Show that "complete" implies "closed" in metric space, but the converse is not true.

Answer

(1) Let $X$ be a metric space, $A \subset X$ be a complete subset, and $p \in X$ be a limit point of $A$. We want to show that $p \in A$. There exists a sequence $a_n$ such that $a_n \rightarrow p$. Note that $a_n$ is a Cauchy sequence, and therefore $a_n \rightarrow a \in A$. Then $p=a \in A$.

(2) Consider $A=\mathbb{Q} \cap [0,2]$ in a metric space $\mathbb{Q}$. Note that $A$ is closed in $\mathbb{Q}$. Consider a sequence $q_n \in A$ converging to $\sqrt{2}$. Then $q_n$ is a Cauchy sequence. However, $q_n$ does not converge in $A$, which implies $A$ is not complete.

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