Mathematics - 2012-01-09 (page 1 of 2)

Asaf Karagila

00:00

@robjohn: Your deletion vote is needed here!

robjohn

@AsafKaragila Do I flag a mod for that?

Asaf Karagila

No... just hit the delete.

You should be able to do that, no?

t.b.

@AsafKaragila It may still be too early.

Asaf Karagila

Oh. Dang.

robjohn

@AsafKaragila I don't have the delete option there.

Asaf Karagila

00:02

Well, I guess we can wait one more day.

robjohn

I'm puny, not a mighty 20K+

t.b.

What's the hurry?

Asaf Karagila

I am an impatient guy!

@robjohn Then this one you can surely vent your rage on! :-P

robjohn

@AsafKaragila You are könig and syphilis was mentioned, so you might be anxious.

t.b.

I'm surprised by the most recent deletion ...

Asaf Karagila

00:04

Which one?

robjohn

@AsafKaragila I've already voted on it and it is closed. Is there something else I can do to it?

Asaf Karagila

@robjohn It's been several days so you should be able and vote for deletion :-P

robjohn

@AsafKaragila There, I voted to delete.

Asaf Karagila

Hah. The lists of closing and deletions are very symmetric.

Same first/middle/last names in both lists!

robjohn

@AsafKaragila Hmm. I guess I deleted a question that was not the question you wanted me to vote on. I still don't see a way to vote on that question.

Asaf Karagila

00:12

@robjohn I suggested both, this one is just too early for you...

robjohn

@AsafKaragila Ah, I see. Timing is everything.

Asaf Karagila

So they say.

QED

I sort of want to ask this on the site

do you think it's a reasonable question

hmm it's maybe too vague

robjohn

@QED this?

The $x^n$ question?

QED

specifically

given distance(x^{n-1},span{1,x,...,x^{n-2}}) >= e, prove a good bound distance(x^n,span{1,x,...,x^{n-1}}) >= f(e)

the problem is I'm totally unspecific about what the bound is

robjohn

00:18

What tools are allowed? Differentiation? Difference Calculus?

QED

everything you've got

but I'll think about it more, I can't ask it as is

robjohn

Suppose $x^n$ were in the span of $\{1,x,x^2,\dots,x^{n-1}\}$...

Then we could write $x^n=a_0+a_1x+a_2x^2+\dots+a_{n-1}x^{n-1}$...

Differentiate $n$ times.

QED

yeah

that's the proof I just gave earlier

except you wrote it in a more beautiful way

robjohn

You can use difference calculus in much the same way.

$\Delta^n$ applied to both sides instead of differentiation.

yoda

hello @AsafKaragila I'm trying to reproduce a bug in chat where room owners cannot explicitly grant write access to 1 rep users. Would you please be willing to do a quick test in this room?

QED

00:22

@robjohn, this isn't what i'm asking about though

robjohn

@QED Sorry, what is the question. I only saw some of the discussion.

Asaf Karagila

@yoda What do I have to do?

QED

we are trying to find error estimates of what's the closest we can approximate x^n using {1,x,...,x^{n-1}}

robjohn

@QED Ah, what norm?

yoda

@AsafKaragila On this page, click on "add user" under "Explicit write access" and add their user id. Hold on, I'll find a 1 rep user for you

QED

00:25

@robjohn, the same one as the thing we were talking about with Srivatsan

robjohn

@QED $L^1([0,1])$ norm?

yoda

@AsafKaragila Here's one such user. So when you add them, it should show their profile and a little orange button underneath (like those on yours and Marc's names). Click that and choose write access

QED

well I was hoping to consider an arbitrary (a,b)

that might make it harder though

Asaf Karagila

Not working, @yoda

yoda

@AsafKaragila Thanks for confirming :)

robjohn

00:30

@Asaf: good job :-)

QED

the proof that alpha^n is algebraic is pretty indirect, so I don't think math.stackexchange.com/questions/97493/… can be done

what degree is it? same degree [or less]?

the degree can't get bigger can it

robjohn

00:46

@QED To use the same idea as with Srivatsan, we would have $n$ equations in $n$ unknowns: $\displaystyle x_{n}^m-x_{n-1}^m+x_{n-2}^m-\dots+(-1)^{n-1}x_1=\tfrac12\text{ for all }m=1\dots n$

At least on the interval [0,1]

Asaf Karagila

Huzzah.

Finished!

robjohn

@AsafKaragila with?

Asaf Karagila

This

@Dylan: Hi.

Dylan Moreland

Hey Asaf.

Asaf Karagila

It only took me ~80 minutes!

robjohn

00:55

@AsafKaragila I don't understand it. It looks nice, though :-)

Asaf Karagila

@robjohn I'm hoping that it will get "peer reviewed" and approved by a couple of the people on MO. I sometimes feel that my answers there tend to get dragged into the technical sides.

robjohn

@AsafKaragila that's the site for it :-)

Asaf Karagila

@robjohn Actually not as much as I do...

Anyway... It's 3am and I am going to bed now!

Goodnight.

Srivatsan

01:23

@robjohn I figured out something.

robjohn

@Srivatsan what's that?

Srivatsan

The correct thing to do for the $x^2 - ax-b$ question is to centre it appropriately, so that the interval is $[-1, 1]$ and not [0,1]. Of course, this is not the original question, but we will fix that later.

OK. So, the new question is to find the best line approximating the parabola in the interval [-1, 1]. Since the function is even, it is natural to expect that the approximation to be even as well.

robjohn

why would the function be even?

Srivatsan

The function is $f(x) = x^2$ over [-1, 1]. That is even, no?

robjohn

yes, but we have $x^2+ax+b$

Srivatsan

01:27

Think of it this way: $x^2$ is the given function and we are finding the best fit line. I am claiming that for this line, $a$ should be zero.

robjohn

That is what one might expect, but I would like to see it proven.

Srivatsan

$$\| f(x) - p(x) \| = \| f(x) - p(-x) \| = \frac{\| f(x) - p(x) \| + \| f(x) - p(-x)\|}{2} \geqslant \left\| f(x) - \frac{p(x) + p(-x)}{2} \right\| .$$

By triangle inequality.

robjohn

Okay, that works.

QED

very nice

Srivatsan

@robjohn So, if the given function is $x^2$, then a best approximation is actually a horizontal line. So we are seeking to minimize $\| x^2 - b \|_1$. This is a well-known problem: the answer is that $b$ is the median of the random variable $U^2$ where $U$ is uniform([-1, 1]).

That works out to be $\frac{1}{4}$. So the best approximation is $p(x) = \frac{1}{4}$.

robjohn

01:34

That works because you know a fact (that I did not know) about a random distribution. Otherwise, one would need to compute that minimum. Nice.

Srivatsan

Yes, you can also compute the minimum directly. Either way, in this case, the problem is very tractable.

@robjohn You recognize that $\| x^2 - b \|_1$ is the average deviation of $x^2$ from $b$, no? I am just saying that the $b$ minimising this average deviation is the median. (Proof is easy by differentiation.)

QED

I'm wondering about if you have $\|x^n - p(x)\| > h$ for all $p$ of degree $< n$.

Srivatsan

What do you mean?

QED

can you find a bound $\| x^{n+1} - p(x) \| > h'$ for all $p$ of degree $< n+1$.

Srivatsan

Not sure.

QED

01:38

h' is in terms of h, n, and the interval

of course we have proved that 0 works, but would be nice to find something stronger

Srivatsan

@robjohn There is one more step: translate this to the original problem. That is easy (make the substitution $2x-1 = y$ or something like that).

QED

we can think of the problem as something like "we can only approximate the derivative of x^{n+1} this well, how well can we approximate the function itself?"

does a good approximation necessarily mean a good approximation of the derivative too?

if the derivatives are too different, the functions will curve away from each other.. but they might not have be as close as possible

My number is 1729 today

t.b.

@QED this number seems to me rather a dull one

Srivatsan

@robjohn: By the way, I have a feeling that in the $x^n$ case, the roots might turn out to be: $\frac{2k-1}{n}$ for $k=1..n$. Just like it was $\frac{1}{4}, \frac{3}{4}$ in the $x^2 +ax+b$ case.

QED

why are you restricting interest to [0,1]?

Srivatsan

01:48

What are the other options?

QED

[a,b]

Srivatsan

No, thank you. I like $[-1, 1]$ better, but $[a, b]$ is needless obfuscation.

QED

how is it obfuscation

Srivatsan

Note that by suitable scaling and shifting, one can transform the solution for $[a, b]$ to any other interval $[a',b']$. So I would like to restrict to the simplest possible interval.

QED

I agree you can scale but I don't see how shifting is possible

t.b.

01:51

what are you guys up to anyway?

Srivatsan

@QED Well, it is in fact possible. =) The idea is that the $\mathrm{error}(f) = \mathrm{error}(f+p)$ for any polynomial $p$ of degree $< n$.

@tb Well, are you asking about the current discussion?

QED

oh very good, I thought about shifting before and convinced myself it wasn't possible somehow

t.b.

yes, I lost the question you're trying to answer somehow.

Srivatsan

@tb This is related to our discussion yesterday. Question is to find the distance of $f(x) = x^n$ from the space of polynomials of degree $< n$. The metric is $L^1([-1, 1])$.

QED

the further we shift to infinity, the closer to linear the polynomials become

Srivatsan

01:56

@QED Indeed it also took me a little bit of time to convince myself... =)

@tb We managed to solve the problem for $n=2$. Hurray!

Benjamin Lim

02:25

hello

t.b.

hi

Benjamin Lim

you're around :D

I've been confused the past few days over irreducibles/primes/units/UFDs/PIDs/etc..

t.b.

@BenjaminLim I heard people say that this happens, yes :)

Benjamin Lim

What do you mean by "you heard..."?

t.b.

I read chat messages aloud to myself :)

@BenjaminLim I fear that you already know more commutative algebra than I'll ever know.

(so wrong person)

Benjamin Lim

02:29

@tb Not possible I am only just starting plus everything is self taught so it's like a hard slog atm

I tend to be bogged down in a heap of details

For example, I am confused when two books give different definitions of a UFD.

Zhen Lin

That's normal, I think. I didn't appreciate commutative algebra the first time I learned it, and I never could remember the details back then.

And then I learned algebraic geometry!

Benjamin Lim

@ZhenLin That's what I am trying to do now :D

I am going to do a reading course on it next semester because my uni does not offer any commutative algebra

and the supervisor is freaking me out by saying he will give me exercises from Atiyah Macdonald

QED

@BenjaminLim, check they are equivalent

Zhen Lin

What you're talking about is very basic and usually not covered under the heading of commutative algebra.

QED

i.e. if one book defines it as A and another as B. Then prove A => B and B => A

Benjamin Lim

02:32

@ZhenLin I know it's not commutative algebra, it's like what you need before tackling it

t.b.

now that we got the label straight, what are the two definitions?

Benjamin Lim

ok, well for a UFD we have two conditions, my problem is with the first one

t.b.

well, marko, ex falso quodlibet

Benjamin Lim

Isaacs: Every non-zero NON-UNIT is a product of irreducible elements

Zhen Lin

Well, some people would probably say that a unit is a product of zero irreducible elements...

Benjamin Lim

02:36

@ZhenLin I've seen some authors do that, why is it the case?

@QED That's what I did when trying to see why the T1 axiom is equivalent to every finite point set being closed. I did not see it in the beginning.

Zhen Lin

Why not? 1 is the empty product.

QED

topology?

t.b.

like 0 is the empty sum.

QED

but there can be units other than 1

-1, i for example

Zhen Lin

Sure, but if $x$ is irreducible, so is $u x$ for any unit $u$.

t.b.

02:38

uniqueness is up to units

Zhen Lin

So this whole factorisation business should be considered modulo units anyway.

QED

oh right, ux and 1/x are irreducibles with product u

no that's not a product of zero irreducible elements

Zhen Lin

If $1/x$ exists then $x$ cannot possibly be irreducible...

Benjamin Lim

Now on the other hand, in C Musili's rings and modules, he defines a UFD to be

(1) Every non-zero element is the product of a unit and irreducible elements

Zhen Lin

That's the correct definition, for pedants.

Benjamin Lim

02:40

does it really matter how Isaacs defines it or how it's defined here?

QED

(1) ?

Benjamin Lim

The first condition in a UFD, not talking about the uniqueness of the factorisation

that's condition (2)

Zhen Lin

Both of those conditions are equivalent. (Easy proof.)

QED

@Srivatsan, we may use [-1,1] for even n and [0,1] for odd n

Srivatsan

What's the use of using [0,1] for odd n?

t.b.

02:53

@Ben: I was rather surprised to see you invest a bounty for that Gauss-Bonnet question. What's your motivation?

Srivatsan

Badge. =)

tb: BTW, are you free now? Can we finish off from where we stopped yesterday?

Benjamin Lim

@Srivatsan It's not that. A friend of mine is doing an honours thesis with Bryan Wang. I was trying to help him out (He saw the question on this site and was interested in it).

t.b.

@Srivatsan I'm there...

@BenjaminLim I see. I had a look at the passage and (as always) Connes's remarks are rather cryptic to me.

Srivatsan

@tb Just a minute. I am making an edit. I will come in a minute.

t.b.

Take your time

robjohn

02:59

@Srivatsan I think I have a proof of that...

I was thinking about it while walking the dog.

Srivatsan

@robjohn Aw, nice.

I'm in the other room for some time. See you in a bit.

Benjamin Lim

@tb I think I am beginning to get it now, slowly

Benjamin Lim

03:18

@ZhenLin Say we take the less pedantic condition of a UFD. Then if I want to show that every irreducible element $a$ is prime in a UFD, I suppose that $a|xy$. Why can I assume that $x,y$ are not units?

Zhen Lin

Because we can divide units away.

Benjamin Lim

I don't understand how this follows from the definition of a unit

Zhen Lin

A unit is an invertible element.

Benjamin Lim

yes

I really don't get it.

Zhen Lin

The point is, if $u$ is a unit, and $a \mid u x$, then $a \mid x$.

Benjamin Lim

03:26

@ZhenLin I get it. If $a| ux$, then $ab = ux$ for some $b$ in the ring, since $u$ is a unit then $x = a(bu^{-1})$ so that $a | x$.

Thanks.

t.b.

05:17

@ZhenLin: are you around?

Potato

Can someone explain why Abel's theorem, giving a criterion for a divisor to be principal, is important?

t.b.

05:37

@Potato seen this?

Potato

@t.b. Woah! That looks like an excellent link!

Zhen Lin

@tb Yes, now I am.

t.b.

Great! I have a somewhat vague question: when you have a monoidal category then you have a notion of algebra. Have you ever seen "algebras" without unit treated in that framework?

What I'm interested in is a categorical formulation of approximate units in Banach algebras and I was never really able to capture that in a nice way categorically

Zhen Lin

Hmmm, no, I'm afraid I haven't.

But does it really make sense? I mean, a monoid object always has a unit...

t.b.

I agree. The thing is that I don't want a unit. It makes perfect sense to have a map $m: A \otimes A \to A$ and require it to be associative, no?

Zhen Lin

05:52

Sure. That would be some kind of semigroup, I guess.

t.b.

Yes. The thing is this: many Banach algebras don't have a natural unit (e.g. compact operators on a Hilbert space) or convolution algebras of non-discrete groups. Adjoining a unit is of course possible but not really the thing I want to do. What suffices for many analytic purposes is an approximate unit.

Zhen Lin

Perhaps there's something in the theory of semigroups which is similar?

t.b.

I wouldn't have seen that. Close but not that close would be local units in the theory of non-unital rings (every finitely generated subring has a unit).

Benjamin Lim

@ZhenLin Thanks for your help :D

t.b.

The thing is that having approximate units is a distinctly analytic one: in the usual formulation you say that you have a net of elements $u_i$ such that $\|u_i a - a\| \to 0$ as well as $\|a u_i - a \| \to 0$ for all $a \in A$.

Zhen Lin

06:00

Hmmm... that does sound difficult to categorify.

t.b.

It's a mess...

Zhen Lin

But, surprising things happen sometimes! Like the fact that compact Hausdorff spaces are monadic over sets...

t.b.

You can see the elements of $A$ as morphisms $a: 1 \to A$, so what you want is a family of morphisms $u_i : 1 \to A$, but I don't know how to express this approximation property nicely.

Zhen Lin

The category of Banach spaces must be enriched over something. Doesn't that give you some wiggle room to express convergence?

t.b.

Well, it's enriched over itself and over topological spaces. But convergence isn't in norm but rather in the strong operator topology.

Zhen Lin

06:06

So approximate units don't converge to the identity operator? Hm.

t.b.

They do but not in norm. You have pointwise convergence, not uniform convergence.

Zhen Lin

If we change the topology on the hom-spaces to pointwise convergence, what breaks?

t.b.

My ability to phrase that abstractly.

Maybe this example helps: think of a locally compact space $X$ and let $A = C_0(X)$ be the algebra of functions vanishing at infinity. You can easily find an approximate unit consisting of functions with compact support just by making the support larger and larger. But for every $u_i$ you'll find a function such that $u_i a = 0$, so $\|u_i a - a\| = \|a\|$.

Zhen Lin

Ah, OK.

Here's what I'm thinking: if we have a commutative non-unital associative algebra $A$, then the exponential transpose of multiplication gives us a map $A \to \textrm{End}(A)$, and this map is injective if multiplication is cancellative. I'm not well-versed in functional analysis, but I imagine this still works if each of these things has the right topology.

Let's say that an operator $A \to A$ is "representable" if it is multiplication by some scalar. These "approximate units" seem to be something akin to "pro-representable" operators.

But you've just told me that the topology of operator convergence is too strong for it to actually be convergence there...

t.b.

Yes, this is a good idea. In fact the presence of approximate units ensures that the map (call it left regular representation) is injective.

But as you noted, the scalars aren't in the norm closure but only in the strong closure of the image.

And what's worse is that with the strong topology you don't get a topological algebra structure on $\operatorname{End}{(A)}$ in general, so things aren't that nice.

Zhen Lin

06:25

Ah. Hmmm...

t.b.

But maybe one has to be just a bit more careful: when I think of $\operatorname{End}{(A)}$ I think of the entire algebra. However, if I'm actually working with the category of Banach spaces and maps of norm at most one, then the set of endomorphisms is bouned, and then I get a topological monoid with respect to the strong operator topology. But it still looks like a major contortion.

Zhen Lin

It wouldn't be so bad if each of the approximate units were such an endomorphism, though.

t.b.

It doesn't hurt me to assume that each $u_i$ has norm $\leq 1$ (so it yields such a morphism). I lose some algebras that people are interested in but none I am interested in.

Anyway: it will not look as nice as I would like it to be. It's a technical condition which is easy and natural enough to state in its original framework, and it bothers me that the abstract theory doesn't give me a neat way to encompass it.

Thanks for listening!

Zhen Lin

No problem.

Amusingly, I just worked out that the correct analogue of the Yoneda embedding here must actually be the double dual embedding.

Nikhil Bellarykar

hi $\forall$

t.b.

06:40

@ZhenLin what do you mean?

Zhen Lin

Well, the Yoneda embedding is $\mathbf{C} \hookrightarrow [\mathbf{C}^\textrm{op}, \textbf{Set}]$

Nikhil Bellarykar

I assume guys here have that fibonacci-with-partition question in mind, dont you all?

t.b.

I don't

Nikhil Bellarykar

ok I will repost it then?

We call an ordered Partition of a positive integer $n$ as the way of writing $n$ as a sum of one or more positive integers, where the order of the sum DOES matter. For example, there are $4$ ordered partitions of $3$, namely $1+1+1, 1+2,2+1,3$

Now suppose we replace the last term of each above partition with a $1$, multiply the terms of each individual partition, then add the results all together. In this manner we get: $(1\cdot 1 \cdot 1 )+ (1 \cdot 1 )+ (2 \cdot 1) +(1)$ respectively, which equals $5$, which curiously is a Fibonacci number! Can someone please explain why this result is a…

t.b.

@ZhenLin yes, in fact $\mathbf{C}^{\operatorname{op}}$ has a nice description: it's isomorphic to the unit ball in the dual with weak topology and continuous maps that preserve convex combinations and zero

Zhen Lin

06:45

@tb: Sure, but what I'm thinking of is lowering the dimension of everything: so replace $\mathbf{C}$ with an object of the category, $\textrm{op}$ with some other endofunctor, and $\textbf{Set}$ with some coseparating object in the category.

In the category of sets this gives us the double powerset embedding $X \to [\mathscr{P}(X), 2]$.

t.b.

@ZhenLin well, that's what I'm doing! I'm taking the ground field which is coseparating by Hahn-Banach and the morphisms are the points in the unit ball in the dual

Nikhil Bellarykar

@tb my question is regarding an 'informative, non-inductive proof' of above thing I posted.

t.b.

(but maybe that's what you were thinking of)

@NikhilBellarykar don't you see that we're talking about something completely different at the moment?

Zhen Lin

Ah. I was thinking of replacing $\textrm{op}$ with $\textrm{Hom}(-, \mathbb{R})$, and $\textbf{Set}$ with $\mathbb{R}$ as well.

Nikhil Bellarykar

@tb true. I will wait

t.b.

06:48

@ZhenLin Yes but now remember that we're only considering morphisms of norm $\leq 1$

Zhen Lin

Ah, right.

I was still thinking about the category of bounded linear maps.

t.b.

Well, the bounded linear maps is what we get when we look at the internal hom

We need to be slightly careful: I'm working with Banach spaces and maps of norm $\leq 1$. Now this category is not enriched over itself but the category of Banach spaces and bounded linear maps is. Problem is: The former is bicomplete, the latter only finitely bicomplete

Zhen Lin

Bleh, I think I've confused myself. Anyway, I must leave now...

t.b.

Maybe I confused you, that's more likely. Thanks again!

Nikhil Bellarykar

07:09

@tb can we talk now?

Potato

Does it make sense to talk of the deRham cohomology of the projective line $\mathbb{P}^1$, where we consider it as a real manifold by taking the complex charts as charts into $\mathbb{R}^2$?

Asaf Karagila

Top of the morrow to you!

Potato

Hello, Asaf!

Asaf Karagila

What's up?

t.b.

@AsafKaragila People are dragging me into your kingdom!

Potato

07:15

@AsafKaragila I'm thinking about the deRham chomology of the projective line, if that even makes sense.

Asaf Karagila

@Potato It doesn't. Why would I care about it?

@tb Step into my web!

t.b.

Projective line of what?

Potato

@AsafKaragila Why doesn't it make sense?

Asaf Karagila

You could have just linked them to one of the many many many many posts in which I blabbed about these strange sets.

@Potato It doesn't make any sense to tell me about it...

Potato

@tb By projective line, I mean $\mathbb{P}^1$.

@AsafKaragila You asked what was up!

Asaf Karagila

07:19

I wonder if anyone has bothered to read what I posted on MO last night and whether or not the vote indicates that the proof is valid.

t.b.

@Potato $\mathbb{P}^1(\mathbb{R})$ or $\mathbb{P}^1(\mathbb{C})$? Anyway: every complex manifold is in particular a real manifold

Asaf Karagila

@tb That's just psychosis. You're saying that imaginary things are real.

Potato

@t.b. Right, but I was wondering if the de Rham cohomology was defined on it, given that it is nonorientable (does that make a difference?) and how to compute it.

Asaf Karagila

@tb May I post one more choice comment on your answer?

t.b.

@AsafKaragila if you give me links to answers where you prove that there is an uncountable set such that every subset is either countable or has countable complement, then yes. Otherwise, this is full-force choice-land.

Asaf Karagila

07:23

@tb Yeah, I was gonna supply some links. I don't mind if you post that comment if you prefer it that way.

t.b.

@AsafKaragila I would add it to the answer (and acknowledge your help!) if you give me the links here :)

Asaf Karagila

No problems, no acknowledgment is needed either ;-)

t.b.

@Potato orientability doesn't enter the definition of de Rham cohomology (for compact connected manifolds it's equivalent to $H^n(M) = \mathbb{R})$). Either way: your question doesn't make sense to me: $\mathbb{P}^1(\mathbb{R}) = S^1$ and $\mathbb{P}^1(\mathbb{C}) = S^2$ are both orientable.

(in general, complex manifolds are orientable)

Potato

@t.b. Indeed. I am being stupid. I was thinking real projective plane and typing the wrong thing.

Of course $P^1$ is orientable, because it's a Riemann surface. I should have caught that.

t.b.

The de Rham cohomology of $\mathbb{P}^2(\mathbb{R})$ is zero except in degree zero where it's $\mathbb{R}$.

(the fundamental group is finite, so $H^1$ is zero and it's non-orientable, so $H^2$ is zero)

Potato

07:30

Ah, ok. Apparently this is an exercise in Bott and Tu, actually.

@t.b. Could you give suggestions for fun things to compute the de Rham cohomology of?

I am quite a fan of Mayer-Vietoris.

t.b.

Have you done the orientable $2$-dimensional surfaces and the $n$-dimensional torus?

Potato

@t.b. Yes to the first. What is the n-dimensional torus?

t.b.

$(S^1)^n$.

Potato

Thank you. I will see what I can do.

t.b.

Compute the Betti numbers.

Asaf Karagila

07:56

@tb The most relevant post that I found is this answer.

And there is this

t.b.

@AsafKaragila Thanks. Very silly question: if the reals are a countable union of countable sets does this hold for every set of reals?

It should, but the argument I'd use for that does seem to use countable choice which it shouldn't!

Asaf Karagila

Of course.

Take the partition of the reals and just intersect with the subsets.

Subsets of countable sets are always countable.

(or finite, of course)

t.b.

I was being silly. I wanted to create a counting function for that set...

Asaf Karagila

:-D

t.b.

08:13

@AsafKaragila Didn't François G. Dorais have an answer on $\sigma$-algebras generated by singletons somewhere?

Asaf Karagila

@tb You want this

t.b.

@AsafKaragila definitely not my day :) Thanks!

t.b.

08:33

@AsafKaragila I chose the lazy way out.

Asaf Karagila

:-P

robjohn