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I am working through a proof that every hilbert space has a orthogonal basis which lies dense in that Hilbert space. In the proof the following is done:

Let $v$ be a vector, and $E \subseteq I$ an arbitrary subset of the (possible uncountable) index set $I$ and consider $$ v_E = \sum_{i \in E} c_i(v) e_i $$ with $c_i(v) = \langle v, e_i \rangle$.

Then in the proof it is concluded that $\sum_{i \in E} |c_i(v)|^2 \le ||v||^2$, and so $$ \sum_{i \in I} |c_i(v)|^2 \le ||v||^2. (*) $$ And then it is said because of that just countable many $c_i(v)$ could be unequal to $0$ and that the series of the $|c_i(v)|^2$ converges.

I don't see why because of (*) the set $\{ c_i(v) | c_i(v) \ne 0 \}$ is countable?

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A remark: the orthogonal basis is not dense itself, it is its linear span that is dense. – Giuseppe Negro Dec 13 '12 at 23:26
thanks! my fault :) – Stefan Dec 13 '12 at 23:32
up vote 3 down vote accepted

Let $I_n$ be the set of all $i$ for which $|c_i(v)| > 1/n$. We have:

$$ \sum_{i \in I_n} |n c_i(v)|^2 \le n^2 \sum_{i \in I} |c_i(v)|^2 < \infty $$

Notice that each term in the LHS sum is larger than $1$. Thus, the number of elements in each $I_n$ must be finite. We conclude that $\bigcup_n I_n = \{i \mid c_i(v) \ne 0\}$ is at most countable.

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First, given an uncountable index set $I$ and nonnegative numbers $d_i$ for $i\in I$, what do we even mean by the expression $\sum_{i\in I} d_i$? Well, we mean the following,

$$ \sum_{i\in I} d_i = \sup_{F\subset I} \sum_{i\in F} d_i, $$ where $F$ is finite (You can give a definition when the $d_i$ are no longer nonnegative, but then the sum isn't necessarily defined, i.e., doesn't converge to anything).

Now that we have this definition, we claim the following proposition.

Proposition. Given any uncountable collection $d_i>0$ of positive numbers indexed by $I$, the sum $\sum_{i\in I} d_i$ is infinite.

Proof. Let $E_0=\{i\in I\mid d_i>1\}$ and $E_n=\{i\in I\mid \frac{1}{n+1}<d_i\le\frac{1}{n}$. Then the $E_n$ are disjoint and their union is $I$. As a result, there is some $k$ for which $E_k$ is infinite (otherwise $I$ would be a countable union of finite sets and therefore countable). Then we can choose arbitrarily large finite subsets $F\subset E_k\subset I$ and for these we have

$$ \sum_{i\in F} d_i \ge \sum_{i\in F} \frac{1}{k+1} = \frac{\vert F\vert}{k+1} $$ Since $k$ is fixed, the right-hand side can become arbitrarily large, and hence the supremum over all finite subsets of $I$ is infinite. #

Now, an application of this proposition is that if the sum over every finite subset of $I$ is uniformly bounded (say by $\Vert v\Vert^2$), then the sum over $I$ is finite and less than or equal to this uniform bound. Hence, the set $\{i\in I\mid d_i>0\}$ is at most countable.

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