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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.

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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 $F_1 = F\cap Y$.

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  • $\begingroup$ Thanks for the examples, it helps me to understand my conceptual problem with the open and closed sets on the relative topology. $\endgroup$ – Richard Clare Mar 22 '17 at 18:13
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It is sufficient to show the metric topology is the subspace topology. This is easily verified.

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