Discussion
Let \((\mathbb{Q}, d)\) be the metric space with d(x, y) = |x-y|. Let \(E = \{p \in \mathbb{Q} : 2 \lt p^{2} \lt 3\}\). Then, the set E is
Answer: A
The set E is closed but not compact.
In \(\mathbb{Q}\), E is the set of rationals in \((-\sqrt{3}, -\sqrt{2}) \cup (\sqrt{2}, \sqrt{3})\). The endpoints are irrational, so no rational lies on them and
\[ \begin{aligned} E = \mathbb{Q} \cap \big(&[-\sqrt{3}, -\sqrt{2}] \\ &\cup [\sqrt{2}, \sqrt{3}]\big) \end{aligned} \]
That is the intersection of \(\mathbb{Q}\) with a closed subset of \(\mathbb{R}\), so E is closed in \(\mathbb{Q}\).
E is not compact: a sequence of rationals in E increasing to \(\sqrt{3}\) has no subsequence converging to a point of E, because its only possible limit, \(\sqrt{3}\), is not rational. Compactness would require E to be complete, and it is not. So 'compact' and 'not closed but compact' fail, and 'neither closed nor compact' fails because E is closed.
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