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.

  • 2
    $\begingroup$ Seems right. But it is hardly readable, for you are mixing $D$, $u$, $\mathbb S$, $b$ up. $\endgroup$ – martini Nov 7 '12 at 14:12
  • $\begingroup$ @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 $\endgroup$ – hmmmm Nov 7 '12 at 14:16
  • 2
    $\begingroup$ See Halmos, Introduction to Hilbert Space and the theory of spectral multiplicity, §16 Dimension, Theorem 1. $\endgroup$ – 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 $|S| \le |K|$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.