Usbrth

I am trying to show that if we have two orthonormal families $a_i_iin K$ and $b_j_jin S$ and these are the basis of some Hilbert Space H, then they have the same cardinality.

So If I suppose that the $a_i_iin K$ 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 $b_i_iin K$ forms a basis we know that and $a_n$ as:

$$a_n=sum_1^infty c_ib_i$$

Now if we let the set $D_n=i$ 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 $linS$ that is not in $D$ then as $b_lin H$ and as the $a_i_iin K$ forms a basis we have that:

$$b_lin overlinelinsum_i=1^infty a_i$$

Then from above we have that:

$$overlinelinsum_i=1^infty a_i=overlinelinb_d$$

If we now consider $$ ||b_l||^2=sum _1^infty c_ilangle b_d_i, b_l rangle =0$$ 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.

Let $(a_i)_i in K$ and $(b_j)_jin S$ 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_jin S_i langle a_i, b_jrangle b_j ]
for some countable subset $S_i subseteq S$. Let $S' := bigcup_iin K S_i$. If there were any $l in S setminus S'$, we have, as $(a_i)$ is a basis,
$$ b_l in H = overlineoperatornamelina_i : i in K $$
on the other hand, for each $i$
$$ a_i in overlineoperatornamelinb_j: j in S_i subseteq
overlineoperatornamelinb_j : j in S' $$
so $b_l in overlineoperatornamelinb_j : j in S'$, hence there is a countable $T subseteq S'$ with $b_l in overlineoperatornamelinb_j : j in T$, giving
$$ |b_l|^2 = sum_j in T|langle b_l, b_jrangle|^2 = 0 $$
Contradiction.

So $S = S'$ and hence
$$ |S| = |S'| = left|bigcup_i in K S_iright| le |K| cdotsup_i |S_i| le |K| cdot aleph_0 = |K| $$
as desired.