Question

Let Y be a subspace of a metric space (X, d). Then show that 𝐹⊆𝑌 is closed in Y if and only if

𝐹=𝑌∩𝐻 for some closed subset H of X.

Answer 1

• Let us suppose that $F\subseteq Y$ is such that there exists a closed set $H\subseteq X$ such that $F=H\cap Y$. Suppose that $(x_n)_{n\in \mathbb{N}}$ is a sequence of elements of $F$ that admits a limit $x$ in $Y$ (i.e. $x_n\to x$ and $x\in Y$). As $F\subseteq H$ and $H$ is closed, we have $x\in H$ (as we can view the sequence $x_n$ as a sequence in $X$). As $x\in H, x\in Y$, we have $x\in H\cap Y = F$. Therefore, $F$ is closed in $Y$.
• Now let us suppose that $F\subseteq Y$ is closed in $Y$. We take $H:= \bar{F}$ and we will prove that indeed $F=Y\cap H$. As $F\subseteq Y$ and $F\subseteq \bar{F}$ we have $F\subseteq Y\cap H$. Now let $x\in Y\cap H$. As $x\in \bar{F}$, there exists a sequence $(x_n)_{n\in \mathbb{N}}$ of elements of $F$ such that $x_n\to x$ in $X$ . As $x\in Y$, we also have that $x_n\to x$ in $Y$. Finally, as $F$ is closed in $Y$, we have $x\in F$ (as it is a limit in $Y$ of elements of $F$). Therefore, $Y\cap H \subseteq F$ and by the double inclusion we conclude that $F=Y\cap H$ for a closed subset $H$ of $X$.