Let $X$ be an infinite dimensional Banach space. Prove that every basis of $X$ is uncountable.
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[Math] Let $X$ be an infinite dimensional Banach space. Prove that every Hamel basis of X is uncountable.
baire-categorybanach-spacesfunctional-analysishamel-basisnormed-spaces
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As David Mitra mentioned in his comment, one such proof can be found in Morrison T.J. Functional Analysis. An Introduction to Banach Space Theory (Wiley, 2000), p.221.
The proof we give is very elementary and avoids the usual argument seen for this fact, which involves either the Baire Category Theorem or the Hahn-Banach Theorem. It is due to the Chinese mathematician Nam-Kiu Tsing (1984).
Proposition 5.1. No infinite-dimensional normed linear space with a countable Hamel basis can be complete.
Proof. Let $(X,\|\cdot\|)$ be a normed linear space with Hamel basis $(e_n)_n$, and note that without loss of generality we can assume $\|e_n\|=1$. Let $S_{n-1}$ denote the linear subspace of $X$ spanned by $\{e_1,e_2,\dots,e_{n-1}\}$, and let $r_n \equiv \inf\{ \|x+e_n\|; x\in S_{n-1}\}$ for any $n\ge2$.
Since $\theta\in S_{n-1}$, it follows that $r_n\le \|0+e_n\| =1$ for all $n\ge2$. Now since $S_{n-1}$ is finite-dimensional, it is complete, and hence closed in $X$. Since $e_n\notin S_{n-1}$, we have that $r_n>0$ for all $n\ge2$. Now define the following scalar sequence $(t_n)_n$ by $t_1=1$, $t_2=\frac13$ and for $n\ge2$, $t_{n+1}=\frac13r_nt_n$. Then note that we have $$0<t_{n+k}\le\left(\frac13\right)^k r_nt_n \le \left(\frac13\right)^{n+k-1}$$ for all $n\ge2$ and $k\in\mathbb N$. Now for each $n\in\mathbb N$, define $u_n=\sum_{i=1}^n t_i e_i$ and note that $(u_n)_n$ is a cauchy sequence in $X$. But also notice that for any element $u=\sum_{i=1}^{m-1} \alpha_i e_i\in X$, we have \begin{align} \| u_{n} - u \| & = \left\| \sum_{i = 1}^{m - 1} (t_{i} - \alpha_{i}) e_{i} + t_{m} e_{m} + \sum_{i = m + 1}^{n} t_{i} e_{i} \right\| \\ & \ge \left\| \sum_{i = 1}^{m - 1} (t_{i} - \alpha_{i}) e_{i} + t_{m} e_{m} \right\| - \left\| \sum_{i = m + 1}^{n} t_{i} e_{i} \right\| \\ & \ge t_{m} \left\| \sum_{i = 1}^{m - 1} \frac{1}{t_{m}} (t_{i} - \alpha_{i}) e_{i} + e_{m} \right\| - \sum_{i = m + 1}^{n} t_{i}\\ & \ge t_{m} r_{m} - \sum_{i = 1}^{n - m} \left( \frac{1}{3} \right)^{i} r_{m} t_{m} \\ & \ge \frac{1}{2} t_{m} r_{m} \end{align} for all $n>m$. But this means that $\|u_n-u\|$ does not go to zero, which implies $(u_n)_n$ does not converge. Hence, $X$ is not complete.
Tsing N.K. [1984]. Infinite dimensional Banach spaces must have uncountable basis—an elementary proof. Amer. Math. Monthly, 96 (5), 505-506. JSTOR
The symbol $\theta$ is used to denote the zero vector of the space $X$.
An "orthonormal basis" is not the same thing as a basis that is orthonormal, and in particular may not even actually be a basis at all. An orthonormal basis for a Hilbert space $H$ is just a maximal orthonormal subset $B\subseteq H$. Such a subset need not actually span $H$, and so it need not be a basis. It is true that we can write an arbitrary element $x\in H$ as the sum $\sum_{b\in B}\langle x,b\rangle b$, but this is an infinite sum in general and so does not show that $x$ is in the span of $B$.
So, a separable infinite dimensional Hilbert space has a countable orthonormal basis, but no such orthonormal basis is actually a basis so this does not contradict your Theorem 1.
Best Answer
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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