Question about a corollary of Bessel inequiality I'm trying to prove: > **Let $(X,<,>)$ be a vector space with inner product and $E \subseteq X$ be a orthonormal subset of $X$. Prove that** > > **$\mathcal A$ ={$x \in E : |<y,x>| \neq0$} it's numerable.** Using the Bessel inequality that says: > **Let $(X,<,>)$ be a vector space with inner product.** > > **If { $x_i$}$_{i=1}^n$ are orthonormal then $\sum_{i=1}^n |<x_i,y>|^2 \le \|y\|^2 $ for every $y\in X$**. I have the suggestion to take $\mathcal A_n=${$x \in E : |<y,x>| 1/n$} for every $n \in \mathbb N$ and aplly Bessel inequiality. I understand why it ends with this, but really don't see how make the $\mathcal A_n$ numerable with Bessel inequality. Thank you.

Let $x\in A_n$. From Bessel inequality (and the fact that being $A_n\subseteq E$ it's an orthornomal subset of X) we obtain

$||y||^2\ge\sum_{x\in A_n} |\langle x,y \rangle|^2\ge \sum_{x \in A_n} \frac{1}{n^2}$

Thus, if $A_n$ was infinite, the RHT would diverge, which is absurd since it is bounded by $||y||^2$. Since $A_n$ is finite, $A=\bigcup_{n\in \mathbb{N}}A_n$ is numerable (being a numerable union of finite sets)