Uncountable orthonormal set in a Hilbert space

Question

Let $H$ be a Hilbert space.

Why if $H$ has an uncountable orthonormal set $ F=\{B_i ; i\in I\} $ then it cannot have a countable basis ?

Answer 1

You can prove (or consider) the reverse, meaning the statement/lemma :

Let $H$ be an infinite dimensional Hilbert space. Then, $H$ has a countable orthonormal basis if and only if $H$ has a dense countable set.

To prove that, I will note some quick hints.

First let us assume that $H$ has a countable orthonormal basis $\{e_n\}$. Then any $x \in H$ can be uniquely written as :

$$x = \sum_{n=1}^\infty \langle x,e_n\rangle e_n$$

But, such $x$ can be written as a linear combination of rationals

$$x = \sum_{n=1}^m q_ne_n$$

since the set $\mathbb Q + \mathbb Qi$ is a countable and dense subset of $\mathbb C$.

Consider now the set :

$$D_n = \bigg\{d_n = \sum_{n=1}^m q_ne_n, \; q_n \in \mathbb Q + \mathbb Qi\bigg\}$$

Then the set $D$ defined as

$$D = \bigcup_{n=1} D_n$$

is countable as a union of countable sets.

Now you need to show that $D$ is also dense. But, note that since the

$$x = \sum_{n=1}^\infty \langle x,e_n\rangle e_n$$

converges, its partial sum serie should also converge, thus

$$\left\| \sum\limits_{n=N+1} \langle x,e_n\rangle e_n\right\|< \frac{\varepsilon}{2}$$

for $N \in \mathbb N$ sufficiently large and $\varepsilon >0$.

Now, keep in mind that $\mathbb Q + \mathbb Qi$ is not only countable, but dense and use the Triangle Inequality and Parseval's Inequality to prove that :

$$\left\| \sum_{n=1}^\infty \langle x,e_n\rangle e_n - \sum_{n=1}^N q_ie_n \right\| < \varepsilon$$

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For the converse, assume that $H$ has a countable dense set $\{h_j\}$ with $j \in \mathbb N$ and that $\{e_i\}$ is an orthonormal basis with $i \in \mathbb N$. Now, try to reach a contradiction by assuming that the orthonormal basis is uncountable.

Answer 2

If $\left \{ e_n:n\in \mathbb N \right \}$ is an orthonormal basis for $\mathcal H$, then $\left \{ \sum_{k=0}^{N}q_k+ir_k:q_k,r_k\in \mathbb Q,N\in \mathbb N \right \}$ ia a countable dense subset.

If $\mathcal H$ has an uncountable orthonormal basis, say, $\left \{ e_{\alpha}:\alpha\in \Lambda \right \}$ and a dense subset $D$, then, since $\|e_{\alpha}-e_{\beta}\|^2=2$ whenever $\alpha \neq \beta$, the balls $B(\alpha,\sqrt2/2)$ are disjoint and so since each of them must contain a point $x_{\alpha}\in D$, the map $\alpha\to x_{\alpha}$ shows that $D$ is uncountable.

Answer 3

I already wrote this in a comment under an answer but that answer was erased, and it seems to be this is the easiest answer, unless I'm missing something basic: a set of orthonormal vectors in any inner product space, and thus also in Hilbert spaces, is linearly independent (reasonably easy proof), and thus in your question $\;F\;$ is a lin. indpendent set, which means $\;\dim H\ge|F|\ge 2^{\aleph_0}\;$ , and since all the basis of a vector space (or Hilbert space) have the same cardinality we're through.

The above remains true taking closures of spans and stuff, as far as I see it...