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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?

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What kind of basis? An algebraic basis? It's important at this point. –  Patrick Da Silva Oct 20 '12 at 17:53

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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:

  • $X=\bigcup\limits_{n=1}^\infty X_n$
  • $X_n$ 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: math.stackexchange.com/questions/854227/finite-dimensional-subspace-normed-vector-space-is-closed)
  • $X_n$ is a proper subspace of $X$, so it has empty interior. See Interior of a Subspace

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:

Other questions and answers on MSE

Online

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
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I have found a proof that a proper subspace ofa normed space has empty interior here: matthewhr.files.wordpress.com/2012/08/… –  DavideZena Aug 15 at 17:56

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