# Let $(X,d)$ be a metric space and $Y \subseteq X$. If $F \subseteq X$ is closed, show that $F\cap Y$ is closed in $(Y,d)$.

Let $(X,d)$ be a metric space and $Y \subseteq X$. Suppose $F \subseteq X$ is closed, show that $F\cap Y$ is closed in $(Y,d)$. Conversely show that if $F_1 \subseteq Y$ is closed in $(Y,d)$, then $\exists F \subset X$ such that $F$ is closed and $F_1 = F\cap Y$.

# Proof:

$[\Longrightarrow]$ We will show that $F\cap Y$ is closed in $(Y,d)$ by proving that $\forall x \in F\cap Y, B^Y(x,\epsilon)\cap(F\cap Y) \neq \emptyset, \mbox { } \forall \epsilon > 0$ Let $x \in F\cap Y$, then $x \in F$ and $x \in Y$. Since F is closed in (X,d), $B^X(x,\epsilon)\cap F \neq \emptyset \mbox { } \forall \epsilon >0.$ Since $Y\subseteq X$ we have that $B^Y(x,\epsilon) = B^X(x,\epsilon)\cap Y$, $\forall \epsilon > 0$, then \begin{align*} B^Y(x,\epsilon)\cap(F\cap Y) &= B^X(x,\epsilon)\cap Y \cap (F\cap Y) \\ & = B^X(x,\epsilon) \cap (F\cap Y) \end{align*} therefore, $x \in B^X(x,\epsilon) \cap (F\cap Y) \Rightarrow B^X(x,\epsilon) \cap (F\cap Y) \neq \emptyset, \forall \epsilon > 0.$

$[\Longleftarrow]$ Let $F_1$ be a closed set in $(Y,d)$, then $F_1 = \overline{F_1} \subseteq Y$ \begin{align*} \overline{F_1} =& \bigcap\{K \subseteq Y | K \mbox{ is closed and } F_1 \subseteq K\} \\ =& \bigcap\{K \subseteq Y | K \mbox{ is closed and } F_1 \subseteq K\}\cap Y. \end{align*} The set $F$ is clearly closed in $(Y,d)$, because is an arbitrary intersection of closed subsets of Y. We can take $F = \bigcap\{K \subseteq Y | K \mbox{ is closed and } F_1 \subseteq K\}.$ This set is closed in $(X,d)$ because is closed in a subset of $X$.

Am I correct? any comments or suggestions? I do not want to prove this using the complements, I know it will be easier.

No, both ways are incorrect.

First is incorrect because what you're trying to prove does not imply what you're supposed to prove. For example the open unit interval $(-1,1)$ in $\mathbb R$ fulfils $\forall x\in (-1,1)\cap\mathbb R, B^\mathbb R(x,\epsilon)\cap((-1,1)\cap\mathbb R) \ne \emptyset$.

The second is incorrect because that a set is closed in a subset does not mean that it's closed in the superset. For example the open unit internval $(-1, 1)$ is closed in $(-1, 1)$, but it's not closed in $\mathbb R$.

If you don't want to use complements you'll probably need to use closures. That a set is closed means that it's equal to it's closure (or that the set is a subset of it's closure).

So what you need to do in the first case is to consider a limit point $x$ of $F\cap Y$ in $(Y,d)$. Such a limit point must be in $Y$ and also it must be that for every $\epsilon>0$ that $B^Y(x,\epsilon)$ must intersect $F\cap Y$ which means that it intersects both $Y$ and $F$ which means that $B^X(x,\epsilon)$ intersects $F$ which means $x\in F$, but since $x\in Y$ we have $x\in F\cap Y$. So $F\cap Y$ contains all of it's limit points and is therefore closed.

The other way around you could use $F$ being the closure of $F_1$ in $(X, d)$. In similar way as before we see that if $F_1$ has a limit point in $(X, d)$ that happens to be in $Y$ it is a limit point in $(Y,d)$ and as seen in $F_1$. This means that it's limitpoints is either in $F_1$ or outside $Y$. We see therefore that $F_1 = F\cap Y$.

• Thanks for the examples, it helps me to understand my conceptual problem with the open and closed sets on the relative topology. – Richard Clare Mar 22 '17 at 18:13

It is sufficient to show the metric topology is the subspace topology. This is easily verified.