Question

As noted by Thierry in his comment, the question isn't posed quite as well as it could be. However, the following example shows that taking duals can result in spaces whose weakly compact subsets can be that are topologically larger than the those of earlier duals. Consider the space $\ell_\infty (S)$ of bounded scalar functions on an infinite set $S$, equipped with the sup supremum norm. First of all, note Note that the Hilbert space $\ell_2 \ell_\infty (S)$ embeds isometrically into has density $\ell_\infty(S)$, but 2^{\vert S\vert}$ (with respect to the norm topology), whereas $\ell_2(R)$ does not embed \ell_\infty (even isomorphically) into 2^{2^{\vert S\vert }})$ has density $2^{2^{2^{\vert S\vert }}}$; in particular, $\ell_\infty (S)$ if and $\vert R\vert > \vert S\vert$ \ell_\infty (2^{2^{\vert S\vert }})$ cannot be isomorphic to one another since every weakly compact subset of $\ell_\infty(S)$ has density character not exceeding $\vert S\vert$ by a result of 2^{2^{2^{\vert S\vert }}}>2^{\vert S\vert }$. However, H. Rosenthal has shown (for the last claim, see p.230 Theorem 5.1 of Injective Banach spaces and the spaces $L^\infty (\mu)$ for a finite measure $\mu$, Acta Math 124, 1970, 205--248). Observing 205--248) that $\ell_\infty(S)^{\ast\ast}$ is isomorphic to $\ell_\infty (2^{2^{\vert S\vert }})$ (again, see Rosenthal's paper, this time Theorem 5.1 on p.234), we have that })$, hence $\ell_\infty(S)$ is not isomorphic to $\ell_\infty(S)^{\ast\ast}$ because $\ell_\infty(S)^{\ast\ast}$ contains \ell_\infty(S)^{\ast\ast}$. The point of this example is that it is a copy counterexample to the OP's question regardless of whether one considers the Hilbert space canonical embedding of $\ell_2 (2^{2^{\vert S\vert}})$, but X$ into $\ell_\infty(S)$ does notX^{\ast\ast}$ or one can consider an arbitrary isomorphism between higher duals.