Discussion
Let A and B be proper dense subsets of a metric space. Then :
Answer: B
The union \(A \cup B\) is dense.
A set is dense when its closure is the whole space. Since \(A \subseteq A \cup B\), the closure of \(A \cup B\) contains the closure of A, which is already the whole space. So the union of a dense set with anything is dense.
The other options fail. In R, the rationals A and the irrationals B are both proper dense subsets, but \(A \cap B = \phi\), which is not dense, and \(A \setminus B = \mathbf{Q}\) is dense here but fails if instead \(B = \mathbf{R} \setminus \{0\}\) and \(A = \mathbf{R} \setminus \{1\}\), where \(A \setminus B = \{0\}\). Neither set need be closed: a closed dense set is the whole space, and both A and B are proper.
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