Infinite-Dimensional Banach Space has Uncountable Dimension

Theorem
Let $\Bbb F \in \set {\R, \C}$.

Let $X$ be an infinite-dimensional Banach space over $\Bbb F$.

Let $B$ be a Hamel basis for $X$.

Then $B$ is uncountable.

Proof
Suppose that $B$ is countable.

Let:


 * $B = \set {e_n : n \in \N}$

For each $n \in \N$, let:


 * $F_n = \span \set {e_k : k \le n}$

Since $\dim F_n = n$, $F_n$ clearly has finite dimension.

We show that:


 * $\ds X = \bigcup_{n \mathop = 1}^\infty F_n$

Let $x \in X$.

Then there exists $e_{n_1}, \ldots, e_{n_k} \in X$ and $\alpha_1, \ldots, \alpha_n \in \Bbb F$ such that:


 * $\ds x = \sum_{j \mathop = 1}^k \alpha_j e_{n_j}$

Setting:


 * $\ds N = \max_{1 \le j \le k} e_{n_j}$

we have that $x \in F_N$, hence:


 * $\ds X = \bigcup_{n \mathop = 1}^\infty F_n$

From Finite Dimensional Subspace of Normed Vector Space is Closed, $F_n$ is closed for each $n \in \N$.

From Interior of Proper Subspace of Normed Vector Space is Empty, $F_n$ has empty interior for each $n \in \N$.

So each $F_n$ is nowhere dense.

Since $X$ is a complete metric space, it is a Baire space.

However, we have written $X$ as a countable union of nowhere dense sets.

This contradicts the Baire Category Theorem.

So $B$ is countable.