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Per definition, a subset $A$ of a metric space $(X,d)$ is relatively compact if and only if its closure $\overline{A}$ is compact. That means that any sequence in $\overline{A}$ contains a subsequence that converges in $\overline{A}$. Does that imply that $\partial A$ is compact?

According to an alternative definition, $A$ is relatively compact if any sequence in $A$ contains a subsequence convergent in $X$. How do you derive from this that $\partial A$ is compact?

In other words, how can you show that both definitions are equivalent?

1 Answers1

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Yes.

The boundary of $A$ is defined as $$ \partial A=\overline{A}\cap\overline{A^c}. $$ So $\partial A$ is a closed subset of a compact set, and hence compact.

Let $\{x_n\}\subset\partial A$. Since $\partial A\subset\overline{A}$ and $\overline{A}$ is compact, then there exists a subsequence $x_{n_k}\to x\in\overline{A}$. However, $\{x_{n_k}\}\subset\partial A$, and since $\partial A$ is closed and $\{x_{n_k}\}$ convergent to $x$, then $x\in \partial A$. Thus every sequence in $\partial A$ possesses a convergent subsequence with limit in $\partial A$. Hence, $\partial A$ is compact.

  • But how do you get there from the alternative definition of relative compactness? –  Dec 02 '20 at 11:10
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    @user1975053 I only use the fact that $\overline{A}$ is compact and $\overline{A^c}$ is closed. – Yiorgos S. Smyrlis Dec 02 '20 at 11:12
  • Ignore the first definition. How can you get from the second definition to the compactness of $\partial A$? –  Dec 02 '20 at 11:17
  • if any sequence in A contains a subsequence convergent in $\overline A$, can there be no sequences in $\partial A$ without an in $\partial A$ convergent subsequence? –  Dec 02 '20 at 11:22
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    @user1975053 See my further explanation. – Yiorgos S. Smyrlis Dec 02 '20 at 11:38
  • My problem is, the compactness of $\overline A$ is not part of the second definition of relative compactness of $A$. How do you get there from this second definition? (I have edited my question accordingly). –  Dec 02 '20 at 12:01