$(M,d)$ be a complete metric space and $N \subset M$ a closed subset $\Rightarrow N$ is complete

Question

$(M,d)$ be a complete metric space and $N \subset M$ a closed subset $\Rightarrow N$ is complete

proof

take a Cauchy sequence in $(N,d)$ then this Cauchy sequence converges in $M$ since M is complete. Now we need to show that this convergence occurs in $N$. Since $N$ is closed, $\cdot\cdot\cdot$

Question Any hint to proceed above reasoning?

Answer 1

Since $N$ is closed, the limit point has to lie in $N$. If not, then you can find some small open ball in $M\setminus N$ containing the limit point. But by definition of a limit point, this implies that...?

Answer 2

If $\left\{x_n \right\}$ is Cauchy sequence in $M$, so it is Cauchy sequence in $N$. Since N is complete, so $\exists x_0 \in N$ such that $x_n \to x_0$. If $x_0 \notin M$, since $N-M$ is opened and $x_0 \in N-M$, so $\exists r >0$ such that $B(x_0, r) \subset N-M$. Therefore $\exists n_0 \in N^* : \forall n > n_0, x_n \in M$, irrationally. So $x_0 \in M$.

Answer 3

Let's Say $N$ is not complete. Then we can say there exist a Cauchy sequence $\{x_n\}$ in $N$ which does not converge in $N$. But it must have a limit point $l$ in $M$ since $M$ is complete. so $l$ does not belong to $N$. Then we can conclude $N$ is not closed. Contrapositively we can say If $N$ is closed in $M$, $N$ is complete