# A hint on how to prove that if $(M,d)$ is complete then $M$ is closed.

I would like to recieve a hint on how to get started with a proof on

$M \subset X$ and $(M,d)$ is complete then $M$ is closed

I highly believe that there are alot of proofs regarding the above, but I'm afraid that if I google it I will be exposed to a solution, and I only want a hint, so that I can get started. ยจ

This is the information I have been given:

• What a metric space is.

• Def. of convergent seq. in metric space.

• Def. of Cauchy seq. in metric space.

• A metric space $(X,d)$ is called complete if every Cachy seq. is convergent.

• If $d'$ is a metric equivalent to $d$ then $(X,d)$ is complete iff $(X,d')$ is complete.

• What an open ball is.

• An closed set is the complement of an open set.

• What an open set is.

• If $M$ is complete then every limit point of $M$ is inside $M$, so $M$ is closed. – Henricus V. Nov 2 '16 at 23:30
• Sorry, but I am not familiar with that definition (We have not gone through it) @HenryW. – Olba12 Nov 2 '16 at 23:32
• I realise that my first draft of my question was bad, but I have made an update. @HenryW. – Olba12 Nov 2 '16 at 23:45

Let's show that the complement $U=X\setminus M$ of $M$ is open. Suppose $x\in U$ and that no open ball with center $x$ is contained in $U$.
In particular, for any integer $n>0$ there is $x_n\in B(x,1/n)$ (the ball with center $x$ and radius $1/n$) such that $x_n\notin U$, that is, $x_n\in M$.
Prove that the sequence $(x_n)$ is Cauchy. Since it is a sequence in $M$, it should converge to a point in $M$. But it converges to $x\notin M$.
Let $x \in X \setminus M$. If $X \setminus M$ is not open, then for each $n = 1, 2, 3, \dots$ $B(x, \frac{1}{n})$ meets $X \setminus M$ at a point other than $x$ (say $x_n$). Thus we have a convergent sequence $\{x_n\}_n$ in $M$ with $\lim_n x_n \notin M$, hence $M$ is not complete.