# The Axiom of Choice and Self-Dual Vector Spaces

There are 5 comments on this post.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 \(\dim V<\infty\).

Some long time ago, after a discussion with Pete L. Clark on math.SE, I set about finding a counterexample. After giving up at first, I found about *automatic continuity*. It turns out that in models like Solovay's model every linear functional from a Banach space to the field (real numbers or complex numbers) is automatically continuous. In such model, if so, any reflexive Banach space is also isomorphic to its double dual.

I began writing a short note with all the relevant theorems and information, hoping to find my first publication there, but alas as I was finishing I saw that this result (and more) was already covered in Schechter's immense book *Handbook of analysis and its foundations*. Now that I have a website, it seems like a good reason to make final adjustments and upload the note.

This is my first note, and any comment or suggestion will be most helpful for future notes (which are coming, I can assure you).

**Summary:**

In models where every set of real numbers has the Baire property it turns out that every linear operator from a Banach space to a separable normed space is automatically continuous. In particular every linear functional is automatically continuous, and therefore the algebraic dual and the topological dual are the same.

Let \(V\) be a vector space over \(\mathbb R\). Denote by \(V^\ast\) the algebraic dual of \(V\), and for a topological vector space denote by \(V^\prime\) the continuous dual. We will show that the following implications are not provable without the axiom of choice:

- \(V\cong V^\ast\) implies that \(\dim V<\infty\);
- \(V\cong V^{\ast\ast}\) by a natural isomorphism if and only if \(\dim V<\infty\);
- If \(V\) is a Banach space, \(V^\prime\) is reflexive if and only if \(V\) is reflexive;
- If \(V\) is a reflexive Banach space, \(W\subseteq V\) is a closed subspace, then \(W\) is also reflexive;
- If \(V^\prime\) is separable then \(V\) is separable.
The Axiom of Choice and Self-Duality of Vector Spaces

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## There are 5 comments on this post.

(Oct 29 2012, 23:56)

That's nice. Have you considered convertin this to HTML? or markdown (using Pandoc)? Then you could post it in its full glory :)

(Oct 30 2012, 00:12 In reply to Peter)

Well, I am not a huge fan of reading things off a website. I am more inclined towards using a PDF reader (even on my iPhone!). Seeing how it's only two and a half pages, I might consider doing that. But generally I don't find this as an appealing option for the moment.

(Oct 30 2012, 02:33 In reply to Asaf Karagila)

Well, a real alternative would be an epub3 file, but that's after you've got some html...

(Oct 30 2012, 02:39 In reply to Peter)

I'm not sure I see the problem with .pdf files. It is, after all, a portable document [format]. Had you said that ten years ago when I was fighting my computer for every bit of free RAM I might had understood that, but nowadays? I don't... I suppose this is a whole other discussion. :)

(Oct 30 2012, 06:22 In reply to Asaf Karagila)

Ha, yes, I remember those day :) Yes, this is a whole other discussion and a long overdue post of mine...