An exercise in functional analysis

[Chevtchenko]

A nice exercise from the elementary theory of Banach spaces: If $X$ is an infinite-dimensional Banach space, show that $X$ must have uncountable algebraic dimension.

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Suppose on the contrary that there exists a sequence $(x_n)$ in $X$ such that $X = span{x_0, x_1, \ldots, x_n, \ldots}$ and put for each $k$ $V_k = span{x_0, \ldots, x_k}$. Of course $V = \cup V_k$, and since each $V_k$ is closed, in view of the Baire Category Theorem there must be a $V_k$ with non-empty interior. Hence $V = V_k$ (contradiction).