TITLE: Showing an orthonormal basis does not exist for an Euclidean space
QUESTION [0 upvotes]: I must construct an example of a non separable Euclidean space which does not have an orthonormal basis. In particular, we know that such a space must be (uncountably)infinite-dimensional. Otherwise, by the
Gram–Schmidt process, we could make the $n$ linearly independent vectors that can be found in the space into an orthonormal system.
So by the mere fact that this is an actual linear space, we definetively must have some set of elements $\{\phi_{\alpha}\}$ which are linearly independent. Hence, we could use the Gram-Schmidt process to make them orthogonal (at least for some subset of them).
My actual question is: when you are required to prove the non-existence of an orthogonal basis, does it suffice to show that the space has uncountably many linearly independent elements? Becuase then you can't really use the Gram-Schmidt process? Or are there Euclidean spaces with an uncountably infinite dimension, which indeed have an orthonormal base?
Happy Sunday!

REPLY [0 votes]: It's not enough. For instance, consider the space of functions $\mathbb R\to\mathbb R$ which vanish at all except finitely many points. The functions which are $0$ everywhere except at a single point, where they are one, form a basis of this space, and there are uncountably many of these (one for every point on the real number line). Now define the inner product
$$\langle f,g\rangle:=\sum_{x\in\mathbb R}f(x)g(x).$$
You'll see that the basis given above is orthonormal with respect to this inner product.
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