This question is motivated by the question link text, which compares the infinite direct sum and the infinite direct product of a ring.
It is well-known that an infinite dimensional vector space is never isomorphic to its dual. More precisely, let $k$ be a field and $I$ be an infinite set. Let $E=k^{(I)}=\oplus_{i \in I} k$ be the $k$-vector space with basis $I$, so that $E^{*}$ can be identified with $k^I = \prod_{i \in I} k$. Then a stronger result asserts that the dimension of $E^{*}$ over $k$ is equal to the cardinality of $k^I$. This is proved in Jacobson, Lectures in Abstract Algebra, Vol. 2, Chap. 9, $\S$ 5 (Jacobson deduces it from a lemma which he attributes to Erdös and Kaplansky). Summarizing, we have
\begin{equation}
\operatorname{dim}_k (k^I) = \operatorname{card} k^I.
\end{equation}
Now, if $V$ is any $k$-vector space, we can ask for the dimension of $V^I$. Does the Erdös-Kaplansky theorem extend to this setting ?
Is it true that for any vector space $V$ and any infinite set $I$, we have $\operatorname{dim} V^I = \operatorname{card} V^I$ ? More generally, given a family of nonzero vector spaces $(V_i)$ indexed by $I$, is it true that $\operatorname{dim} \prod_{i \in I} V_i = \prod_{i \in I} \operatorname{card} V_i$ ?
If $V$ is isomorphic to $k^J$ for some set $J$, then the result holds as a consequence of Erdös-Kaplansky. In the general case, we have $V \cong k^{(J)}$, and we can assume that $J$ is infinite. In this case I run into difficulties in computing the dimension of $V^I$. I can only prove that $\operatorname{dim} V^I \geq \operatorname{card} k^I \cdot \operatorname{card} J$.
Best Answer
The answer to both questions is yes.
As a preliminary, let's prove that for any infinite-dimensional vector space $V$, that
Proof: Since $card(k) \leq card(V)$ and $\dim V \leq card(V)$, the inequality
$$card(k) \cdot \dim V \leq card(V)^2 = card(V)$$
is obvious. On the other hand, any element of $V$ is uniquely of the form $\sum_{j \in J} a_j e_j$ for some finite subset $J$ of (an indexing set of) a basis $B$ and all $a_j$ nonzero. So an upper bound of $card(V)$ is $card(P_{fin}(B)) \sup_{j \in P_{fin}(B)} card(k)^j$. If $B$ is infinite, then $card(P_{fin}(B)) = card(B) = \dim(V)$, and for all finite $j$ we have $card(k^j) \leq card(k)$ if $k$ is infinite, and $card(k^j) \leq \aleph_0$ if $k$ is finite, and either way we have
$$card(V) \leq \dim V \cdot \max\{card(k), \aleph_0\} \leq \dim V \cdot card(k)$$
as desired. $\Box$
The rest is now easy. Suppose $I$ is an infinite set, and suppose without loss of generality that $V_i$ is nontrivial for all $i \in I$. Put $V = \prod_{i \in I} V_i$. We have
$$\dim V \geq \dim k^I = card(k)^I \geq card(k)$$
where the equality is due to Erdos and Kaplansky. Therefore
$$\dim(V) = \dim(V)^2 \geq \dim V \cdot card(k) = card(V) = \prod_i card(V_i)$$
by the lemma above.