TITLE: Hilbert space has countable basis
QUESTION [5 upvotes]: Let $H$ be Hilbert space. I want to show that if $H$ has a countable orthonormal basis, then every orthonormal basis for $H$ must be countable. I spent almost a day, but could not solve this problem. I am trying to solve it with some kind of contradiction. I started by assuming that there exist an uncountable basis. Then I want to derive something crazy but cannot find it. Could you give me some hints or suggestions?

REPLY [2 votes]: $\newcommand{gae}[1]{\newcommand{#1}{\operatorname{#1}}}\gae{span}$I assume you mean Hilbert basis. Consider this lemma:

Let $(X,d)$ be a metric space, and let $A\subseteq X$ be a dense subset. Then, every dense subset $B$ of $X$ has a dense subset of cardinality at most $\lvert A\rvert$.

Proof: If $X$ is finite, then $X$ is discrete and $A=B=X$. If $X$ is infinite, then $A$ and $B$ must be infinite. Moreover, the set $S=\{B(a,q)\,:\, a\in A\land q\in\Bbb Q_+\}$ is a basis of $X$. Since $A$ is infinite, $\lvert S\rvert\le \lvert A\times \Bbb Q\rvert=\lvert A\rvert$. For all $D\in S$, select some $b_D\in B$, given by density of $B$, such that $b_D\in D$. The set $\{b_D\,:\, D\in S\}$ intersects all the elements of a basis, and therefore it is dense. Moreover its cardinality is at most $\lvert S\rvert\le\lvert A\rvert$.

Now, let $\mathcal B_1,\mathcal B_2\subseteq H$ be two subsets of orthonormal vectors each of which generates a dense subspace. Also, let $\kappa_i=\lvert\mathcal B_i\rvert$ and $\kappa_1\le\kappa_2$. If $\kappa_1<\aleph_0$, then proving $\kappa_2=\kappa_1$ is elementary linear algebra. If $\kappa_1\ge\aleph_0$, then call $\span_{\Bbb Q}(\mathcal B_1)$ the $\Bbb Q$-linear subspace generated by $\mathcal B_1$. It is clear that $\overline{\span_{\Bbb Q}(\mathcal B_1)}=\overline{\span_{\Bbb R}(\mathcal B_1)}= H$ and that $\kappa_1=\lvert \span_{\Bbb Q}(\mathcal B_1)\rvert$. By the above lemma, $ \span_{\Bbb R}(\mathcal B_2)$ must have a dense subset $G$ of cardinality at most $\kappa_1$. However, since every vector in $\span_{\Bbb R}(\mathcal B_2)$ is linear combination of finitely many elements of $\mathcal B_2$, there is a subset $G'\subseteq\mathcal B_2$ of cardinality at most $\lvert G\rvert$ such that $\span_{\Bbb R}(G')\supseteq G$. However, $$\mathcal B_2\setminus G'\subseteq (\span_{\Bbb R}(G'))^\perp=\left(\overline{\span_{\Bbb R}(G')}\right)^\perp\subseteq\left(\overline G\right)^\perp=H^\perp=\{0\}$$
which means that $\mathcal B_2=G'$ and therefore $\kappa_2=\lvert G'\rvert\le\lvert G\rvert\le \kappa_1$.