Exercise 16.2

Proof. First, we have that

by the definition of subspace topologies. So for any $V\in T_Y$ we have that $V=Y\cap U$ where $U\in T$. Then also $U\in T^{\prime }$ since $T^{\prime }$ is finer than $T$. This shows that $V\in T_Y^{\prime }$ since $V=Y\cap U$ where $U\in T^{\prime }$. Hence $T_Y^{\prime }$ is finer than $T_Y$ since $V$ was arbitrary.

To show that it is not necessarily strictly finer, consider the sets $X=\left\{a,b,c\right\}$ and $Y=\left\{a,b\right\}$ so that clearly $Y\subset X$. Consider also the topologies

on $X$ so that clearly $T^{\prime }$ is strictly finer than $T$. This results in the subspace topologies

which are clearly the same so that $T_Y^{\prime }$ is not strictly finer than $T_Y^{\prime }$, noting that it is technically still finer. However, if we instead have the topologies

then

so that $T_Y^{\prime }$ is strictly finer than $T_Y$. Thus we can say nothing about the strictness of the relation of the subspace topologies. □