# Let $X$ be an infinite dimensional Banach space. Prove that every Hamel basis of X is uncountable.

Let $X$ be an infinite dimensional Banach space. Prove that every basis of $X$ is uncountable.

Can anyone help how can I solve the above problem?

It seems that the proof using the Baire category theorem can be found in several places on this site, but none of those questions is an exact duplicate of this one. Therefore I’m posting a CW-answer, so that this question is not left unanswered.

We assume that a Banach space $$X$$ has a countable basis $$\{v_n; n\in\mathbb N\}$$. Let us denote $$X_n=[v_1,\dots,v_n]$$.

Then we have:

So we see that $$\operatorname{Int} \overline{X_n} = \operatorname{Int} X_n=\emptyset$$, which means that $$X_n$$ is nowhere dense. So $$X$$ is a countable union of nowhere dense subsets, which contradicts the Baire category theorem.

Some further references:

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Books

• Corollary 5.23 in Infinite Dimensional Analysis: A Hitchhiker’s Guide by Charalambos D. Aliprantis, Kim C. Border.
• A Short Course on Banach Space Theory By N. L. Carothers, p.25
• Exercise 1.81 in Banach Space Theory: The Basis for Linear and Nonlinear Analysis by Marián Fabian, Petr Habala, Petr Hájek, Vicente Montesinos, Václav Zizler