always the case that a basis for a topology can be replaced by a strict subset, which is still a basis for the same topology.
And even though all open sets must be unions of basis elements, there are usually many different collections of basis element that union up to a given open set.

[fierce lily]

I am just confused about why we need to emphasize finitely many, isn't it possible that there is an infinite sequence xn?

[alpine nest]

The point is that for at most finitely n you can say that the x_n's are in V

Whether or not there are infinitely many x_n's or not is irrelevant

There are inifnitely many n's though

But only for finitely many of those can you say that x_n is in V

So it's impossible for the limit of the x_n's to be y

[slender glen]

is that not just how convergence is defined

[opaque zodiac]

What is an easy example of a 1-site whose category of infinity-sheaves fails Whitehead's principle (which I think is the same thing as failing to be hypercomplete?)?

@dim perch I need you

Anyway I was told such a thing exists

[dim perch]

so far be it from me to have convenient examples of pathological toposes

[opaque zodiac]

Is failing WP considered pathological?

The HoTT book talks about this possibility positively, iirc

[dim perch]

but I know that HTT counterexample 6.5.4.6 is a non-hypercomplete sheaf topos on a really explicit topological space

[dim perch]

uh

perhaps I'm just a Whitehead's Theorem Enjoyer

I assume "whitehead's principle" means that infinitely connected morphisms are equivalences/the topos-theoretic homotopy groups detect equivalences

it's at least much harder to do homotopy theory without this

[opaque zodiac]

based mid-pdf link

more people need to know that this is possible

Huh I think someone told me Deligne's theorem did work for infinity-toposes

oh I should have literally read the paragraph above the one I was looking at

Wait so does Lurie not use the term 'hypercomplete'?

Also you're probably not the right person to ask, but do you know an example where the internal logic of the infinity-topos is classical?

This is supposed to be possible too

[dim perch]

lurie does in fact use the term 'hypercomplete;' Shv(X)^ is referred to as the hyprcompletion of Shv(X) for instance

[opaque zodiac]

Ah

[dim perch]

tbh I don't know about the internal logic in an ∞-topos

[opaque zodiac]

Wait so the hypercompletion isn't always hypercomplete?

[dim perch]

but my understanding is that every n-topos is uniquely an n-category which is an ∞-topos

so

perhaps Set

[opaque zodiac]

no but I'm wanting an example that looks like a model of HoTT

so you know all the ∞-shit and stuff

[dim perch]

The hypercompletion is a localization from toposes to hypercomplete toposes; the argument above is that Shv(X) is not fixed by this localization, so it must not have been hypercomplete in the first place

i.e. Shv(X)^ is hypercomplete, and Shv(X) ~= Shv(X)^ iff Shv(X) is hypercomplete

[opaque zodiac]

ahh

ah

okay so a Boolean one would I think be one where it's sheaves over a Stonean space (i.e., extremally disconnected profinite space)

[dim perch]

ok fine, but I don't really know what it means to be classical in this sense... I'd guess that if spaces aren't classical enough, then it'd be hard to give examples which are not truncated in some sense

[opaque zodiac]

I don't know if there is an example of that form

[opaque zodiac]

No so Spaces™️ satisfies LEM

pretty sure

since a singleton is extremally disconnected

[dim perch]

is this like

not a measure that S (i.e. spaces) is classical?

[opaque zodiac]

uhh

wait

it is classical?

I'm confused by your question

but that might be my fault

[opaque zodiac]

Also, for the record, the last sentence of this picture is just fucking bullshit

We're literally talking about a non-hypercomplete ∞-topos built out of sets right now

[dim perch]

I'm kind of confused as to what the goal of our conversation is

[opaque zodiac]

sorry

[dim perch]

I was attempting to anwer the question of whether there's an example where the internal logic is classical