# Is there an infinite dimensional inner product space without an orthogonal Hamel basis?

I want to know if there exists an (real or complex) vector space $$X$$ with infinite dimension and an inner product $$\langle\cdot,\cdot\rangle$$ such that there is no orthogonal Hamel (algebraic) basis of $$X$$. I do not seek conditions on to the topological aspects of completeness nor an example Schauder Basis or a Maximal Orthogonal System in Hilbert Spaces. I am aware that this is not usually discussed in the scenario of infinite dimensional vector spaces, but this is exactly what I'm curious about.

I know how to prove that every inner product space of Hamel dimension $$\aleph_0$$ has an orthonormal Hamel basis using the Gram-Schimidt process but this proof does not work for the uncountable case.

I also know that a maximal orthogonal set of non-zero vectors is not always a Hamel basis, thus a simple application of Zorn's Lemma does not seem to solve the problem. Finally, I have seen many other similar questions here but none of them provide an answer for the question posed here.

Let $$X$$ be an infinite dimensional Hilbert space. Assume by contradiction that $$\mathcal{B}=\{e_k\}_{k\in K}$$ is an orthonormal Hamel basis of $$X.$$ Fix a sequence $$\{e_{k_j}\}_{j=1}^\infty$$ of distinct elements of $$\mathcal{B}.$$ The series $$\sum_{j=1}^\infty 2^{-j}e_{k_j}$$ is absolutely convergent, thus it represents an element $$x_0\in X.$$ By assumption $$x_0$$ is a finite linear combination of the elements in $$\mathcal{B}.$$ Thus $$\langle x_0,e_k\rangle \neq 0$$ for at most finitely many $$k\in K.$$ However $$\langle x_0,e_{k_j}\rangle =2^{-j}$$ for any $$j\in\mathbb{N},$$ which leads to a contradiction.

• I was hoping to not use anything related to the completeness, but I was wrong. Thanks for the answer. Commented Jun 7, 2023 at 0:27
• You welcome. The completeness is essential as the linear span of an infinite orthonormal system is an orthonormal basis for that span. Commented Jun 7, 2023 at 7:41
• "a complete...Hilbert space"? Are there any noncomplete ones? Commented Jun 7, 2023 at 12:57
• @leftaroundabout Thanks, a misprint. Commented Jun 7, 2023 at 13:12

A slightly different argument to see that an infinite-dimensional separable Hilbert space is a counterexample.

Claim: Let $$V$$ be any separable inner product space (not necessarily complete). Then any orthonormal set $$S \subset V$$ is at most countable.

Proof: For orthonormal vectors $$x,y \in S$$, we have $$\|x-y\|^2 = \langle x-y, x-y \rangle = \|x\|^2 + \|y\|^2 - 2 \langle x,y \rangle = 1 + 1 - 0$$ so that $$\|x-y\| = \sqrt{2}$$. (This is essentially the Pythagorean theorem.) Let $$D$$ be our countable dense set. For every $$x \in S$$ we can choose $$x' \in D$$ with $$\|x-x'\| < \sqrt{2}/2$$. By the triangle inequality, the elements $$x' : x \in S$$ are distinct. So the map $$x \mapsto x'$$ is a 1-1 map of $$S$$ into $$D$$, hence $$S$$ is at most countable.

Couple this with the Baire category argument mentioned in Green Park's answer that in any infinite-dimensional Hilbert space, a Hamel basis is uncountable. We conclude that in an infinite-dimensional separable Hilbert space, there cannot be an orthonormal Hamel basis.

Let $$V$$ be an infinitely dimensional separable Hilbert space, and let $$\langle., .\rangle$$ be its inner product. Equip it with the topology induced by the inner product. Then the inner product is a continuous function from $$V\times V$$ into $$\mathbb K=\mathbb R$$ or $$\mathbb C$$. We will use:

1. Finite dimentional closed subspaces are closed and have empty interior.
2. A Hamel basis of $$V$$ cannot be countable, or $$V$$ could be written as a countable union of finite dimensional subspaces, contradicting Baire's Theorem for complete metrizable spaces.
3. Since $$V$$ is metrizable and separable, $$V$$ is hereditarely separable.

Now assume by contradiction that $$V$$ has an orthogonal Hamel basis (or, more generally, an uncountable orthogonal system of nonzero vectors), $$B$$. By 2. $$B$$ is separable. Let $$S\subseteq B$$ be a countable set such that $$B\subseteq \overline S$$. By 1., there exists $$x\in B\setminus S$$. Since $$V$$ is first-countable (since it is metrizable), there exists a sequence $$(e_n: n\in \mathbb N)$$ of elements of $$S$$ converging to $$x$$.

Since every pair of distinct elements of $$B$$ are orthogonal, $$\langle x, e_n\rangle=0$$ for every $$n\in \mathbb N$$. Then by continuity, $$0=\langle x, \lim_{n\rightarrow \infty} e_n\rangle=\langle x, x\rangle$$, thus $$x\in V$$, contradicting the fact that $$x\neq 0$$.

This argument can be adapted for any separable inner product space with uncountable Hamel dimension.