Orthonormal Subset of Hilbert Space Extends to Basis

Theorem
Let $H$ be a Hilbert space, and let $S$ be an orthonormal subset of $H$.

Then there exists a basis for $H$ that contains $S$ as a subset.

Proof
Consider the set $\OO$ of orthonormal subsets $S'$ of $H$ that contain $S$:


 * $\OO := \set{ S' \subseteq H: S \subseteq S', \text{$S'$ is orthonormal} }$

In particular, $S \in \OO$ so that $\OO$ is non-empty.

Give $\OO$ the subset ordering.

For a chain $\CC \subseteq \OO$, we assert that $\bigcup \CC \in \OO$.

For each $c \in \bigcup \CC$, there is a $C \in \CC$ such that $c \in C$.

Hence $\norm c = 1$.

Moreover, for $c, c' \in \bigcup \CC$, there is a $C \in \CC$ such that $c, c' \in C$ because $\CC$ is a chain.

Then since $C \in \OO$, it follows that $c \perp c'$.

Hence $\bigcup \CC \in \OO$.

Therefore, the conditions of Zorn's Lemma apply to $\OO$.

Let $O$ be a maximal element of $\OO$.

Then $O$ is a maximal orthonormal subset of $H$.

That is, $O$ is a basis containing $S$.