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 $\dim V<\infty$. 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:

  1. $V\cong V^\ast$ implies that $\dim V<\infty$;
  2. $V\cong V^{\ast\ast}$ by a natural isomorphism if and only if $\dim V<\infty$;
  3. If $V$ is a Banach space, $V^\prime$ is reflexive if and only if $V$ is reflexive;
  4. If $V$ is a reflexive Banach space, $W\subseteq V$ is a closed subspace, then $W$ is also reflexive;
  5. If $V^\prime$ is separable then $V$ is separable.

    The Axiom of Choice and Self-Duality of Vector Spaces

First Post

Well… This is my first post on this blog, and I have absolutely no idea how to start it.

Should I make it about myself? about my life? about my academic status? How I about I tell cool stories from my life, perhaps inebriated adventures? army experiences? Maybe I should write about mathematics. Perhaps some nice proof or some nice theorem?

Well, it seems that in a most self-referential way I already began my first post and I am writing about it in itself. Now that we have done that, let me do all of the above.

My name is Asaf Karagila. I live in Israel, and have lived here my whole life. I just finished my M.Sc. in Ben-Gurion University under the supervision of Uri Abraham. My thesis was titled Vector Spaces and Antichains of Cardinals in Models of Set Theory. It extends a classical construction by Läuchli of a vector space which has no basis. I also presented a paper by Feldman and Orhon in which they generalize Hartogs theorem, I slightly improve upon their original proof and I extend their results.

Now, and for the next couple of years, I am a Ph.D. student in the Hebrew University in Jerusalem, supervised by Menachem Magidor. There is no concrete topic right now, and I going to have a busy year full of studying. It’s a good thing, I hope.

All those things will find their way in details to this site (as well as my thesis itself once I am finished with minor corrections and improvements).

To finish, here is a bit of mathematics.

——-$\newcommand{\Seq}{\operatorname{Seq}}$

Definition. We say that $A$ is Dedekind-finite if $B\subseteq A$ and $|A|=|B|$ then $A=B$.

Proposition.$A$ is Dedekind-finite if and only if $\aleph_0\nleq|A|$
Proof. Exercise.

Assuming the axiom of choice the above proposition shows that a set is Dedekind-finite if and only if it is finite (i.e. has an injective function into a finite ordinal). However without the axiom of choice it is consistent to have infinite Dedekind-finite sets.

For a set $A$, let $\Seq(A)=\{f\colon n\to A\mid f\text{ injective}\}$.

Lemma. Let $A$ be a Dedekind-finite set, then $\Seq(A)$ is also Dedekind-finite.
Proof. If $\Seq(A)$ was not Dedekind-finite it would have a countably infinite subset $B$. Each sequence in $B$ defines a finite subset of $A$, their union is a union of enumerated sets and therefore can be well-ordered. We assumed that $B$ is infinite and therefore the union must be infinite, since a finite set can only have finitely many different enumerations.

Therefore $\bigcup_{f\in B}\rng(f)$ is a countably infinite subset of $A$, in contradiction to the assumption that $A$ is Dedekind-finite. $\square$

Theorem. (Tarski) Suppose that there exists an infinite Dedekind-finite set, then there exists a set $\cal A$ of Dedekind-finite sets such that $(\cal A,\subseteq)$ is order isomorphic to $(\mathbb R,\lt)$.

Proof. Let $A$ be an infinite Dedekind-finite set, and let $S=\Seq(A)$. Let $\{A_r\mid r\in\mathbb R\}\subseteq\power(\omega)$ be a chain of order-type $(\mathbb R,\lt)$ (e.g., fix an enumeration of the rationals, and use it to define $A_r$ as the indices of the rationals in the Dedekind cut of $r$). Let $S_r=\{f\in S\mid |f|\in A_r\}$, and $\mathcal A=\{S_r\mid r\in\mathbb R\}$. If $r\lt t$ then $A_r\subsetneq A_t$ and therefore $S_r\subsetneq S_t$, and since $S$ is Dedekind-finite we have that $S_r$ is Dedekind-finite as well. Furthermore, by Dedekind-finiteness if $r\neq t$ we have that $|S_r|\neq|S_t|$. $\square$

Note that the axiom of choice was not used anywhere in the above proofs. The lemma used the fact that we took a countable union over enumerated sets; and the theorem used the fact that we can find a chain of order-type $(\mathbb R,\lt)$ in $\power(\omega)$ which is again a choice-free proof.

Corollaries. Suppose that there exists an infinite Dedekind-finite set, then

  1. there are at least $2^{\aleph_0}$ Dedekind-finite sets of different cardinalities;
  2. there is an infinite descending sequence of cardinals.

Ah, how strange can be the universe when the axiom of choice is absent!