Ytukyg

Let $H$ be a Hilbert space.

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

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 :

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

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 :

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

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

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

converges, its partial sum serie should also converge, thus

$$left| sumlimits_{n=N+1} langle x,e_nrangle e_nright|< frac{varepsilon}{2}$$

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 :

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

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.

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

If $mathcal H$ has an uncountable orthonormal basis, say, $left { e_{alpha}:alphain Lambda right }$ 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 $alphato x_{alpha}$ shows that $D$ is uncountable.

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 $;dim Hge|F|ge 2^{aleph_0};$ , 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...