Exercise 16.1

Proof. Let $T$ be the topology on $X$ and $T_Y$ be the subspace topology that $Y$ inherits from $X$. Also let $T_A$ and $T_A^{\prime }$ be the topologies that $A$ inherits as a subspace of $Y$ and $X$, respectively. Therefore we must show that $T_A=T_A^{\prime }$. Now, by definition of subspace topologies we have that,

Now suppose that $W\in T_A$ so that $W=A\cap V$ for some $V\in T_Y$. Then we have that $V=Y\cap U$ for some $U\in T$, and hence

since we have that $A\cap Y=A$ since $A\subset Y$. Since $U\in T$ this clearly shows that $W\in T_A^{\prime }$ so that $T_A\subset T_A^{\prime }$ since $W$ was arbitrary.

Then, for any $W\in T_A^{\prime }$, we have that $W=A\cap U$ for some $U\in T$. Let $V=Y\cap U$ so that clearly $V\in T_Y$. Then as before we have that $A=A\cap Y$ since $A\subset Y$ so that

and thus $W\in T_A$ since $V\in T_Y$. Since $W$ was arbitrary this shows that $T_A^{\prime }\subset T_A$, which completes the proof that $T_A=T_A^{\prime }$. □