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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 $E$.

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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.

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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 $1$.

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?

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