Discussion
Let X be a separable inner product space and Y be any orthonormal set in X. Then :
Answer: D
Y is countable.
Distinct elements of an orthonormal set are at distance \(\sqrt{2}\): \(\|e - f\|^{2} = \|e\|^{2} + \|f\|^{2} = 2\). The open balls of radius \(\sqrt{2}/2\) around them are therefore disjoint. A separable space has a countable dense set, and each of these disjoint balls must contain a point of it, so there can be only countably many of them.
An orthonormal set is always linearly independent, so A is false. It need not be infinite: in a finite dimensional space it has at most \(n\) elements, so B fails. And in an infinite dimensional space an orthonormal basis is not a Hamel basis, since not every vector is a finite combination of its elements (C).
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