Theorem - Every Hilbert space $H \neq \{0\}$ has an orthonormal basis.

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Proof : As could be expected, the proof makes use of Zorn's Lemma. Let $\mathcal{O}$ be the set of all orthonormal sets of $H$ . It is clear that $\mathcal{O}$ is non-empty since the set $\{x\}$ is in $\mathcal{O}$ , where $x$ is an element of $H$ such that $\|x\| = 1$ .

The elements of $\mathcal{O}$ can be ordered by inclusion, and each chain $\mathcal{C}$ in $\mathcal{O}$ has an upper bound, given by the union of all elements of $\mathcal{C}$ . Thus, Zorn's Lemma assures the existence of a maximal element $B$ in $\mathcal{O}$ . We claim that $B$ is an orthonormal basis of $H$ .

It is clear that $B$ is an orthonormal set, as it belongs to $\mathcal{O}$ . It remains to see that the linear span of $B$ is dense in $H$ .

Let $\overline{\mathrm{span}\,B}$ denote the closure of the span of $B$ . Suppose $\overline{\mathrm{span}\,B} \neq H$ . By the orthogonal decomposition theorem we know that

Hence, $\overline{\mathrm{span}\,B} = H$ , and this finishes the proof. $\square$