# Showing the basis of a Hilbert Space have the same cardinality

I am trying to show that if we have two orthnormal 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 carnality.

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

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Seems right. But it is hardly readable, for you are mixing $D$, $u$, $\mathbb S$, $b$ up. –  martini Nov 7 '12 at 14:12
@martini oh dear, just looked it over, I stopped halfway through and then continued on and used different notation by mistake, sorry . thanks for the comment –  hmmmm Nov 7 '12 at 14:16
See Halmos, Introduction to Hilbert Space and the theory of spectral multiplicity, §16 Dimension, Theorem 1. –  vesszabo Nov 7 '12 at 20:16
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 $|K| \le |S|$ 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 $S' := \bigcup_{i\in K} S_j$. If there were any $l \in S \setminus S'$, we have, as $(a_i)$ is a basis, $$b_l \in H = \overline{\operatorname{lin}\{a_i : i \in S\}}$$ over the otherhand, 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 wished.