As noted previously, we can restrict our attention to Banach spaces.

A Banach space $X$ is reflexive if the canonical embedding into its double-dual is onto.

Now let $X$ be any Banach space. We define $X^{(\alpha)}$ for all ordinals $\alpha$ as follows:
$X^{(0)}=X$, $X^{(\alpha+1)}=(X^{(\alpha)})^\prime$, and for limit ordinals $\delta$ let $X^{(\delta)}$ be the direct limit of the $X^{(\alpha)}$, $\alpha<\delta$, $\alpha$ is even.

For a given space $\alpha$, we can ask whether $X^{(\alpha)}$ will ever become reflexive.

The key fact is that a closed subspace of a reflexive space is again reflexive.
(It was pointed out before that $X$ is reflexive iff its dual is. But for the following discussion we have to think about subspaces of reflexive spaces.)
This implies that if $X$ is not reflexive, then the sequence $X^{(\alpha)}$, $\alpha$ ordinal, will never stabilize, i.e., no $X^{(\alpha)}$ will be reflexive, like in the $\ell^\infty$ example above (which can be iterated through all the ordinals).

I find this fact rather striking. It was pointed out to me by either Dirk Werner or Ehrhard Behrends. I don't remember exactly who.

Now remarkably, this needs the axiom of choice (which comes from the use of the Hahn-Banach Theorem in the proof). I have convinced myself a while ago that if you don't have a nontrivial atomless finitely additive probability measure on $\mathbb N$ (which can happen if AC fails),
then the sequence for $X=c_0$ is $c_0$, $\ell^1$, $\ell^\infty$, $\ell^1$ and so on.

I.e., $\ell^1$ and $\ell^\infty$ are both reflexive in this situation and duals of each other.
This is because without the nontrivial measure, every functional on $\ell^\infty$ that vanishes on $c_0$ vanishes on all of $\ell^\infty$.
If I remember correctly, there is some remark about this in Solovay's paper on the model of set theory in which all sets of reals are measurable. In this model there is no nontrivial
finite, finitely additive, atomless measure on the integers.

isomorphisms. This distinction causes some subtle problems; unfortunately, I can't think of an especially short and illuminating example right now. – Thierry Zell Nov 15 '10 at 19:34