Let S be a subset of a Hilbert space H. Prove that Span(S) is dense in H if and only ifS⊥ = 0.

Sure, here is the proof:

First, let's define some terms:

• Span(S) is the smallest subspace of H that contains S.
• S⊥ is the orthogonal complement of S, which is the set of all vectors in H that are orthogonal to every vector in S.
• A subset of a topological space is said to be dense if its closure is the whole space.

Now, let's prove the statement in two directions:

1. Suppose Span(S) is dense in H. This means that the closure of Span(S) is H. Since the orthogonal complement of Span(S) is the same as the orthogonal complement of S (because Span(S) contains S and is contained in any subspace containing S), we have S⊥ = (Span(S))⊥. Now, if x is in S⊥, then x is orthogonal to every vector in Span(S), and hence to every vector in H (because Span(S) is dense in H). Therefore, x must be the zero vector, so S⊥ = {0}.

2. Conversely, suppose S⊥ = {0}. This means that the only vector in H that is orthogonal to every vector in S is the zero vector. Therefore, the orthogonal complement of Span(S) is also {0}, because Span(S) contains S and is contained in any subspace containing S. Now, the orthogonal complement of a subspace is closed, so the closure of Span(S) is the whole space H (because its orthogonal complement is {0}). Therefore, Span(S) is dense in H.

So, we have proved that Span(S) is dense in H if and only if S⊥ = {0}.