Let $\{e_k : k\in K \}$ be an orthonormal set in a Hilbert space $H$, where $K$ is some arbitrary set.
Let $x \in H$.
Show that for each $n\in \mathbb{N}$, $F_n = \{ k\in K: |\langle e_k, x\rangle|^2 \geq \frac{1}{n}\}$ is finite. Estimate the size of $F_n$,
Show $\{k \in K : |\langle e_k, x\rangle| \ne 0 \}$ is at most countable.
Things I know:
By definition of an orthonormal set, if $e_k \ne x$, then $\langle e_k, x\rangle=0$. Also we have $\| e_k\|=1 $.
Also, the $\{k \in K : |\langle e_k, x\rangle| \ne 0 \}$ is the $(\cup F_n)^c$
From here I'm not sure what to do.
Any help is appreciated!
Thanks!
Probably you mean $\langle e_i,e_j\rangle=0$ for $i\ne j$, an arbitrary $x\in H$ need not be orthogonal to the basis elements.
For any vector $x\in H$ and finite subset $I\subseteq K$ let $x_I$ denote the orthogonal projection of $x$ to ${\rm span}(e_i:i\in I)$, that is, $$x_I=\sum_{i\in I}\langle x,e_i\rangle e_i$$ and because it's a projection of $x$, we have $\|x_I\|\le\|x\|$.
Now, if $I\subseteq F_n$, then $$\|x\|^2\ge\|x_I\|^2=\sum_{i\in I}|\langle x,e_i\rangle|^2\ge\frac{|I|}n$$ which implies $|I|\le n\,\|x\|^2$.
$\{k\in K:\langle x,e_k\rangle\ne 0\}=\bigcup_{n\in\Bbb N}F_n$ and not the complement of the union.
A countable union of finite sets is countable.