Dimension of $\prod_{i\in I} \mathbb{C}$ Let $I$ be a set and consider the complex vector space
$$V:=\prod_{i\in I} \mathbb{C}=\{(x_i)_{i\in I}: x_i\in \mathbb{C}, \sup_i |x_i|<\infty\}$$ with the pointwise operations.
Is it true that $\dim(V)= |I|$? If $I$ is finite, this is obvious. If $I$ is infinite, then $V$ contains the algebraic direct sum of copies of the complex numbers so that $\dim(V)\ge |I|$. Is there equality? If not, cab we explicitly describe the dimension of $V$ as an infinite cardinal?
 A: Edit: The argument given here before is not correct. Below is a corrected argument.$\newcommand{\restr}{\upharpoonright}$

No, for infinite $I$, the dimension is $2^{\lvert I\rvert}$.
Lemma: Let $K$ be any field and let $f_1,\ldots, f_n\in K^I$ be linearly independent. Then there is a finite $I_0\subseteq I$ such that their restrictions $f_j\restr_{I_0}$ are linearly independent.
Proof: Suppose not. Let $m\leq n$ be maximal dimension of the subspace of $K^{I_0}$ spanned by the corresponding restriction, for arbitrary finite $I_0\subseteq I$. We may assume (by induction) that $m=n-1$ and that the restrictions of $f_1\restr_{I_0},\ldots, f_{n-1}\restr_{I_0}$ to some fixed finite $I_0\subseteq I$ are linearly independent. Then (by linear independence and hypothesis) there are unique $\alpha_j\in K$ such that $f_n\restr_{I_0}=\sum_{j=1}^{n-1} \alpha_jf_j\restr_{I_0}$. But for any $i\notin I_0$, for $I_1=I_0\cup \{i\}$, $f_n\restr_{I_1}$ is a linear combination of $f_j\restr_{I_1}$. By uniquness of $\alpha_j$, this means that we must have $f_n(i)=\sum_{j<n}\alpha_j f_j(i)$. Since $i$ is arbitrary, we must have $f_n=\sum_{j<n}\alpha_jf_j$, contradicting the independence of $f_1,\ldots, f_n$. $\square$
Now, suppose $f_1,\ldots, f_n\in \mathbf Q^I$ are $\mathbf Q$-independent. Then by lemma there is a finite set $I_0\subseteq I$ such that $f_1\restr_{I_0},\ldots, f_n\restr_{I_0}$ are $\mathbf Q$-independent. But $\mathbf Q^{I_0}$ is finite-dimensional, so linear independence is witnessed by non-vanishing of a determinant (a maximal minor of the matrix given by the coordinates of $f_j\restr_{I_0}$ with respect to the standard basis). But then the same determinant does not vanish over $\mathbf C$, so $f_1\restr_{I_0},\ldots, f_n\restr_{I_0}$ have to be linearly independent over $\mathbf C$.
It follows that any basis of $\mathbf Q^I$ is linearly independent in $\mathbf C^I$. Since $\mathbf Q^I$ clearly has rational dimension $2^{\lvert I\rvert}$ (by an easy cardinality argument), and likewise, the complex dimension of $\mathbf C^I$ is bounded above by the same, the conclusion follows.
