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

Question

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?

Reason that the point of convergence of that sequence in $M$ is a limit point of $N$ — Kaj Hansen, Jun 01 '17 at 10:31
hii@Daschin if u accept my answer plzz click on the tick button. I am short of reputations.. It's a request.. — anonymous, Jun 01 '17 at 11:15
See also Showing that if a subset of a complete metric space is closed, it is also complete and Proof that a subspace $A$ of a complete metric space $X$ is complete iff $A$ is closed. — Martin Sleziak, Jun 01 '17 at 11:15

score 0 · Answer 1 · answered Jun 01 '17 at 10:34

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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...?

answered Jun 01 '17 at 10:34

user363120

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a set $A $is closed $\Rightarrow$ any limit points of A lies in A. Is it trivial fact? – Daschin Jun 01 '17 at 10:36
@Daschin What is your definition of a closed subset of a metric space? – user363120 Jun 01 '17 at 10:44
when its complement($M\setminus A$) is open, we call it closed – Daschin Jun 01 '17 at 10:46
@Daschin When does a sequence $(x_n)_n$ in $M$ converge to a point $x\in M$? – user363120 Jun 01 '17 at 10:50
when for $n > N$, if $d(x_n , x)<e$ $\forall$ $e\gt0$. – Daschin Jun 01 '17 at 10:57
@Daschin No, if for every $e>0$ there exists some $N$ such that $d(x_n,x) < e$ for all $n> N$. What can you conclude if the open ball in my answer is of the form $B_e(x)={y\in M: d(x,y)< e}$? – user363120 Jun 01 '17 at 11:08

Vũ Nguyễn Đình · Answer 2 · 2017-06-01T11:00:21.553

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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 $x_0 \in M$.

edited Jun 01 '17 at 11:00

answered Jun 01 '17 at 10:36

Vũ Nguyễn Đình

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one can't use the fact N is complete before prove it – Daschin Jun 01 '17 at 10:37
@ Daschin I edited my answer. – Vũ Nguyễn Đình Jun 01 '17 at 11:01

score 0 · Answer 3 · edited Jun 01 '17 at 11:16

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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

edited Jun 01 '17 at 11:16

Martin Sleziak

53,687

answered Jun 01 '17 at 10:53

anonymous

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

3 Answers3