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The answer to both questions is yes.

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

• Lemma: $card(V) = card(k) \cdot \dim V$

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

$$\dim(V) = \dim(V)^2 \geq \dim V \cdot card(k) = card(V)$$ card(V) = \prod_i card(V_i)$$by the lemma above. 1 The answer to both questions is yes. As a preliminary, let's prove that for any infinite-dimensional vector space V, that • Lemma: card(V) = card(k) \cdot \dim V Proof: Since card(k) \leq card(V) and \dim V \leq card(V), the inequality$$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)

by the lemma above.