Banach Spaces - How can $B,B',B'', B''', B'''',B''''',\ldots$ behave?

(ZFC)

Let $ \big\langle B,+,\cdot, \:\: \|\cdot\| \:\: \big\rangle $ be a Banach space.

Define $ \mathbf{B} \; = \;\big\langle B,+,\cdot, \:\: \|\cdot\| \:\: \big\rangle $.

Define $\: \mathbf{B}_0 = \mathbf{B} \:$. For all non-negative integers $n$,
define $\mathbf{B}_{n+1}$ to be the Banach space that is the continuous dual of $\mathbf{B}_n$.

Define the relation $\:\sim\:$ on $\:\{0,1,2,3,4,5,\ldots\}\:$ by
$m\sim n \:$ if and only if $\: \mathbf{B}_m$ is isometrically isomorphic to $\mathbf{B}_n$.

$\sim\:$ is obviously an equivalence relation.

What can the quotient of $\:\{0,1,2,3,4,5,\ldots\}\:$ by $\:\sim\:$ be?

The only thing I know about this is that $\:\{\{0,1,2,3,4,5,\ldots\}\}\:$
and $\:\{\{0,2,4,6,8,\ldots\},\{1,3,5,7,9,\ldots\}\}\:$ are both possible.

Solution 1:

This is not a complete answer, but rather an attempt to draw attention to the relevant subproblems.

In his comment @AlexanderThumm made the following observation: if $m>n$ and $m\sim n$ then $n+k(m-n)\sim n$ for all nonnegative $k$. This follows from the observation that if $X$ and $Y$ are isometric Banach spaces then so are $X'$ and $Y'$.

It's a simple inference from this that either $\mathbf{B}_m \not\cong \mathbf{B}_n$ whenever $m\neq n$, or there are integers $N$ and $n\leq N$ such that $\mathbf{B}_i$ are pairwise nonisometric for $1\leq i\leq N$ and $\mathbf{B}_{N+1}\cong\mathbf{B}_{n}$, $\mathbf{B}_{N+2}\cong\mathbf{B}_{n+1}$, and so on.

It remains to show that each of these situations is possible, or to rule some of them out.

The first case (all $\mathbf{B}_n$ distinct) can happen, considering the sequence $c_0, \ell^1, \ell^\infty,\ldots$. (It's well known that none of these spaces is reflexive, but we need a better argument to rule out the existence of any isometric isomorphism.) For an arbitrary infinite set $S$ consider the space $\ell^\infty(S)$ of bounded functions on $S$, with the uniform norm. This space has cardinality $2^{|S|}$. It is apparently known however (as cited here) that $\ell^\infty(S)''$ is isomorphic to $\ell^\infty(2^{2^S})$, so $\ell^\infty(S)''$ is not isomorphic to $\ell^\infty(S)$ since $2^{2^{2^{|S|}}}>2^{|S|}$. The same argument shows that the $2k$th dual of $\ell^\infty(S)$ is not isomorphic to $\ell^\infty(S)$. It follows from this that no two of the sequence $c_0,\ell^1,\ell^\infty(\mathbf{N}),\ldots$ are isomorphic: if $\mathbf{B}_m\sim\mathbf{B}_n$, say, then $\mathbf{B}_{2m}\sim\mathbf{B}_{2n}$, so we have a contradiction.

I don't have anything to contribute to the other problems, other than to highlight some interesting first cases: