# 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 $S$.

$\blacksquare$

## Axiom of Choice

This theorem depends on the Axiom of Choice, by way of Zorn's Lemma.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.