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

Question

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?

Answer 1

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 $X_n=[v_1,\dots,v_n]$, i.e., the linear span of the first $n$ vectors.

Then we have:

• $X=\bigcup\limits_{n=1}^\infty X_n$
• $X_n$ is a finite-dimensional subspace of $X$, hence it is closed. (Every finite-dimensional normed space is complete, see PlanetMath. A complete subspace of a normed space is closed. See also: Finite-dimensional subspace normed vector space is closed)
• $X_n$ is a proper subspace of $X$, so it has empty interior. See Every proper subspace of a normed vector space has empty interior

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.

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Some further references:

Other questions and answers on MSE

• Is there an easy example of a vector space which can not be endowed with the structure of a Banach space
• Your favourite application of the Baire Category Theorem
• Two problems: When a countinuous bijection is a homeomorphism? Possible cardinalities of Hamel bases?
• In the question Cardinality of a Hamel basis of $\ell_1(\mathbb{R})$ you can learn even more - that the cardinality of the Hamel basis is at least $\mathfrak c=2^{\aleph_0}$.

Online

• Banach spaces of infinite dimension do not have a countable Hamel basis at PlanetMath
• uncountable Hamel basis - post by Henno Bradsma from Ask an Analyst (Wayback Machine)
• Blog post Uncountability of Hamel Basis for Banach Space II from Matt Rosenzweig's Blog

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

Answer 2

One easy proof might the that following:

If possible let hamel basis of $X$ is countable and the basis is $\{v_1,\cdots,v_n,\cdots\}.$ WLOG we can assume $\|x_n\|=1.$ Now, take $x=\sum_0^\infty \frac{x_n}{2^n}$ which is not linearly combination of the above bases, but it is in $X$ by completeness of $X$, a contradiction.