Question

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, which of the following statements is TRUE?

• A. P is TRUE and Q is FALSE
• B. Both P and Q are TRUE
• C. Both P and Q are FALSE
• D. P is FALSE and Q is TRUE

Hint:

In topology, compactness can be lost when the topology is coarser, and Hausdorff property is violated when there aren't enough open sets to separate points.

Answer 1

The Correct Option is B

Step 1: Analyzing Statement P: The subspace \( ([0, 1], T_1) \) is compact in the Euclidean topology. If \( T_1 \) is a proper subset of \( T_2 \), then \( T_2 \) might introduce more open sets, possibly causing the space to lose compactness. Hence, statement P is true: If \( T_1 \) is a proper subset of \( T_2 \), \( ([0, 1], T_2) \) is not compact. Step 2: Analyzing Statement Q: For \( T_2 \) to be a proper subset of \( T_1 \), it means \( T_2 \) has fewer open sets than \( T_1 \). The subspace \( ([0, 1], T_2) \) might not be Hausdorff because the lack of sufficient open sets could prevent the separation of points. Hence, statement Q is also true: If \( T_2 \) is a proper subset of \( T_1 \), \( ([0, 1], T_2) \) is not Hausdorff. Thus, the correct answer is (B) Both P and Q are TRUE.