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 $X$.

  • $\begingroup$ Unless I am mistaken, my textbook claims that a sequence converging to a point in the subspace does not necessarily imply that the subspace is closed. Is there something about the fact that it is retract that allows this? $\endgroup$ – mrose Sep 19 '17 at 23:56
  • $\begingroup$ @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/… $\endgroup$ – John Griffin Sep 20 '17 at 0:02
  • $\begingroup$ I'm sorry, topology confuses me quite a bit. The textbook says it is not true "for a general topological space." Is this wrong? $\endgroup$ – mrose Sep 20 '17 at 0:13
  • $\begingroup$ It's correct. A metric space is not a general topological space. For a general topological space you need a different notion of convergence, like nets. But for metric spaces, sequences suffice. $\endgroup$ – John Griffin Sep 20 '17 at 0:16
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    $\begingroup$ It's starting to make sense. Thank you $\endgroup$ – mrose Sep 20 '17 at 0:18

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