I'm reading the Algebra book by Knapp and he mentions in passing that an orthonormal set in an infinite dimension vector space is "never large enough" to be a vector-space basis (i.e. that every vector can be written as a finite sum of vectors from the basis; such bases do exist for infinite dimension vector spaces by virtue of the axiom of choice, but usually one works with orthonormal sets for which infinite sums yield all the elements of the vector space).
So my question is – how can we prove this claim? (That an orthonormal set in an infinite dimension vector space is not a vector space basis).
Edit (by Jonas Meyer): Knapp says in a footnote on page 92 of Basic algebra:
In the infinite-dimensional theory the term "orthonormal basis" is used for an orthonormal set that spans $V$ when limits of finite sums are allowed, in addition to finite sums themselves; when $V$ is infinite-dimensional, an orthonormal basis is never large enough to be a vector-space basis.
Thus, without explicitly using the word, Knapp is referring only to complete infinite-dimensional inner product spaces.
Best Answer
Take any infinite dimensional inner product space $V$ and any orthonormal sequence $(w_n : n \in \mathbb{N})$. Let $W$ be the subspace generated by this sequence. Then $W$ is certainly an infinite dimensional vector space (because it has an infinite independent subset). Also $W$ has an orthonormal basis, because the inner product on $W$ is inherited from $V$ and thus $(w_n)$ is still an orthonormal sequence in $W$. This means that the theorem you have suggested, "an orthonormal set in an infinite dimension vector space is not a vector space basis", is not true.
What I believe might be true is that no infinite dimensional complete inner product space has a orthonormal basis. This is the question that Andrey Rekalo addressed in another answer.