Sachchidanand Prasad

16-05-2024

    Problem: 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
    • P is true and Q is false
    • Both P and Q are true
    • Both P and Q are false
    • P is false and Q is true
    topology compactness GATE-2022
Solution: Consider the statement P. We will show that P is true. Let $T_1 \subsetneq T_2$ and $[0,1]$ is compact with respect to $T_2$. Consider the identity map \[ I : \left( [0,1], T_2 \right) \to \left( [0,1], T_1 \right). \] Since $T_1 \subset T_2$, the identity map $I$ is continuous. Let $U \in T_2$. Then the complement of $U$, say $K = [0,1]\setminus U$ is closed. Since $[0,1]$ is compact with respect to $T_2$, the set $K$ is also compact in $T_2$. Hence, $I(K) = K$ is also compact with respect to $T_1$. Since $[0,1]$ is Hausdorff in $T_1$, so $I(K)=K$ is closed. Therefore, $[0,1]\setminus K = U$ is open in $T_1$. Thus, $T_2 \subseteq T_1 $, a contradiction.
We now consider the statement Q. We claim that Q is true. Let us suppose that $T_2 \subsetneq T_1$ and $([0,1],T_2)$ is Hausdorff. Then the identity map \[ ([0,1], T_1) \rightarrow ([0,1], T_2) \] is continuous. By the same argument given in the previous statement we conclude that identity map is a homomorphism and hence $T_2 = T_1$, a contradiction.
We note that here one property we used (with proof) for both the statements. If $f :(X,\mathcal{T} _1) \rightarrow (Y, \mathcal{T} _2)$ is a bijection with $X$ being compact and $Y$ being Hausdorff, then $f$ is a bijection.