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$4.9$ Let $A$ be a subset of a metric space $S$. If $A$ is complete, prove that $A$ is closed. Prove that converse also holds if $S$ is complete.

For the first part, I assumed $\{ a_n\}$ to a Cauchy sequence in $A$. And since $\{a_n\}$ converges in $A$, the limit point of $\{a_n\}$ also lies in $A$ making $A$ closed.

I need proof of the converse and example that it does not hold if $S$ is not complete.

Martin Sleziak
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Mula Ko Saag
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1 Answers1

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I would argue that you need to flesh out the first part a bit. But, leaving that aside, let's look at the converse.

You want to show that if $S$ is a complete metric space and $A\subseteq S$ is closed, then $A$ is complete. So, naturally, you want to consider a Cauchy sequence $(a_n)_{n\in\mathbb{N}}$ of elements in $A$. Note that a sequence in $A$ is also a sequence in $S$, and $S$ is complete; also note that $A$ contains all of its limit points. You should be able to put those together to finish the result.

As far as a counterexample: think about $S=\mathbb{Q}$, with its usual metric inherited from $\mathbb{R}$. Can you think of any sets that are closed in $\mathbb{Q}$ but which have possible limit points (from $\mathbb{R}$) missing?

Nick Peterson
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