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

## 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.

## 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.

## 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.

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 $$\dim V<\infty$$.