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:

Let $x \in X$.

By Expression of Vector as Linear Combination from Basis is Unique/General Result, there is a unique finite subset $C_x$ of $R \times Y$ such that:


 * $\displaystyle x = \sum_{(r, v) \in C_x} r \cdot v$ and
 * $\forall (r,v) \in C_x: r \ne 0_R$

Then we define $\Phi(x) := \operatorname{Img}(C_x)$.

We next show that $\langle \Phi(x) \rangle_{x \in X}$ satisfies the marriage condition. That is, for every finite subset $F$ of $X$, $|F| \le |\bigcup \Phi(F)|$:

Since $X$ is a basis, it is linearly independent.

By Subset of Linearly Independent Set is Linearly Independent, $F$ is also linearly independent.

By the definition of $\Phi$, $F \subseteq \operatorname{span}\bigcup \Phi(F)$.

By Finite Union of Finite Sets is Finite, $\bigcup \Phi(F)$ is finite.

Thus by Linearly Independent Subset of Finitely Generated Vector Space, $|F| \le |\bigcup \Phi(F)|$.

By Hall's Marriage Theorem/General Set, there is an injection from $X$ into $Y$.

Precisely the same argument shows that there is an injection from $Y$ into $X$.

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

Also see

 * Bases of Finitely Generated Vector Space