# 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 $(Y,d)$.

Book: Functions of Complex Variable 2nd ed. Conway GTM Book

Chapter 2: Topology in $\mathbb{C}$, Section: 2.1

We already proof this in exercise 8 using the word "open" instead of "closed".

Now we want to prove that:

Exercise 9: 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 $F_1 = F\cap Y$.

and I want to use the previous problem to solve this one and this is my try:

$\textbf{Proof:}$

$[\Longrightarrow]$ Since $F$ is closed in $(X,d)$, by definition

$F^c = X\setminus F \mbox{ is open in } (X,d)$

By previous exercise, we can take $F^c = G$, then $G\cap Y$ is open in $(Y,d)$.

Now $G\cap Y \mbox { open in } (Y,d) \Rightarrow (X\setminus F)\cap Y \mbox{ is open in (Y,d) }$ Hence,

$(X\setminus F)^c \cap Y = F \cap Y$ is closed in $(Y,d)$.

$[\Longleftarrow]$

Let $F_1 \subseteq Y$ be closed in $(Y,d)$, then $G_1 = Y\setminus F_1$ is open in $(Y,d)$. By previous exercise, $\exists G \subseteq X$ such that $G$ is open in $(X,d)$ $G_1 = G \cap Y$ Let $F = X\setminus G$, this set is closed in $(X,d)$, and

$Y\setminus F_1 = (X\setminus F)\cap Y$

is open on $(Y,d)$ and then

$F_1 = F \cap Y.$

• Looks correct to me. Although it took me a while to understand how you derived the second-to-last line, and how the last line followed from it. If this is for an assignment, I'd add some additional steps and/or explain a little more clearly at those points. – WB-man Mar 28 '17 at 18:19