# Every Hilbert space has an orthonomal basis - using Zorn's Lemma

The problem is to prove that every Hilbert space has a orthonormal basis. We are given Zorn's Lemma, which is taken as an axiom of set theory:

Lemma If X is a nonempty partially ordered set with the property that every totally ordered subset of X has an upper bound in X, then X has a maximal element.

Given a orthonormal set $E$ in a Hilbert space $H$, it is apparently possible to show that $H$ has an orthonormal basis containing $E$.

I tried to reason as follows: Suppose $E$ is a finite set of $n$ elements. Then one can number the elements of $E$ to create a totally ordered set of orthonormal elements. Then the span $<E>$ can be identified with $R^n$, where each element $v = v_1 e_1 + v_2 e_2 + ... + v_n e_n$ is identified with the vector $(v_1, v_2, ..., v_n)$. On $R^n$ we have a total order, namely the "lexigraphical order" $(x_1,x_2,...,x_n) \leq (y_1,y_2,...,y_n)$ if $x_1 < y_1$ or if $x_1 = y_1$ and $x_2 < y_2$ or if $x_1 = y_1$, $x_2 = y_2$ and $x_3 < y_3$ and so on. Hence $E$ is a totally ordered subset of $H$ and $H$ is a partially ordered set. However, this set doesn't seem to have an upper bound. The set $E$ does have an upper bound. If we define a total order on $E$ only, then X is a partially ordered set satisfying the criteria so X has a maximal element?

This is as far as I got, and I am not sure the entire argument is correct. I don't see how what kind of maximal element I am seeking, since the orthonormal basis of a Hilbert space can have countably infinity number of elements.

• The ordering you want to consider is inclusion of orthonormal bases. – user17794 Oct 21 '12 at 16:56
• The orthonormal basis of a Hilbert space can have an arbitrary number of elements, possibly a lot more than countably infinitely many. – tomasz Oct 21 '12 at 17:06
• Adding a bit to Tim Duff's comment: in mathematical practice, when applying Zorn's lemma, it is almost always the case that the partial ordering is the inclusion, and the upper bound of a chain is the union (see Tukey's lemma ). – tomasz Oct 21 '12 at 17:14

First I should remark that there is absolutely no need to appeal to Zorn's lemma in the case of a finite dimensional vector space.

I should also add that when talking about about a basis for a Hilbert space one has to distinguish between an algebraic space, which is a linearly independent space whose linear span is everything; and the topological basis whose span is dense in the space. The algebraic space is known as a Hamel basis whereas the topological one is known as a Schauder basis.

The use of Zorn's lemma here is quite standard, but I suppose that you have yet to see many uses of Zorn's lemma, which is why you find this to be a difficult task.

Zorn's lemma asserts that every partially ordered set in which every chain has an upper bound has a maximal element. So to use it we need to come up with a partially ordered set that has the wanted property.

In our case, this should be sets of orthonormal vectors, ordered by inclusion. If we show that the increasing union of such sets is a set of orthonormal vectors then we have shown that every chain is bounded.

The key in doing such things is to note that if $\{E_i\mid i\in I\}$ is a chain (namely for $i<j$, $E_i\subseteq E_j$) and $E=\bigcup_{i\in I}E_i$, then if $E$ had two vectors which are not orthogonal then there would be some $i\in I$ such that both these vectors are in $E_i$; since our assumption was that the elements of $E_i$ are pairwise orthogonal, this cannot happen. Similarly for normality.

Now use Zorn's lemma and assume that $E$ is a maximal element in this order. If it isn't a basis, find out a way for $E$ to be extended, which will contradict its maximality. Therefore $E$ has to be a basis, and all its elements are pairwise orthogonal, and have norm $1$.

• @Peter: No, Zorn's lemma is non-constructive. The fact the increasing union gives an orthonormal set simply shows that given a chain it has an upper bound. Therefore by Zorn there is a maximal element. Secondly, note I used $I$ as an index set. I haven't said anywhere that $I$ is finite, countably infinite, or otherwise. It's just any index set, so regardless to how long the chain is, its union is an upper bound. – Asaf Karagila Oct 22 '12 at 16:31
• @Asaf: my small misunderstanding: in second paragraph, second line, it should be "algebraic basis, which is linearly independent set...." – Groups Oct 1 '15 at 5:33
• thanks you also for distinguishing first the term basis in topological and algebraic contexts. – Groups Oct 1 '15 at 6:33
• @Alex: That this is the same argument as in all these other cases. Being orthonormal is a property whose negation can be expressed by a finite linear combination. Even if the chain is infinite, the union is still orthonormal, as any counterexample of that would appear in a finite, and thus bounded, stage of the chain. – Asaf Karagila May 18 '18 at 22:55
• It seems completeness is not used in this argument, so that it would work for any inner product space (not necessarily a Hilbert space)? – Matthew Kvalheim Mar 9 at 18:12

I don't believe your argument leads anywhere fruitful. How is $H$ partially ordered that is consistent with the total order on $E$ (or $<E>$)?

Let me nudge you in the right direction: The partially ordered set $X$ you wanna apply Zorn to is the set of all orthonormal subsets of $H$, with $\subseteq$ inducing the partial order. Do you see how any totally-ordered subset of $X$ has an upper bound and how you might proceed?

• The lexicographic order on $H$ seen as ${\bf R}^\kappa$ for some $\kappa$ is a total order extending the order on $E$, if done right. However, as you have said, it hardly brings about anything fruitful with regard to the theorem. – tomasz Oct 21 '12 at 17:11