Is the statement, if $E$ is closed in $X$, then $E$ is closed in $Y$, true?

Question

Let, $(Y,d)$ be a metric space and let $(X,d)$ be a subspace of $Y$. Since $(X,d)$ and $(Y,d)$ are metric spaces, we have that: Equivalence of Definitions of Closed Sets

Let, $\{x_n\}_{n \ge 1} \subseteq E$ such that $\lim_{n \rightarrow \infty } x_n = c$, then $c \in E$ by definition of $E$ being closed in $X$. Therefore, $E$ is closed in $Y$. Something about my proof does not seem right.

Thank you.

It may happen that $c\in Y\setminus X$, i.e., the sequence is not convergent in $X$. — Hagen von Eitzen, Jun 20 '21 at 17:32
Pick $Y=\mathbb{R}$, $X=E=(0,2)$ to see that the statement is not true. — Severin Schraven, Jun 20 '21 at 17:36
Let X be an open, not closed, subset of Y and E= X. Then E is closed in X but not in Y — user247327, Jun 20 '21 at 17:38

score 0 · Accepted Answer · answered Jun 20 '21 at 18:10

0

Any closed subset $E$ of $X$ takes the form $C\cap X$, where $C$ is a closed set in Y. If $X$ is closed in $Y$, then definitely $E$ is closed in $Y$. If $X$ is not closed, $X$ itself is a closed subset of $X$ but not closed in $Y$.

answered Jun 20 '21 at 18:10

Yathi

1,854

Is the statement, if $E$ is closed in $X$, then $E$ is closed in $Y$, true?

1 Answers1