TITLE: The Space $C(\Omega,\mathbb{R})$ has a Predual?
QUESTION [9 upvotes]: Let $\Omega$ be a compact metric space and $C(\Omega,\mathbb{R})$ the space of borel continuous real valued functions. I would like to know if there is any real Banach space $V$ such that its dual space (topological)is exactly $C(\Omega,\mathbb{R})$. 
My main interest is when $\Omega$ is an infinite cartesian product as for example,
$\Omega=E^{\mathbb{Z}^d}$, where $E$ is a compact metric space. 
If the answer for this case is also negative are we in a better situation if $E$ is a finite set ?
Thanks for any comment or reference.
Edition. Remove the superfluous hypothesis pointed out by Philip.

REPLY [10 votes]: Firstly, let me just point out that every compact metric space is automatically separable.
Secondly, note that $c_0$ does not embed into any separable dual space, hence neither does any Banach space containing a subspace isomorphic to $c_0$. Since every $C(K)$ space ($K$ compact Hausdorff) contains a subspace isomorphic to $c_0$, no $C(K)$ space embeds isomorphically into a separable dual space. In particular, since metrizability of $K$ is equivalent to $C(K)$ being norm separable, the answer to your question is always no.
For a reference for all of the above claims, look up $C(K)$ and $c_0$ in the index of Albiac and Kalton's book Topics in Banach Space Theory.