Bases of Vector Space have Equal Cardinality

Theorem
Let $V$ be a vector space.

Let $X$ and $Y$ be bases of $V$.

Then $X$ and $Y$ are equinumerous.

Proof
We will first prove that there is an injection from $X$ to $Y$.

Let $\Phi:X \to \mathcal P(Y)$ be defined thus:

For each $x \in X$, express $x$ as a sum of non-zero products of elements of $Y$.

This can be done in exactly one way, up to rearrangement.

Then define $\Phi(x)$ as the set of all elements of $y$ in the expression of $x$.

Let $\mathcal F$ be the set of all partial functions $f$ from $X$ to $Y$ such that:


 * $f$ is one-to-one.
 * If $(x,y) \in f$, then $y \in \Phi(x)$.

Then for any $x \in X$, $\{ f(x): f \in \mathcal F \land x \in \operatorname{Dom}(f) \}$ is finite.

$\mathcal F$ has finite character:

Let $f \in \mathcal F$.

Let $F$ be a finite subset of $X$.

Then $f \restriction F$ is obviously in $\mathcal F$.

Suppose instead that $f$ is a partial function from $X$ to $Y$ and its restriction to each finite subset of $\operatorname{Dom}(f)$ is in $\mathcal F$.

Then $f$ is one-to-one: Let $p,q \in X$ with $p \ne q$.

Then $\{p, q\}$ is a finite subset of $\operatorname{Dom}(f)$, so $f \restriction \{p, q\} \in \mathcal F$.

Thus $f(p) \ne f(q)$.

If $(x,y) \in f$, then $(x,y) \in f \restriction \{x\}$, so $y \in \Phi(x)$.

Thus $f \in \mathcal F$.

So we see that $f$ has finite character.

Let $F$ be a finite subset of $X$.

By the Cowen-Engeler Lemma, $\mathcal F$ has an element $\phi$ whose domain is $X$ (that is, it is a mapping from $X$ to $Y$.

But a one-to-one mapping is an injection, so $\phi$ is an injection from $X$ to $Y$.

Precisely the same argument shows that there is an injection $\psi$ from $Y$ to $X$.

Thus by the Cantor-Bernstein-Schröder Theorem, $X$ and $Y$ are equinumerous.