Discussion
Let X be a metric space and \(Y, Z \subset X\) such that \(Y \cap Z = \phi\). Then :
Answer: D
The distance is positive if one of Y or Z is compact and the other is closed in X.
Suppose Y is compact, Z closed and disjoint. The function \(y \mapsto d(y, Z)\) is continuous and positive on Y (positive because Z is closed and \(y \notin Z\)), so it attains a positive minimum on compact Y. That minimum is \(d(Y, Z)\).
The weaker conditions fail. Disjoint closed sets can be at distance 0: the hyperbola branch \(xy = 1\) and the x-axis in \(\mathbf{R}^{2}\). Compactness of one set alone is not enough either: \(Y = \{0\}\) and \(Z = (0, 1]\) are disjoint, Y is compact, and \(d(Y, Z) = 0\).
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