This question is motivated by the question [link text](https://mathoverflow.net/questions/49507/can-i-a-be-isomorphic-to-i-a-for-infinite-i), 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 $\operatorname{dim} V^I \geq \operatorname{card} k^I \cdot \operatorname{card} J$.



Dimension of infinite product of vector spaces
The answer to both questions is yes.


As a preliminary, let's prove that for any infinite-dimensional vector space $V$, that


* **Lemma:** $|V| = |k| \cdot \dim V$


where $|X|$ denotes the cardinality of a set $X$.


Proof: Since $|k| \leq |V|$ and $\dim V \leq |V|$, the inequality


$$|k| \cdot \dim V \leq |V|^2 = |V|$$


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 $|V|$ is $|P\_{fin}(B)| \cdot \sup\_{j \in P\_{fin}(B)} |k|^j$. If $B$ is infinite, then $|P\_{fin}(B)| = |B| = \dim(V)$, and for all finite $j$ we have $|k^j| \leq |k|$ if $k$ is infinite, and $|k^j| \leq \aleph\_0$ if $k$ is finite, and either way we have


$$|V| \leq \dim V \cdot \max\{|k|, \aleph\_0\} \leq \dim V \cdot |k|$$


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


$$\dim V \geq \dim k^I = |k|^I \geq |k|$$


where the equality is due to Erdos and Kaplansky. Therefore


$$\dim(V) = \dim(V)^2 \geq \dim V \cdot |k| = |V| = \prod\_i |V\_i|$$


by the lemma above.


