Showing the basis of a Hilbert Space have the same cardinality I am trying to show that if we have two orthonormal families $\{a_i\}_{i\in K}$ and $\{b_j\}_{j\in S}$ and these are the basis of some Hilbert Space H, then they have the same cardinality.
So If I suppose that the $\{a_i\}_{i\in K}$ is countable, i.e., that $K$ is countable and that $S$ is uncountable, then we want to show that this leads to a contradiction.
As the $\{b_i\}_{i\in K}$ forms a basis we know that and $a_n$ as:
$$a_n=\sum_{1}^{\infty} c_ib_i$$
Now if we let the set $D_n=\{i|\mbox{for i in the sum of}\ a_n\}$ and then take:
$D:=\bigcup_n^{\infty} D_n$ then this is the set of all indices and it is a countable set as it is the countable union of countable sets.
Take $l\in{S}$ that is not in $D$ then as $b_l\in H$ and as the $\{a_i\}_{i\in K}$ forms a basis we have that:
$$b_l\in \overline{lin\{\sum_{i=1}^{\infty} a_i\}}$$
Then from above we have that:
$$\overline{lin\{\sum_{i=1}^{\infty} a_i\}}=\overline{lin\{b_d|d\in D\}}$$
If we now consider $$ ||b_l||^2=\sum _1^{\infty} c_i\langle b_{d_i}, b_l \rangle =0$$ so we have the contradiction.
So is the above proof correct and can we generalise this further to different cardinalities? Does H have to be a Hilbert space for this to be true?
Thanks very much for any help.
 A: Let $(a_i)_{i \in K}$ and $(b_j)_{j\in S}$ two bases of a Hilbert space $H$. Following your idea, we will show that $|S| \le |K|$ and by symmetry conclude $|S| = |K|$. Note that for finite dimensional spaces everything we are up to is well known from basic linear algebra. So we will suppose $\aleph_0 \le |K|, |S|$ in what follows.
Let $i \in K$, then we can write, as the $(b_j)$ form a basis 
\[ a_i = \sum_{j\in S_i} \langle a_i, b_j\rangle b_j \]
for some countable subset $S_i \subseteq S$. Let $S' := \bigcup_{i\in K} S_i$. If there were any $l \in S \setminus S'$, we have, as $(a_i)$ is a basis, 
$$ b_l \in H = \overline{\operatorname{lin}\{a_i : i \in K\}} $$
on the other hand, for each $i$
$$ a_i \in \overline{\operatorname{lin}\{b_j: j \in S_i\}} \subseteq 
   \overline{\operatorname{lin}\{b_j : j \in S'\}} $$
so $b_l \in  \overline{\operatorname{lin}\{b_j : j \in S'\}}$, hence there is a countable $T \subseteq S'$ with $b_l \in \overline{\operatorname{lin}\{b_j : j \in T\}}$, giving
$$ \|b_l\|^2 = \sum_{j \in T}|\langle b_l, b_j\rangle|^2 = 0 $$
Contradiction.
So $S = S'$ and hence 
$$ |S| = |S'| = \left|\bigcup_{i \in K} S_i\right| \le |K| \cdot\sup_i |S_i| \le |K| \cdot \aleph_0 = |K| $$
as desired.
