Discussion
Let \(\tau_{1}\) be the usual topology on \(\mathbb{R}\). Let \(\tau_{2}\) be the topology on \(\mathbb{R}\) generated by \(\mathcal{B} = \{[a,b) \subset \mathbb{R} : -\infty \lt a \lt b \lt \infty\}\). Then the set \(\{x \in \mathbb{R} : 4\sin^{2} x \le 1\} \cup \{\frac{\pi}{2}\}\) is
Answer: C
The set is closed in both \((\mathbb{R}, \tau_{1})\) and \((\mathbb{R}, \tau_{2})\).
\(A = \{x : 4\sin^{2} x \le 1\}\) is the preimage of the closed set \((-\infty, 1]\) under the continuous function \(4\sin^{2} x\), so it is closed in the usual topology. Adding the single point \(\frac{\pi}{2}\) (where \(4\sin^{2} x = 4\), so it was not in \(A\)) keeps the set closed, since a finite union of closed sets is closed.
\(\tau_{2}\) is the lower-limit (Sorgenfrey) topology, which is finer than \(\tau_{1}\): every usual open set is a union of intervals \([a, b)\). A set closed in \(\tau_{1}\) has a \(\tau_{1}\)-open complement, which is also \(\tau_{2}\)-open, so the set is closed in \(\tau_{2}\) too.
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