Exercise 17.12

Proof. Suppose that $X$ is a Hausdorff space and that $Y$ is a subset of $X$. Consider any two distinct points $y_1$ and $y_2$ in $Y$ so that of course also $y_1,y_2\in X$. Then there are neighborhoods $U_1$ and $U_2$ of $y_1$ and $y_2$, respectively, that are disjoint since $X$ is Hausdorff. Since $U_1$ is open in $X$, we have that $V_1=U_1\cap Y$ is open in $Y$ by the definition of a subspace topology. Clearly also $V_1$ contains $y_1$ since $y_1\in U_1$ and $y_1\in Y$. Similarly $V_2=U_2\cap Y$ is an open set of $Y$ that contains $y_2$. Then, for any $x\in V_1$ clearly $x\in U_1$ so that $x\notin U_2$ since $U_1$ and $U_2$ are disjoint. Then $x\notin U_2\cap Y=V_2$. Since $x$ was arbitrary, this shows that $V_1$ and $V_2$ are disjoint, which then shows that $Y$ is a Hausdorff space as desired. □