# Retract subspace of metric space is closed [closed]

Let $(X,\varrho)$ be a metric space, and $Y\subseteq{X}$ a retract subspace of $X$. Show that $Y$ is closed in $X$.

Let $(a_n)$ be a sequence in $A$ that converges to a point $x\in X$. Since $A$ is a retract subspace, there is a continuous map $r:X\to A$ such that $r(a)=a$ for every $a\in A$. By continuity, we have $a_n=r(a_n)\to r(x)$. By uniqueness of limits in a metric space, we have $x=r(x)\in A$. Therefore $A$ is closed.
By replacing sequences by nets, this can be generalized to any Hausdorff space $X$.
• @mrose No it's a general topological property. A set $A$ is closed in a metric space (or more generally, a first countable space) iff whenever a sequence $(a_n)$ in $A$ converges to a point $x\in X$, then $x\in A$. See here: math.stackexchange.com/questions/760594/… – John Griffin Sep 20 '17 at 0:02