Asaf Karagila
I don't have much choice...

Question on Dually Dedekind-infinite sets

I got an email a few days ago from Lucas Polymeris, a Chilean student, who asked me a very nice question. I want to go through the journey I took from that question to the answer.

Before we get to the question, let's review some definitions.

Zornian Functional Analysis coming to arXiv!

Back in autumn 2015 I took a functional analysis course with Prof. Matania Ben-Artzi, and he let me write a term paper about uses of the axiom of choice in functional analysis for my final grade. One year later, in October 2016, I finally posted the note here. It then received some feedback from some people, and about a year after that I posted a small revision.

Earlier this week I suggested my note as a source for the proof that the Baire Category Theorem is equivalent to Dependent Choice. After doing that, I stumbled upon an errata by Theo Bühler and Dietmar A. Salamon to their Functional Analysis book, which refers to my write-up.

Countable sets of reals

One of the classic results of Sierpinski is that if there are as many countable sets of reals as there are reals, then there is a set which is not Lebesgue measurable. (You can find a wonderful discussion on MathOverflow.)

This is fact is used in the paradoxical decomposition theorems (which I often enjoy bringing up as a counter-argument to bad arguments that the Banach–Tarski paradox implies we need to accept that all sets are measurable as an axiom):

Cohen's Oddity

We all know and love Cohen's first model where the axiom of choice fails. It is the O.G. symmetric extension. But Cohen didn't invent the idea on his own, he used Fraenkel's ideas from his work on set theory with atoms and permutation models. The two results, however, are significantly different.

Fraenkel's construction does not affect sets of ordinals, in particular the real numbers can still be well-ordered in his models. Cohen's work, however, directly breaks that. The Dedekind-finite set added is a set of reals. In particular, the reals cannot be well-ordered no more.

Definable Models Without Choice

Suppose that a parameter formula defines an inner model. Does that inner model satisfy choice?

Well, obviously, if choice failed then the answer is no, just by taking $$x=x$$. But what if we remove that option. Namely, if the inner model is not the entire universe, then choice holds.

What a long strange trip it's been...

As some of you may have noticed, I don't use this blog to write about my papers in the "traditional way" math bloggers summarize and explain their recent work. I think my papers are prosaic enough to do that on their own. I do use this blog as an outlet when I have to complain about the arduous toil of being a mathematician (which has an immensely bright light side, of course, so in the big picture I'm quite happy with it).

This morning I woke up to see that my paper about the Bristol model was announced on arXiv. But unbeknownst to the common arXiv follower, this also marks the end of my thesis. The Hebrew University is kind enough to allow you to just stitch a bunch of your papers (along with an added introduction) and call it a thesis. And by "stitch" I mean literally. If they were published, you're even allowed to use the published .pdf (on the condition that no copyright infringement occurs).

Zornian Functional Analysis or: How I Learned to Stop Worrying and Love the Axiom of Choice

Back in the fall semester of 2015-2016 I had taken a course in functional analysis. One of the reasons I wanted to take that course (other than needing the credits to finish my Ph.D.) is that I was always curious about the functional analytic results related to the axiom of choice, and my functional analysis wasn't strong enough to sift through these papers.

I was very happy when the professor, Matania Ben-Artzi, allowed me to write a final paper about the usage of the axiom of choice in the course, instead of taking an exam.

Cofinality and the axiom of choice

What is cofinality of a[n infinite] cardinal? If we think about the cardinals as ordinals, as we should in the case the axiom of choice holds, then the cofinality of a cardinal is just the smallest cardinality of an unbounded set. It can be thought of as the least ordinal from which there is an unbounded function into our cardinal. Or it could be thought as the smallest cardinality of a partition whose parts are all "small".

Not assuming the axiom of choice the definition of cofinality remains the same, if we restrict ourselves to ordinals and $$\aleph$$ numbers. But why should we? There is a rich world out there, new colors that were not on the choice-y rainbow from before. So anything which is inherently based on the ordering properties of the ordinals should not be considered as the definition of an ordinal. So first let's recall the two ways we can order cardinals without choice.

When the box means nothing

When assuming the axiom of choice the product topology and box the topology are quite different when considering infinite products. For example the Tychonoff product of countably many sets of three elements is compact, metrizable an all in all a very nice space. On the other hand, the box product is not separable or second countable at all.

But without the axiom of choice the world is indeed a strange place. This was posted as answer on math.SE earlier today.

On the Partition Principle

Last Wednesday I gave a talk about the Partition Principle in our students seminar. This talk covered the historical background of the oldest open problem in set theory, and two proofs that for a long time I avoided learning. I promised to post a summary of the talk here. So here it is. The historical data was taken from the paper by Banaschewski and Moore, "The dual Cantor-Bernstein theorem and the partition principle." (MR1072073) as well Moore's wonderful book "Zermelo’s Axiom of Choice" (which has a Dover reprint!).

To Colloops a cardinal

This is nothing new, but it's a choice-y way of thinking about it. Which is really what I enjoy doing.

Definition. Let $$V$$ be a model of $$\ZFC$$, and $$\PP\in V$$ be a notion of forcing. We say that a cardinal $$\kappa$$ is "colloopsed" by $$\PP$$ (to $$\mu$$) if every $$V$$-generic filter $$G$$ adds a bijection from $$\mu$$ onto $$\kappa$$, but there is an intermediate $$N\subseteq V[G]$$ satisfying $$\ZF$$ in which there is no such bijection, but there is one for each $$\lambda\lt\kappa$$.

Anti-anti Banach-Tarski arguments

Many people, more often than not these are people from analysis or worse (read: physicists, which in general are not bad, but I am bothered when they think they have a say in how theoretical mathematics should be done), pseudo-mathematical, non-mathematical, philosophical communities, and from time to time actual mathematicians, would say ridiculous things like "We need to omit the axiom of choice, and keep only Dependent Choice, since the axiom of choice is a source for constant bookkeeping in the form of non-measurable sets".

People often like to cite the paradoxical decomposition of the unit sphere given by Banach-Tarski. "Yes, it doesn't make any sense, therefore the axiom of choice needs to be omitted".

No uniform ultrafilters

Earlier this morning I received an email question from Yair Hayut. Is it consistent without the axiom of choice, of course, that there are free ultrafilters on the natural numbers but none on the real numbers?

Well, of course that the answer is negative. If $$\cal U$$ is a free ultrafilter on $$\omega$$ then $$\{X\subseteq\mathcal P(\omega)\mid X\cap\omega\in\cal U\}$$ is a free ultrafilter on $$\mathcal P(\omega)$$. But that doesn't mean that the question should be trivialized. What Yair asked was actually slightly subtler than that: is it consistent that there are free ultrafilters on $$\omega$$, but no uniform ultrafilters on the real numbers?

Equivalent to the axiom of choice that I didn't know about

First I must apologize. I wanted to write a second post about forcing and preserving choice principles (I gave a nice talk in the student seminar about a week after the previous post), and I had a lot of things to say. I just ended up not writing it, and for absolutely no good reason. And somehow things continued that way and I felt more and more awkward to post anything because of that, but the vicious cycle must break somewhere.

I recently tried to figure out the consequence of some forcing in $$\ZF$$. This has led me to the following statement:

Strong chain conditions and preservation of choice principles

I recently returned from a wonderful week in Italy, where I attended the Young Set Theorist 2013 conference. I met a lot of new people, some old acquaintances, baffled people with oversized pickles, and most importantly shared and learned some great ideas.

One of the nicer things I'd done was to work with Thomas Johnstone on some preservation theorem related to forcing and choice principles (see also this announcement by Victoria Gitman). In order to clean up a bit the proof, I'll introduce a new definition which is going to slightly extend the ideas originally discussed in Italy. So without further jibber jabber, let's talk mathematics.

Provable Equality Of Exponentiation

It's an almost trivial theorem of cardinal arithmetics in $$\ZF$$ that given four cardinals, $$\frak p,q,r,s$$ such that $$\frak p<q,\ r<s$$ we have $$\frak p^r\leq q^s$$.

In a recent question on math.SE some user has asked whether or not we always have a strict inequality. Everyone sufficiently familiar with the basics of independence results would know that it is consistent to have $$2^{\aleph_0}=2^{\aleph_1}=\aleph_2$$, in which case taking $$\mathfrak{p=r}=\aleph_0,\ \mathfrak{q=s}=\aleph_1$$ gives us equality. But it's also trivial to see that we can always pick cardinals whose difference is large enough to keep the inequality true.

Choice Principles: What are they?

What does the phrase "$$\varphi$$ is a choice principle" mean? This is something that I have spent quite a lot of my time thinking about. Directly and indirectly. What are choice principles as we know them? And who gets to decide?

For a set theorist, at least a "classical" set theorist (working within the confines of $$\ZF$$ and its extensions to $$\ZFC$$ and so on), a choice principle can aptly be defined as "Sentence $$\varphi$$ in the language of set theory which is provable from $$\ZFC$$ but independent from $$\ZF$$". Indeed that is how I think of choice principles, and how I referred to them in my masters thesis (albeit I prefaced that definition by pointing out its naivety).

Infinite dimensions and the axiom of choice

In a recent math.SE question, Thomas Andrews asked whether or not the existence of an infinite linearly independent set in a vector space which is not finitely generated requires the axiom of choice.

The answer is positive. It does require the axiom of choice. The counterexample is due to Läuchli who constructed a model in which there was a vector space which was not finitely generated, but every proper subspace is finitely generated. Given such vector space it is obvious that no infinite set can be linearly independent.

Vector Spaces and Antichains of Cardinals in Models of Set Theory

I finally uploaded my M.Sc. thesis titled “Vector Spaces and Antichains of Cardinals in Models of Set Theory”.

There are several changed from the printed and submitted version, but those are minor. The Papers page lists them.

The Philosophy of Cardinality: Pathologies or not?

What are numbers? For the layman numbers are those things we use for counting and measuring. The complex numbers are on the edge of being numbers, but that's only because they are taught in high-schools and many people still consider them imaginary (despite them having some reasonably applicative uses).

But a mathematician knows that a number is basically a notion which represents a quantity. We have so many numbers that I don't even know where to begin if I wanted to list them. Luckily most of the readers (I suppose) are mathematicians and so I don't have to.

The Axiom of Choice and Self-Dual Vector Spaces

I have uploaded a note titled The Axiom of Choice and Self-Duality of Vector Spaces. Here is a short summary and background.

It is a well known fact (in $$\ZFC$$ at least) that if $$V$$ is a vector space, and $$V^\ast$$ is the algebraic dual of $$V$$ then $$V\cong V^{\ast\ast}$$ if and only if $$\dim V<\infty$$.