Orthogonal projection onto the closure of a subspace spanned by a set of orthonormal vectors Suppose $H$ is a Hilbert space and $\{e_1,\cdots\}$ is a set of infinitely many orthonormal vectors. Then let $M$ be the closure of the subspace spanned by these orthonormal vectors.
I know that for $f\in H$, the orthogonal projection of $f$ onto $M$ is given by 
$$\sum_i \langle f,e_i \rangle e_i$$
However, How can I justify that the infinite sum converges?
My way is to extend these orthonormal vectors to an orthonormal basis. So we have $\{e_i\} \cup \{h_j\}$ as an orthonormal basis. Then $f=\sum_i \langle f,e_i \rangle e_i + \sum_j \langle f,h_j \rangle h_j$. So that the concerned infinite sum must converge. However, extending these vectors to an orthonormal basis will use Zorn's Lemma. Is there any other argument to justify the convergence of $\sum_i \langle f,e_i \rangle e_i$?
 A: Say the family of orthonormal vectors is indexed by some set $I$. For any finite $F \subseteq I$ we have $$0 \leq \bigg\lVert f - \sum_{i \in F} \langle f, e_i \rangle e_i \bigg\rVert^2 = \lVert f \rVert^2 - 2\sum_{i \in F} |\langle f, e_i \rangle|^2 + \sum_{i \in F} |\langle f, e_i \rangle|^2 = \lVert f \rVert^2 - \sum_{i \in F} |\langle f, e_i \rangle|^2.$$
So, $\sum_{i \in F} |\langle f, e_i \rangle|^2 \leq \lVert f \rVert^2$ for all finite $F \subseteq I$. Thus $\sum_{i \in I} |\langle f, e_i \rangle|^2 \leq \lVert f \rVert^2 < \infty$. In particular, the set 
$$\{i \in I : \langle f, e_i \rangle \neq 0\} = \bigcup_{n=1}^\infty \{i \in I : |\langle f, e_i \rangle| > 1/n\}$$
is countable since for each $n$ the set $\{i \in I : |\langle f, e_i \rangle | > 1/n\}$ is finite. Let $(e_j)_{j=1}^\infty$ be an enumeration of those $e_i$'s for which $\langle f, e_i \rangle \neq 0$. Then for $m \leq k < \infty$,
$$\bigg\lVert \sum_{j=m}^k \langle f, e_i \rangle e_j \bigg\rVert^2 = \sum_{j=m}^k |\langle f, e_j \rangle|^2 \to 0 \text{ as } k,m \to \infty.$$
In other words, the partial sums of $\sum_{j=1}^\infty \langle f,e_j \rangle e_j$ are Cauchy, so the series converges by completeness of $H$.
