Prove the following statement with a counterexample: if $E$ is a compact metric space and $K$ is a closed subset of $E$ then $K$ is compact. demonstration. If we take a sequence $x_n$ of elements of $K$, since $E$ is compact, there must be a sub-succession of $x_n$ that is convergent. but since $K$ is closed, this sub-succession must converge to an element in $K$. which proves that $K$ is closed. b) consider $X$ a complete metric space and $S \subseteq X$. Show that $S$ is complete if and only if $S$ is closed in $X$. as shown? considering that $S$ is closed if and only if for every sequence of elements of $S$ convergent to a point $x \in X$ we have that $x \in S$, together with the fact that every convergent sequence is Cauchy.
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If $X$ is complete, then $S$ is complete iff it's closed in $X$. Though your question is not clear from your statements, your proof is correct – Richard Chen Oct 27 '21 at 12:55
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for part b, is that answer sufficient? some other way to prove it? – angelabayona Oct 27 '21 at 13:21
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In part b), you can argue as follows: If $S$ is complete, then given $x \in \overline{S}$, $x = \lim x_n$ with $(x_n)$ in $S$ in particular by the fact if $(x_n)$ is convergent, it is Cauchy, then since $S$ is complete, $x_n$ converges to an element of $S$, by uniqueness of the limit, that element must be $x$, so $x$ is in $S$. On the other hand suppose $S$ is closed on $X$. Let $(x_n)$ be a Cauchy sequence on $S$. So $(x_n)$ is a Cauchy sequence on $X$, so it is convergent on $X$, say for an element $x$, but since $x_n$ belongs to $S$ for all $n$ and $S$ is closed, so $x \in S$, in short every Cauchy sequence in $S$ converges to an element of $S$, so $S$ is complete.
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