12:07 AM
@MikeMiller ah yeah thats a good remark. We did prove that $2\chi(M) = \chi(\partial M)$ for odd-dimensional manifolds but I did not think of plugging in an empty boundary

the good ol' doubling

0

Define $X^{\{n\}}:=\{A\subseteq X:|A|=n\}$, the set of subsets of cardinality $n$. If $X$ is a topological space, $X^{\{n\}}$ can be given a topology by considering it to be a quotient of $X^n$ minus the extended diagonal. Define a continuous $n$-partition of a space $X$ to be a partition of $X$ ...

Dumb question: Does the statement "P and Q unless R" translate to "If not R, then either not P or not Q"?

12:26 AM
Can all three be false? @user193319

Oh, should it be "If R, then either not P or not Q"?

Same question
(false implies true)

Yes, I agree false implies true, but I'm not sure I get your point.

Is "False and False unless False" valid?

I suppose not.

12:33 AM
I'm honestly not sure what "unless" is logically, myself
Is "True unless True" valid? I'm not sure
If it's not, then "unless" is the same as exclusive or (and therefore symmetric: A unless B is the same as B unless A). I think.

Probably best to simplify it to "P unless R". no reason to hassle with the conjunction

Actually, it would be symmetric even if True unless True were valid

P unless R reads as "not R implies P" to me

it's the case of R false which isn't so clear
if R is false, then "not R implies P" is vacuously true

12:37 AM
but "P unless 2+2=5" sounds an awful lot like just P

True implies False is not valid
"not False implies P" isn't a tautology

hmm, i got that backwards
trying that again

This light is on, unless that light on.
Can we say exactly one light is on?

R = F => "not R -> P" = "T->P"="P"
R=T => "F->P" is vacuously true
i guess it comes down to: If R is true, does that require P be false? Or merely that P's truth is irrelevant
which in everyday language seems pretty ambiguous

how do you tell if two compactifications are the same?

12:42 AM
> The University of Michigan won’t have a football season this fall unless all students are able to be back on campus for classes.

yeah. is the students being back on campus sufficient or merely necessary

oh, two compactifications are the same if there exists a homeomorphism between the two from what I understand

I don't think that's the sensible notion
compactifications of a space X are specific embeddings of the form X->Y, the right notion of morphism between those should be given by commutative triangles

Top 5 google news:
'Don't Upgrade Your Camera Gear Unless It's Limiting You'
WhatsApp will disable messages unless users agree to new privacy policy
No one is safe unless everyone is safe
Deion Sanders was robbed on Sunday, unless he wasn’t
The 5 Worst Video Game Cities To Live In (Unless You Want To Suffer)

1:13 AM
is there a map that sends the conformal compactification of $\Bbb R^2$ to the conformal compactification of $\Bbb R^{1,1}$ if you treat both spaces as topological vector spaces? I think yes because $\Bbb R^2$ and $\Bbb R^{1,1}$ are equivalent when treated as vector spaces/topological vector spaces
I think I should understand the isomorphism between the two (without the compactificatins)

Something like that

Yeah it is
@Astyx cool

1:33 AM
so if it's a conformal compactification then no can't treat them as vector spaces because in a vector space there's no notion of angle

1:54 AM
1 sec let me edit little bit of the above it is very clear.
@Astyx So given homogenous polynomial F of degree d that defines a bundle called hyperplane bundle, which is a line bundle L(d). We can easily define a section by the following steps first define holomorphic function $f_j : U_j \rightarrow \mathbb{C}$ given by $f_j = F / z_j^d$
one can see easily these f_j induce a section.
sorry I want to discuss these things as it makes it clear in my brain
@TedShifrin maybe you know this but does hyperplane bundle or weakly positive go into detecting twists of a vector bundle?

2:20 AM
I don't understand your question. One usually uses it to make “canonical” twists.

@TedShifrin what is canonical twists?
is that related to adjunction formula ?

Not what I meant.
I'm referring to things like Kodaira vanishing/embedding. You twist by powers of the positive line bundle (which ultimately becomes — more or less — the hyperplane bundle).
I meant canonical in the usual sense, not canonical bundle sense.

I see that is very cool

2:51 AM
I have this problem and it seems obvious but I'm not really sure why the commutative condition is needed on R. Define $IM = \{\sum_{1}^{n}r_ix_i | r_i \in I, x_i \in M, n \geq 1 \}$ where I is a left ideal of a ring R and M is an R-module. If R is commutative $M/IM$ is an $R/I$-module.

3:11 AM
@robjohn true

3:34 AM
cc^ @Semiclassical

6 hours later…
9:25 AM
I just derived $$\left(1+\frac1n\right)^{\sqrt{n(n+1)}}\le e\le\left(1+\frac1n\right)^{n+\frac12}$$ for an answer I wrote. I have never seen the lower bound before.

nice

$$n+\frac12-\sqrt{n(n+1)}=\frac{\frac14}{n+\frac12+\sqrt{n(n+1)}}$$
so the exponents get very close; $\le\frac1{8n}$

coolio

is there any obstruction to generalizing $\chi(X \# Y) = \chi(X) + \chi(Y) - 2$ (for surfaces) to $\chi(X \# Y) = \chi(X) + \chi(Y) + 1 + 3\cdot (-1)^{n-1}$ for $n$-dimensional manifolds
basically I wanna cut out a an n-disk out of each manifold, which leaves behind a sphere. but the CW structure might be messed up

Wot
What is $\chi(S^{n-1})$ man
$\chi(A \cup B) = \chi(A) + \chi(B) - \chi(A \cap B)$ always

9:41 AM
2 0 2 0 2 0 2

Mayer Vietoris
@user2103480 so thats what you wanna subtract out

oooff I added in that formula but meant to subtract

1+3*(-1)^(n-1) wtf
such a complicated way to write 0 if odd 2 if even
didnt even recognize they were the same numbers

well you get two extra (n-1) cells so I gotta add $2(-1)^{n-1}$ and then I subtract $1 + (-1)^{n-1}$

u wot m8
S^n has a CW structure with a single 0 cell single n cell
1 + (-1)^n
wtf is 1 + 3*(-1)^n lmfao
theyre not even the same numbers
hilarious
Where are you getting extra (n-1) cells from

9:46 AM
@BalarkaSen I meant to cut out an open n-disk such that that the boundary contains a 0-cell in each manifold

Ok? I don't understand your computation
Can you write it out for me
You should get that for any sufficiently nice decomposition $X = A \cup B$ of your space $\chi(X) = \chi(A) + \chi(B) - \chi(A \cap B)$
So for two $n$-manifolds $M, N$, $\chi(M \# N) = \chi(M - pt) + \chi(N - pt) - \chi(S^{n-1})$

then in my head there'd be an extra (n-1)-cell in each manifold so the characteristic of the resulting cell complex is $\chi(X) + (-1)^{n-1}$
or $\chi(Y)$, respectively

I assume you're computing $\chi(M - pt)$

& the complexes intersection would be a sphere, with one 0-cell and one (n-1)-cell
@BalarkaSen yup that's the goal

$M - pt$ is $M$ with one $n$-cell removed, I don't see the extra $(n-1)$-cell.

9:53 AM
Ok thank you, then I better know a CW structure if I want to compute this generally
This also explains why it worked so easily for surfaces, one just threw away two 2-cells

You don't need to impose a CW structure to compute these things
But you can

it's all exam preparation so any strategy better be quick

it is quick, but you need to think
you cannot expect to get a list of quick strategies that you can plug and chug without putting any thought to it
$M = (M - D^n) \cup_{S^{n-1}} D^n$, so $\chi(M) = \chi(M - D^n) + \chi(D^n) - \chi(S^{n-1}) = \chi(M - D^n) + 1 - \chi(S^{n-1})$. So $\chi(M - pt) = \chi(M - D^n) = \chi(M) + \chi(S^{n-1}) - 1$. Adding everything up, you get $\chi(M \# N) = \chi(M) + \chi(N) + \chi(S^{n-1}) - 2 = \chi(M) + \chi(N) - \chi(S^n)$
And that is your final formula

I just realized that it is not true that in a compactly generated space a set is Borel iff its intersection with every compact set is, in fact this seems to fail already for $F_\sigma$ sets. Makes sense in hindsight but I never thought about it

10:09 AM
But wait that is exactly what I meant when I said "add $2(-1)^{n-1}$ and then I subtract $1 + (-1)^{n-1}$". The result is $\chi(M) + \chi(N) -1 + (-1)^{n-1}$
And thanks for the more formal writeup, that really helps

But I did not understand what $(n-1)$-cells meant.

Ah I meant the boundaries of the disks. Completely fair to complain about the lack of precision

Ah OK I got you now
That is right

I blame it on the sheer number of topics. I'm mad at the fact that we did tangent spaces, transverse intersections and intersection forms all in the second-last week (this weeks the last), and that the exam is next week and this is still relevant for it

Wow, that is totally not algebraic topology

10:14 AM
Like how does he expect us to do any of that even semi-formally with any degree of assurance that we're not messing up. Maybe Z_2 coefficients so that we do not have to care about orientations

His sketch of Lefschetz duality was garbage anyway
He suddenly went to the smooth category
Just do things properly or don't teach the material
If you do AT slowly you can do it in a very self-contained, organized way
I learnt AT before I learnt multivariable calculus, no joke

yeah it sounds fair, except maybe some explicit calculations in spheres or projective spaces

I feel like your instructor is using way too much. You can do all computations in a self-contained way at the cost of drawing more diagrams.

Are we talking about the same lecturer. I mean I probably agree since I do not really understand most of the things in the last weeks anyways

It is OK if you actually know what he's using but people taking an AT course should not have background in differential geometry
Oh, is this not part of your Topology-2 course?

10:18 AM
It is

Which is supposed to be just AT right

and he is using way too much. But I didnt show any lefschetz stuff yet, no? It was last friday's topic
@BalarkaSen Hm well he himself said it's too quick. Apparently the department wants him to cover:

Sorry I meant Alexander duality maybe

ah yes yeah it was alexander duality

got it

10:24 AM
the lecture itself is really good. Still it's hard to actually solve problems yourself since the fine details of many things - such as those connected sums - we do not know. So the ideas might stand, and the execution still fails
@AlessandroCodenotti .... does this fail in any real world example?

Is $\Bbb R$ times an uncountable discrete space a real world example?
But I'm not even sure whether it holds in infinite dim Banach spaces right now

In $\Bbb R$ itself this should only fail using choice
but I'm not sure on which topological assumptions this hinges. Probably in separable spaces it's the same?
(it's always the same in separable spaces)
proof that $\Bbb R$ is the unique separable space. wheres my fields

10:42 AM
@user2103480 no it holds in $\Bbb R$
It works in all $\sigma$-compact spaces

10:54 AM
okay that is better than expected

11:05 AM
That's easy, if $A\cap K_i$ is Borel for a sequence $K_i$ of compact sets that cover $X$, then also $A=\bigcup_i(A\cap K_i)$ is Borel

11:46 AM
Meh $\Bbb N^{\Bbb N}$ is compactly generated, separable and still not sigma-compact
not nice
And any infinite-dimensional NVS with the norm-topology also not

12:17 PM
All reasonable spaces are compactly generated

ok algebraic topologist

Both first countable and LCH imply compactly generated, so almost all spaces are
iirc being compactly generated is the same as being a quotient of a LCH spaces

12:42 PM
Did you know that, in a sense, the sphere has $2$ mod $n$ many points for all $n$? It's true!
4

Let $X$ be a manifold and suppose $f:X\to X^{\{n\}}$ is a continuous $n$-partition. Let $p:X\to Y$ be the quotient map associated to this partition of $X$. I claim that in fact $p$ is a covering map. (Conversely, it is easy to see that any partition that comes from an $n$-sheeted covering map ...

(Short answer: it's because the Euler characteristic of the sphere is 2)

Hello Guys..

1:04 PM
How would you tackle the following integral $\int Ae^{-Ct}e^{i(Dt-Et)}dt$? The integration limits are from negative to positive infinity and $i$ denotes the imaginary number. $t$ is real and $A,C,D,E$ are real constants.

1:15 PM
I'm looking for the fourier transform $X\left(\omega\right) = \int_{-\infty}^{\infty}{x\left(t\right)e^{-i\omega t} dt}$

Hello, I have a question regarding Markov chains. Can anyone take a look? Thanks.
0

I am reading E-Li-Vanden-Eijnden's Applied Stochastic Analysis, and the author claims: The recurrence relation of $\boldsymbol{\mu}_n=\boldsymbol{\mu}_{n-1}\boldsymbol{P}$, where $\boldsymbol{\mu}_n$ is the distribution of $X_n$, and $\boldsymbol{P}$ is the transition matrix, can be written as ...

1:58 PM
@AkivaWeinberger Does your topology on $X^{n}$ agree with the subspace topology of $X^{n}$ as a subspace of the space of all closed sets with the Vietoris topology?

I don't know the Vietoris topology, but probably

It has as basis $\{F\mid F\cap U_i\neq\varnothing,F\subseteq\bigcup U_i\}$ where $U_i$ range over open sets of $X$ and $i$ over $\Bbb N$
Forget about it, your spaces are very nice, I´m just talking about the topology on compact subsets induced by the Hausdorff metric

2:51 PM
@BalarkaSen quote our lecturer "if I ask in the exam 'compute the homology of the torus via mayer-vietoris' and you say that H_0 is Z since it is path connected, I have to subtract a point since you're not working on the exercise as I proposed"

sounds fair to me; you can avoid many computations by doing trickery

Ugh I can already sense that I will get ridiculous point subtractions

@BalarkaSen that´s true, but I feel like that you should do so

once one understands the rote machine, yeah

2:53 PM
the keyboard I have in my office makes the wonkiest apostrophes and I don´t know how to avoid it

dont put apostrophes

Good idea. I will try to avoid them

@BalarkaSen I mean, it's still a result we are given way before MV comes up

i bet H_0(T^2) is just an exaggerated example

I think not

2:55 PM
i mean like you can avoid computing anything in a certain Hatcher exercise where he wants you to compute simplicial homology of a lens space
by using poincare duality

@BalarkaSen for that we'd have to prove that these are orientable first, which would probably be a hassle

its not, anyway that misses the point i was making
one should just do certain computations. at least be absolutely sure they can compute

Sure, sure, but these kinds of strict rules make it really hard to gauge what level of detail is expected
especially since the lecturers own style and computations in exercise sessions were quite mixed in level of detail

if they want you to do a MV computation just run MV
i always avoid H_0 because i am never sure what happens there but why should you be confused just because i am confused

yeah, but c'mon

2:58 PM
its a good skill to do algebra

H_0 is number of components
that's so basic it should be usable at any time
avoiding Poincaré duality can be reaosnable, because that's a big gun, knowing H_0 isn't

yeah fair, but is "computing everything except the zero stuff since thats just components" really not showing proficiency in MV?

yeah maybe but eg i dont know how the nonreduced MV leads up to H_0
why shouldnt you too

you mean the boundary map?
or just generally how I can use this to compute the stuff

yeah i just dont know how the last bit of the long exact sequences go, H_0 confuses me
but to an AT student it shouldnt

3:01 PM
it's definitely good practice and I will spent some time today just computing zeroth homologies I've normally waved away

nothing should be confusing more or less

I mean, you will have to use that H_0 is number of components at some point
at least for a point

the lecturer should then clearly state which homologies we are allowed to take for granted since we calculated a lot of em in the lecture

im just saying learn as much as you can in a way that it doesnt confuse you later

If $\phi: X\to X/W$ is a quotient map where $X$ is a top. sp. and $W$ is a subspace. Then if $U\subset X$ is a subspace, $\phi(U)$ and $U/W$ are homeomorphic?

3:02 PM
i have learnt many things which i have realized over time i dont understand because i get confused
anyway exam should be easy
its not the place to expose your confusions

whenever you decompose a connected space into two connected subspaces, you can start M-V at 0->H_1->...
this is a computation you do once and that's it

your arrows are going the other way.

uh H_1?
cothorgott

I've been doing too much cohomology

i am not sure what you said is true, the boundary map is H_1(X) -> H_0(A \cap B), and A \cap B can be disconnected

3:08 PM
Hi guys wonder if anyones around for a vague silly question. If I let a density (say on $\mathbb{R}$) evolve/flow according to a vector field $v$ i.e $\partial_t \rho = div (\rho v)$ is there a way to curve the space $\mathbb{R}$ so that in the curved space the density doesnt appear to change?

oh, I also forgot to say that should be connected

I also dont follow why the inclusion is necessarily injective. It should be injective if the intersection is connected though

I'm not paying enough attention

but thats exactly why torus is a relevant example
you break it into two annuli, the intersection is disconnected

Yeah it was good to see for general computation

3:09 PM
oh, if that's what you wanna do

one should not get confused, thats all my point is
anyway

I do the torus by puncturing

i know everyone here can do these things
@Thorgott cool yeah

@BalarkaSen don't say that, alessandro will come up with some absurd example

given enough time you can compute homology of any simplicial complex because its totally algorithmic

3:12 PM
I think it's relevant to try computing the cohomology ring of RP^n in Z_2 via intersection forms

i am assuming everyone here is a turing machine

pure combinatorics

but anyway exams should be easy and routine
whoever does not believe this is a lunatic

I am seeing these exercises in a whole new light
(a) Compute the cohomology groups with arbitrary coeffcients of $S^{n}$ in two ways:
- via the long exact sequence of a pair in cohomology, and
- via the Mayer-Vietoris sequence for cohomology.
(b) Compute the cohomology groups of all closed surfaces via the Mayer-Vietoris sequence for cohomology.

lol

3:16 PM
idgi
LES for (a) seems convoluted
what do they want you to do? (S^n,S^{n-1})?

(S^n, D^n) surely
lol no

Can I now also not use that $H^0(X;G) = Hom(H_0(X),G)$ and I must compute this via the LES
tfw no universal coefficient theorem

lol yeah RIP

ah ok yeah, they want the quotient

i mean (D^n, S^(n-1)) yeah

3:18 PM
same thing as M-V pretty much, but ok

What does "of a pair" mean ?

yeah I got you

does that mean you cannot use uniformization when doing (b)
lol

We can use classification of closed surfaces if that is what you mean
everything else would be absurd

yeah i meant that
this guy is hilarious

3:21 PM
no man

if he really wanted to do this he should have taught way less

compute the homology of all surfaces on a priori grounds
solve topology
bonus exercise: compute the homology of all 3-manifolds

@Thorgott incidentally that was exercise 4

maybe user can do this exercise

not me, but maybe TheReal_user

3:24 PM
take two solid surfaces $B_g$ of genus $g$ (the natural 3-manifolds which bound $\Sigma_g$, the surface of genus $g$), let $f : \Sigma_g \to \Sigma_g$ be a diffeomorphism. compute, only using Mayer-Vietoris, homology of $M = B_g \cup_f B_g$

Thx will try

dont take it too seriously thats equivalent to thorgott's bonus exercise
might be fun to try tho

shit I really have no sense of whats hard and whats an easy trick

really?
are all 3-manifolds obtained that way

yeah

3:26 PM
wew

thats why their homology is algorithmic

Is "the crushing theorem" something in english terminology? (the one giving the LES of a pair)

is that just a weird topologists way of describing quotients

also the understanding of 3-manifolds boils down to understanding the isotopy class of $f$ as a self-diffeo of $\Sigma_g$
by isotopy extension theorem
so you just need to understand $\text{MCG}(\Sigma_g)$ really well
this is the 3D topology -> 2D topology reduction

I mean, $H_n(M,\Sigma_g)\cong H_n(B_g,\Sigma_g)^2$, so it remains to analyze $H_n(M)\rightarrow H_n(M,\Sigma_g)$, which comes down to analyzing the boundary maps in the LES
thanks for reminding me I still have to prove isotopy extension

3:31 PM
really all you need to understand is $H_1$
MV gets you $H_1(M) \cong \Bbb Z^{2g}/stuff$
stuff comes from understanding what $f$ does to the generators of $H_1(B_g)$ drawn on top of $\Sigma_g = \partial B_g$
i dont think you can say anything more general than this
you can even compute $\pi_1(M)$ as some amalgamation of two copies of $F_g$ over a surface group
you have to understand that $f$; that basically captures all the info a 3-manifold carries
look up heegaard diagrams
i have never learnt how to draw those tbh

3:49 PM
our lecturer said ideas from the proof of poincaré duality might come up in the exam, and I'm a bit frightened by that. What would the core methods be? Dual cells, cap products and cohomology with compact support?
Seems a bit nasty to include dual cells, and cohomology with compact support might crop up for noncompact manifolds, but I'd be pretty lost if I need to compute some limit of groups

I did not learn the proof by dual cells
I do some sort of induction on charts
compactly supported cohomology is good to know a little better
compute $H_c^k(\Bbb R^n)$, $H_c^k(S^n \times \Bbb R)$, $H_c^k(M \setminus \{p_1, \cdots, p_n\})$ where $M$ is a compact manifold, and $H_c^k(\Sigma_\infty)$ where $\Sigma_\infty$ denotes surface of genus $\infty$.

Thanks!
(for the suggestions)

i mean one-ended surface of genus infinity
but ok thats the hardest one anyway

we called this the "loch ness monster surface"

haha yeah
Dennis Sullivan and Tony Phillips were the ones who coined this terminology I think
fun people

4:38 PM
@Thorgott orientable

ah, that's of course necessary
is every sphere bundle the sphere bundle associated to some disk bundle? I can prove this in TOP, but dunno about DIFF.

on homotopy groups you're asking if pi_k Diff(D^{n+1}) -> pi_k Diff(S^n), restriction to boundary, is a surjective homomorphism for all k, n
which I am certain is false but I don't remember a c/e
Should already be false at the level of pi_0
Which corresponds to bundles over the circle

5:04 PM
alright, I feared as much

Ugh, since I now know that I have to do these things in detail: what's the quickest way to compute cellular cohomology, assuming I'm given the boundary maps in the usual chain complex. I guess it's "compute kernels and images of coboundary maps explicitly and then translate these by explicitly using the isomorphism $\mathrm{Hom}(\Bbb Z^n, G) \cong G^n$"
working with the quotients of the hom-groups directly seems horrible

nossa mano

I don't think I understand, if you know the maps in the complex, homology is literally just ker/im

cohomology

oh wait, you're saying you know the chain complex defining homology, but want to compute cohomology
yeah ok, if you wanna avoid universal coefficients, dualize the complex
remember that if your groups are given as Z^n and the maps as matrices, dualization is literally just transposition

5:15 PM
its not like I want to avoid it
@Thorgott perfect that's useful
thank you!

everybody say "thank you thorgott"

thank you thorgott

tbh I'd like to avoid that pseudonym completely

Hey, a few years ago I stumbled across a blog post written probably by a mod on stackexchange stating that the quality of a community decreases as its popularity increases. It was liked somewhere on meta math stackexchnage regard to allowing low quality question. Does any one have that link?

@Thorgott aka thunderjahwe

5:22 PM
where can I find that bounded measurable functions can be embedded as a weakly closed subspace of dual space to finite signed measures?

between which kinds of spaces? Arbitrary measurable into $\Bbb R$ or some normed space?

measurable into R, yes
we assume some measurable space is given

@Jakobian Nice avatar. Where is this from?

we embedd bounded measurable functions into the dual space of signed measures, by defining $l_f(\varphi) = \int f(x)\varphi(\mathrm{d}x)$

I'm looking for the function $X\left(\omega\right) = \int_{-\infty}^{\infty}{x\left(t\right)e^{-i\omega t}dt}$, where $x(t)=Ae^{Ct}$ for $t\geq 0$ and $0$ otherwise. $C$ is a constant. Can this Fourier transform be found somewhere in the tables given here (en.m.wikipedia.org/wiki/…) and if so which?

5:35 PM
Isn't this Riesz-Markov-Kakutani? Or was that for $C(X)^*$?

I am wondering how to prove that this gives a weakly closed subspace. All the author gave is that if $l_{f_n}\to l$, then $f(x) = \lim f_n(x) = \lim l_{f_n}(\delta_x) = l(\delta_x)$
and now it's supposed to be that $l = l_f$
convergence in the weak sense
@BalarkaSen I don't remember

Well, if $l$ and $l_f$ agree on the Dirac mass measures, they have to agree on all measures, right? Dirac measures span a dense subspace of all the measures in the weak-$*$ sense

lol

5:40 PM
Now we know your neighborhood, @Astyx.

lol

futuristic indeed

People don't realize how ahead France really is

0

Let $X$ be a cell complex that is a union of two contractible subcomplex whose intersection is contractible. Then $X$ is contractible. I found some solution: Let $X = U\cup V$ with $W = U\cap V$, where $U,V,W$ are all contractible subcomplexes. First consider the quotient map $q:X\to X/W=:X_1$ ...

@Jakobian f(x)?

5:43 PM
Is there anyone can answer this questions?

This assumes that the functions are defined on $\Bbb R$?
Since the dirac measure must be defined on $\Bbb R$

@schn I guess this is 205 in the linked table?

Ah okay no sorry I messed up, the signed measures are just on the measurable space

dirac measure can be defined on any space
you just grab a point, no

yeh yeh

5:52 PM
@BalarkaSen idk. But there's also question whetever f is bounded in the first place

That is valid.
I don't know why it's bounded
But if it is, what I said works

6:24 PM
So the set of all coinfinite subsets of the naturals... should be uncountable?
Probably a diagonal argument
They are obviously in correspondence with infinite subsets so no need for the co
Yeah kinda obvious now

@BalarkaSen do you want a trivial question

7:10 PM
@LeakyNun Wasn't that it?
2

perhaps

7:24 PM
@Thorgott just dualizing by transposition sped up the computations by at least 300%
thank you generalized linear algebra
was still instructive to do it one time explicitly

7:45 PM
nice

does zeroth relative cohomology give me 0 for path connected spaces as well

8:32 PM
@Astyx I hope you don't live under that waterfall/drainage pipe.

What are peoples thoughts on Eric Weinstein ?
Out of interest
there must be some differential geometry people here

@BigSocks If by coinfinite you mean its complement is infinite, then the even integers are coinfinite and the power set of the even integers is uncountable. (Each element of the power set is also coinfinite)

8:50 PM
neat trick, thank you