TITLE: Proof that two bases of a vector space have the same cardinality in the infinite-dimensional case.
QUESTION [33 upvotes]: I am having a difficulty setting up the proof of the fact that two bases of a vector space have the same cardinality for the infinite-dimensional case. In particular, let $V$ be a vector space over a field $K$ and let $\left\{v_i\right\}_{i \in I}$ be a basis where $I$ is infinite countable. Let $\left\{u_j\right\}_{j \in J}$ be another basis. Then $J$ must be infinite countable as well. Any ideas on how to approach the proof?

REPLY [12 votes]: Once you have the necessary facts about infinite sets, the argument is very much like that used in the finite-dimensional case. The two crucial pieces of information are (1) that if $I$ is an infinite set of cardinality $\kappa$, say, then $I$ has $\kappa$ finite subsets, and (2) that if $|J|>\kappa$, and $J$ is expressed as the union of $\kappa$ subsets, then at least one of those subsets must be infinite.
Let $B_1 = \{v_i:i\in I \}$ and $B_2 = \{u_j:j \in J \}$, and suppose that $|J|>|I| = \kappa$. Each $u_j \in B_2$ can be written as a linear combination of some finite subset of $B_1$, say $u_j = \sum\limits_{i \in F_j}k_{ji}v_i$, where $F_j$ is a finite subset of $I$. For each finite $F \subseteq I$ let $J_F = \{j \in J:F_j = F\}$; clearly $J$ is the union of these sets $J_F$. But by (1) above $I$ has only $\kappa$ finite subsets, and $|J|>\kappa$, so by (2) above there must be some finite $F \subseteq I$ such that $J_F$ is infinite.
To simplify the notation, let $F = \{i_1,i_2,\dots,i_n\}$, and for $\mathcal{l}=1,2,\dots,n$ let $v_\mathcal{l} = v_{i_\mathcal{l}}$; then every vector $u_j$ with $j \in J_F$ is a linear combination of the vectors $v_1,v_2,\dots,v_n$. In other words, $\{u_j:j \in J_F\} \subseteq \operatorname{span}\{v_1,v_2,\dots,v_n\}$, and of course $\{u_j:j \in J_F\}$, being a subset of the basis $B_2$, is linearly independent. But $\operatorname{span}\{v_1,v_2,\dots,v_n\}$ is of dimension $n$ over $K$, so any set of more than $n$ vectors in $\operatorname{span}\{v_1,v_2,\dots,v_n\}$ must be linearly dependent, and we have a contradiction. It follows that we must have $|J| \le |I|$. By symmetry (or by the same argument with the rôles of $I$ and $J$ interchanged), $|I| \le |J|$, and hence $|I|=|J|$.