Hamel Basis in Infinite dimensional Banach Space without Baire Category Theorem

Question

Prove that every Hamel basis in an infinite dimensional Banach Space is uncountable without using Baire Category Theory. We are assuming axiom that vector space dimension (if exist) is well-defined.

Note: Axiom that vector space dimension(if exist) is well-defined is independent of DC. See Sizes of bases of vector spaces without the axiom of choice at MathOverflow.

Answer 1

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 $X$.