Completeness and closedness

Proposition. Let $(X,d)$ be a metric space .

1. If $E\subset X$ is complete , then $E$ is closed in $X$.
2. If $X$ is complete and $E\subset X$ is closed, then $E$ is complete (with the restricted metric).

Context. This gives a practical way to recognize complete sets: inside a complete ambient space, “closed” and “complete” coincide.

Proof sketch.

1. Take a sequence $(x_n)\subset E$ converging in $X$ to $x$. Any convergent sequence is Cauchy , so $(x_n)$ is Cauchy in $E$. By completeness of $E$, it converges in $E$ to some $y\in E$. By uniqueness of limits , $x=y\in E$, hence $E$ is closed.
2. Let $(x_n)$ be Cauchy in $E$. Then it is Cauchy in $X$ and hence converges in $X$ (since $X$ is complete) to some $x\in X$. Because $E$ is closed and $(x_n)\subset E$ converges to $x$, we have $x\in E$. Thus $(x_n)$ converges in $E$.