Question

Prove that a Hilbert space is seperableiff every ortho normal set in

H is countable.

Answer 1

1. If H is separable, every orthonormal set in H is countable

2. If H contains an orthonormal sequence which is total in H, then H is separable.

Proof:

1.

Let B be a countable dense subset of H. Let M be an uncountable orthonormal set. Then the distance between any two elements x, y ∈ M is √ 2 since

$||x-y||^2=||x||^2+||y||^2=2$

For each x ∈ M, we define N_x as the ball centered at x with radius √ 2/3. Then the N_x are disjoint. Since B is dense, for each x, there is b_x ∈ B such that b_x ∈ N_x. Since N_x are disjoint, the collection {b_x} is an uncountable subset of B. This is a contradiction.

2.

Let A be the set of all linear combinations

$γ _1^{(n) } e_1 + . . . +γ _n^{(n) } e_n , n = 1, 2,...$

where the coefficients γ are complex rational (γ = a + ib, both a and b rational) or just rational (if the underlying field was real). Think of A as the rational-span of (e_k).

Then A is the countable union of countable sets. We claim it is dense in H. Fix any x ∈ H and any

$\varepsilon$ > 0. Then since the sequence (e_n) is total in H, there is some point

$y = \sum \langle e_k,x\rangle e_k$

such that $||k_x − y_k|| < \varepsilon/2$ , and then se use the triangle inequality and the density of the rationals to see that there is a nearby point v such that $||x − v ||<\varepsilon$