8. Let (X, d) be a metric space and let S ⊆ X. Prove that dS (x) = 0 if and only if x ∈ S

The statement is asking to prove that the distance from a point x to a set S in a metric space (X, d) is zero if and only if x is an element of S.

Here's a step-by-step proof:

1. Assume that x is an element of S. By definition of the distance function dS(x), we have dS(x) = inf{d(x, y) : y ∈ S}. Since x is in S, we have d(x, x) = 0 (since the distance from any point to itself is zero in a metric space). Therefore, dS(x) ≤ 0. But dS(x) is always nonnegative, so we must have dS(x) = 0.

2. Conversely, assume that dS(x) = 0. This means that for every ε > 0, there exists y in S such that d(x, y) < ε. In particular, for ε = 1/n for each positive integer n, there exists y_n in S such that d(x, y_n) < 1/n. This implies that the sequence {y_n} converges to x. Since S is closed (as it is a subset of a metric space), it contains all its limit points. Therefore, x must be in S.

Therefore, we have shown that dS(x) = 0 if and only if x is in S.