Discussion
Consider \(([0,1], T_{1})\), where \(T_{1}\) is the subspace topology induced by the Euclidean topology on \(\mathbb{R}\), and let \(T_{2}\) be any topology on [0, 1]. Consider the following statements:
P : If \(T_{1}\) is a proper subset of \(T_{2}\), then \(([0,1], T_{2})\) is not compact.
Q : If \(T_{2}\) is a proper subset of \(T_{1}\), then \(([0,1], T_{2})\) is not Hausdorff.
Then
Answer: B
The answer is Both P and Q are TRUE.
Both follow from one fact: a continuous bijection from a compact space onto a Hausdorff space is a homeomorphism. \(([0,1], T_{1})\) is compact and Hausdorff.
P: if \(T_{2}\) strictly contains \(T_{1}\) and \(([0,1], T_{2})\) were compact, the identity from \(T_{2}\) to \(T_{1}\) would be a continuous bijection from a compact space to a Hausdorff one, hence a homeomorphism, forcing \(T_{2} = T_{1}\). So \(T_{2}\) is not compact.
Q: if \(T_{2}\) is strictly contained in \(T_{1}\) and were Hausdorff, the identity from \(T_{1}\) to \(T_{2}\) would be a homeomorphism for the same reason, forcing equality. So \(T_{2}\) is not Hausdorff.
A compact Hausdorff topology is both maximal compact and minimal Hausdorff, so every option calling P or Q false is wrong.
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