Mathematics

11:00 AM

I'll just say that the only ultrafilters which are not free are principal ultrafilters. I.e., ultrafilters of the form $\mathcal U_a=\{A\subseteq X; a\in A\}$.

That is, we fix one point $a$ and take all sets containing $a$.

Alessandro Codenotti

@MartinSleziak $a\in A$ rather than $a\in X$

Martin Sleziak

Thanks. I've edited the typo.

If you take principal utlrafilter, then for every subset $A\subseteq X$ there is $B\in\mathcal U_a$ such that $A=^* B$.

You can simply take $B=A\cup\{a\}$.

Secret

Homotopy type theory

In mathematical logic and computer science, homotopy type theory (HoTT ) refers to various lines of development of intuitionistic type theory, based on the interpretation of types as objects to which the intuition of (abstract) homotopy theory applies. This includes, among other lines of work, the construction of homotopical and higher-categorical models for such type theories; the use of type theory as a logic (or internal language) for abstract homotopy theory and higher category theory; the development of mathematics within a type-theoretic foundation (including both previously existing ...

This is actually pretty visual

you turn everything in type theory into paths and points

I am guessing, the homotopies of A->B->C->A and A->B->C will be distinct

because one is a cycle

Balarka Sen

11:15 AM

Is there a finitely additive measure on $\mathcal{P}([0, 1])$ which extends the Lebesgue measure on the Borel subsets of $[0, 1]$?

@Alessandro @MartinSleziak

Martin Sleziak

@BalarkaSen Do you require it to be translation invariant?

Balarka Sen

No, I don't think so. Isn't it Borel's theorem that there does not exist a countably additive measure on the full algebra $\mathcal{P}([0, 1])$ such that $\mu([a, b]) = b- a$? I admittedly don't know the proof though

With the translation invariant assumption it's not possible because of the Vitali set business, as I recall

Martin Sleziak

Yes, if we have translation invariant, $\sigma$-additive and $\mu([a,b])=b-a$, then the argument showing that Vitali's set is not measurable goes through.

WP: Banach measure

> Stefan Banach showed that it is possible to define a Banach measure for the Euclidean plane, consistent with the usual Lebesgue measure. The existence of this measure proves the impossibility of a Banach–Tarski paradox in two dimensions:

> it is not possible to decompose a two-dimensional set of finite Lebesgue measure into finitely many sets that can be reassembled into a set with a different measure, because this would violate the properties of the Banach measure that extends the Lebesgue measure

Balarka Sen

Wow huh

So I assume this Banach measure is even translation-invariant

Alessandro Codenotti

See here for some details (exercise 2.15 in particular) spot.colorado.edu/~baggett/funcchap2.pdf

Martin Sleziak

11:22 AM

I do not know details about this. (I have only copied what I found.) I wished that Wikipedia article would be written more clearly.

Balarka Sen

Because you'd need that to argue the impossibility of Banach-Tarski in $\Bbb R^2$, I guess

So Banach-Tarski says there's no finitely additive translation-invariant measure extending the Lebesgue measure to the full sigma algebra for $\Bbb R^3$!

Martin Sleziak

Alessandro's link will be probably better source than my Wikipedia article.

Balarka Sen

Thanks, @Alessandro, having a look

Martin Sleziak

That's a nice observation. And this one is explicitly stated on Wikipedia: Banach–Tarski paradox.

> In fact, the Banach–Tarski paradox demonstrates that it is impossible to find a finitely-additive measure (or a Banach measure) defined on all subsets of an Euclidean space of three (and greater) dimensions that is invariant with respect to Euclidean motions and takes the value one on a unit cube. In his later work, Tarski showed that, conversely, non-existence of paradoxical decompositions of this type implies the existence of a finitely-additive invariant measure.

Balarka Sen

Yeah I never noticed this.

I vaguely remember this from some notes on amenability @Alessandro sent me once

But those are about paradoxical decompositions of countable groups and finitely-additive invariant measures on them

Martin Sleziak

11:29 AM

Amenability is one of the things which seem interesting, but I never got around to learn a bit more about this topic. (Beyond the question where the name comes from: Why is “Amenable Group” a pun?)

Alessandro Codenotti

Right, having a paradoxical decompositions and being amenable are opposite, so Banach-Tarski works in $\Bbb R^3$ because you can find a free group in the group or rotations and translations iirc

Amenability is weird in the sense that it's studied by people working in very different areas

ÍgjøgnumMeg

Wait, is $\frac{1}{2} \in \Bbb Z[\frac{1 + \sqrt{p}}{2}]$

Alessandro Codenotti

At least for p=1

ÍgjøgnumMeg

lol

Alessandro Codenotti

Actually not even

I misread your brackets (I'm on my phone)

ÍgjøgnumMeg

11:35 AM

Hmm that's annoying

Balarka Sen

@BalarkaSen Alright so Alessandro's exercises point out that this is just Banach-Tarski. Let $X = \mathcal{B}([0, 1])$ and let $Y \subset X$ be the subspace of Lebesgue measurable functions of bounded support. Then $I : Y \to \Bbb R$ defined by $I(f) = \int f$ is a linear functional which we essentially want to extend to $X$. Sup norm of $f$ is a sublinear function which dominates $I$

Then we get an extension $\tilde{I} : X \to \Bbb R$, and define $\mu(E) = \tilde{I}(\chi_E)$

ÍgjøgnumMeg

Oh no it doesn't matter, crisis avoided

Balarka Sen

The notes does a little better by making it translation invariant but using some general statement for Hahn-Banach

Cool

Sebastian Nielsen

11:49 AM

How do I solve this equation for $f$?
$$\lim_{n\to \infty} \frac{f(n)}{f(n+1)}2^n = 4$$

Do you know what this type of equation is called? So that I can look it up.

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$Mathematics$ Mathematics