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This is Prob. 10, Sec. 3.5, in the book Introductory Functional Analysis With Applications by Erwine Kreyszig:

Let $X$ be an inner product space, let $M$ be an uncountable orthonormal subset of $X$, let $x \in X$ such that $x$ is not the zero vector $\mathbf{0}_X$ in $X$, and let $S(x)$ be the subset of $M$ defined as follows: $$ S(x) \colon= \left\{ \ v \in M \ \colon \ \langle x, v \rangle \neq 0 \ \right\}. $$ Then how to show that this set $S(x)$ is at most countable?

And, this is Prob. 8, Sec. 3.4, in the same book:

Let $\left( e_k \right)_{k \in \mathbb{N} }$ be an orthonormal sequence in an inner product space $X$, let $x \in X$, let $m \in \mathbb{N}$, let $A_m(x)$ be the subset of $\mathbb{N}$ defined as follows: $$ A_m(x) \colon= \left\{ \ k \in \mathbb{N} \ \colon \ \left\lvert \left\langle x, e_k \right\rangle \right\rvert > \frac{1}{m} \ \right\}, $$ and let $n_m$ be the cardinality of $A_m(x)$. Then $$ n_m \leq m^2 \lVert x \rVert^2. $$

The proof of this result involves the Bessel's inequality and goes as follows:

As $\left( e_k \right)_{k \in \mathbb{N}}$ is an orthonormal sequence in the inner product space $X$, so by Theorem 3.4-6 (Bessel Inequality) in Kreyszig, for any $x \in X$, the series $\sum \left\lvert \left\langle x, e_k \right\rangle \right\rvert^2$ converges in $\mathbb{R}$, and $$ \sum_{k=1}^\infty \left\lvert \left\langle x, e_k \right\rangle \right\rvert^2 \leq \lVert x \rVert^2. $$

Let $x\in X$, let $m$ be a given natural number, and let $A_m(x)$ be ths subset of $\mathbb{N}$ given by $$ A_m(x) \colon= \left\{ \ k \in \mathbb{N} \ \colon \ \left\lvert \left\langle x, e_k \right\rangle \right\rvert > \frac{1}{m} \ \right\}. $$

Let $n_m$ denote the cardinality of the set $A_m(x)$ (which is the same as the number of elements in the set $A_m(x)$ if $A_m(x)$ is finite ), where $$n_m \in \{ \ 0 \ \} \cup \mathbb{N} \cup \{ \ \aleph_0 \ \},$$ where $\aleph_0$ (pronounced ``aleph null'') denotes the cardinality of the set $\mathbb{N}$ of natural numbers, because the set $A_m(x)$ can be empty, non-empty but finite, or countably infinite. Furthermore, as $A_m(x) \subset \mathbb{N}$ and as $\mathbb{N}$ is countable, so $A_m(x)$ cannot be uncountable.

If $x = \mathbf{0}_X$, the zero vector in $X$, then $$ \left\langle x, e_k \right\rangle = 0 $$ for all $k \in \mathbb{N}$, and so the set $A_m(x)$ is empty, and therefore $$ n_m = 0 = m^2 \cdot 0 = m^2 \lVert x \rVert^2. $$

So let's suppose that $x$ is not the zero vector in $X$, and suppose also that $n_m \geq m^2 \lVert x \rVert^2$. Then as $$ \left\lvert \left\langle x, e_k \right\rangle \right\rvert^2 \geq 0 $$ for all $k \in \mathbb{N}$ and as $$ \left\lvert \left\langle x, e_k \right\rangle \right\rvert^2 > \frac{1}{m^2} $$ for all $k \in A_m(x)$, so we note that \begin{align*} \sum_{k=1}^\infty \left\lvert \left\langle x, e_k \right\rangle \right\rvert^2 &= \sum_{k \in \mathbb{N} } \left\lvert \left\langle x, e_k \right\rangle \right\rvert^2 \\ &= \sum_{k \in A_m(x) } \left\lvert \left\langle x, e_k \right\rangle \right\rvert^2 + \sum_{k \in \mathbb{N} - A_m(x) } \left\lvert \left\langle x, e_k \right\rangle \right\rvert^2 \\ &\geq \sum_{k \in A_m(x) } \left\lvert \left\langle x, e_k \right\rangle \right\rvert^2 \\\ &> \frac{n_m}{m^2} \\ &\geq \frac{m^2 \lVert x \rVert^2 }{m^2} \\ &= \lVert x \rVert^2, \end{align*} which contradicts the Bessel's inequality. Hence we must have $$ n_m < m^2 \lVert x \rVert^2, $$ as required.

In the above calculation, we have used the equality $$ \sum_{k=1}^\infty \left\lvert \left\langle x, e_k \right\rangle \right\rvert^2 = \sum_{k \in \mathbb{N} } \left\lvert \left\langle x, e_k \right\rangle \right\rvert^2. $$ This is because of Theorem 3.55 in the book \emph{Principles of Mathematical Analysis} by Walter Rudin, 3rd edition, which says that if a series of complex numbers converges absolutely, then, by altering the order of the terms of that series in any way whatsoever, we obtain a series that also converges absolutely and has the same sum as the sum of the original series.

Is this proof correct? If so, then can we use the latter result to establish the former?

If not, then where have I erred in this proof? And, how to give an independent proof of the former result?

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    $\begingroup$ Can't the countability of $S(x)$ be shown more directly using Bessel's inequality? For the left hand side of $\sum_{v \in M} \left\lvert \left\langle x, v \right\rangle \right\rvert^2$ to be finite, the number of non-zero terms needs to be countable. $\endgroup$ – md2perpe Jun 29 '17 at 12:46
  • $\begingroup$ @md2perpe can you please present your argument in a detailed proof, using the notation similar to mine? What you say is intuitively clear, but I'm unable to come up with a rigorous way of presenting it. $\endgroup$ – Saaqib Mahmood Jun 29 '17 at 14:11
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For a arbitrary set $A$ of non-negative numbers we define $$\sum_{a \in A} a = \sup \{ \sum_{a \in F} a \mid F \subset A \text{ finite} \}$$

For every finite (even countable) $F \subset M$ we know that $$\sum_{v \in F} | \langle x, v \rangle |^2 \leq \|v\|^2$$ and therefore also $$\sum_{v \in M} | \langle x, v \rangle |^2 = \sup_{F \text{ finite}} \sum_{v \in F} | \langle x, v \rangle |^2 \leq \|v\|^2$$

Now it's well-known that a sum of non-negative values can only be finite if the number of non-zero terms is at most countable. Otherwise there would be an infinite number of terms greater than $\epsilon$ for some $\epsilon>0$. But that would make the sum greater than $n \epsilon$ for every $n \in \mathbb N$, and so the sum would be infinite.

See also for example the following questions about uncountable sums:

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Let $\{ e_{\alpha} \}_{\alpha \in \Lambda}$ be an orthonormal subset of an inner product space $X$. Let $x\in X$ be given, and define $$ E_n = \{ \alpha \in \Lambda : |(x,e_{\alpha})| \ge 1/n \},\;\; n=1,2,3,\cdots. $$ Then it is easy to argue that $$ \bigcup_{n=1}^{\infty} E_n = \{ \alpha : (x,e_{\alpha}) \ne 0 \}. $$ Indeed, if $(x,e_{\alpha})\ne 0$ for some $\alpha$, then $|(x,e_{\alpha})| \ge 1/n$ for some $n$, which puts $\alpha \in E_n$ and, hence, in the union on the left; conversely, $E_n \subseteq \{ \alpha : (x,e_{\alpha}) \ne 0 \}$ clearly holds for all $n$. By Bessel's inequality, each set $E_n$ must empty or finite because $\{ \alpha_1, \alpha_2, \cdots, \alpha_N \}\subseteq E_n$ implies $\|x\|^2 \ge \sum_{k=1}^{N}|(x,e_{\alpha_k})|^2 \ge N/n$, which bounds $N$ by $n\|x\|^2$. So each $E_n$ is finite or empty, which makes their union empty, finite, or countably infinite.

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