Complete metric space proof

[Incand]

Homework Statement

Let $E$ be a metric subspace to $M$. Show that $E$ is closed in $M$ if $E$ is complete. Show the converse if $M$ is complete.

Homework Equations

A set $E$ is closed if every limit point is part of $E$.
We denote the set of all limit points $E'$.

A point in $p\in M$ is a limit point to $E\subseteq M$ if $\forall \epsilon > 0$ $\exists q \in E \cap B(p,\epsilon)$

The Attempt at a Solution

We want to show that $E' \subseteq E$. Take $p \in E'$ then clearly we can choose
$p_n \in B(p,1/n)$ so that $p_n \in E$.
But then for all $\epsilon > 0$ $d(p_n,p_m)\le d(p_n,p)+ d(p_m,p)<\epsilon$ for $n,m \ge N_\epsilon = 2/\epsilon$ i.e. $(p_n)$ is a cauchy sequence.

But then it must converge to some $\lambda \in E$. However $\lambda = p$ since $d(p,\lambda) = d(p_n,p)+d(p_n,\lambda)< \epsilon$

Is this a correct proof? Should I use a similar approach to the second part?

 

[micromass]

But then for all $\epsilon > 0$ $d(p_n,p_m)\le d(p_n,p)+ d(p_m,p)<\epsilon$ for $n,m \ge N_\epsilon = 2/\epsilon$

Can you elaborate?

 

[Incand]

Can you elaborate?

I tried to expand the argument a bit, hopefully this works better:

We have $d(p_n,p)<1/n$ where $p_n \in E$. By the triangle inequality we have
$d(p_n,p_m) \le d(p_n,p)+d(p_m,p) < 1/n+1/m \le 2/N_\epsilon = \epsilon$ for $n,m\ge N_\epsilon$.
So $\forall \epsilon > 0$ we have $d(p_n,p_m) < \epsilon$ for $n,m \ge N_\epsilon = 2/\epsilon$.

Btw. I'm unclear about if using $1/n$ is neccesary at all. It seemed like a good idea at the time but perhaps starting from $d(p_n,p)< \epsilon/2$ we get $n,m$ can take any values in $Z^+$ instead,.

 

[micromass]

I tried to expand the argument a bit, hopefully this works better:

We have $d(p_n,p)<1/n$ where $p_n \in E$. By the triangle inequality we have
$d(p_n,p_m) \le d(p_n,p)+d(p_m,p) < 1/n+1/m \le 2/N_\epsilon = \epsilon$ for $n,m\ge N_\epsilon$.
So $\forall \epsilon > 0$ we have $d(p_n,p_m) < \epsilon$ for $n,m \ge N_\epsilon = 2/\epsilon$.

Seems ok. The only minor concern is that $2/\varepsilon$ is not necessarily an integer, so you might not be able to put it equal to $N_\varepsilon$.

Btw. I'm unclear about if using $1/n$ is neccesary at all. It seemed like a good idea at the time but perhaps starting from $d(p_n,p)< \epsilon/2$ we get $n,m$ can take any values in $Z^+$ instead,.

I think it is definitely necessary, I don't see how you can do without.

 

[Incand]

So choosing $N_\epsilon = floor(1+2/\epsilon)$ should do it.

I think it is definitely necessary, I don't see how you can do without.

Thanks, that's good to know!Moving on to the next part I take it I'm meant to show that
$M$ complete, $E$ closed $\Longrightarrow$ $E$ complete.

Let's consider an arbitrary CS $(p_n)$ in $E$. Since $M$ is complete it converges to a point $p\in M$.
This means that $\forall \epsilon > 0$ $\exists N$ so that $d(p_n,p)< \epsilon$ $\forall n \ge N$.
If $p_n = p$ for some $n$ we have $p\in E$. If $p_n \ne p$ it means that $p$ is a limit point of $E$ since $p_n \in E$ and $\epsilon$ is arbitrary small. And since $E' \subseteq E$ since $E$ is closed we have that $E$ is complete.

 

[micromass]

So choosing $N_\epsilon = floor(1+2/\epsilon)$ should do it.


Thanks, that's good to know!Moving on to the next part I take it I'm meant to show that
$M$ complete, $E$ closed $\Longrightarrow$ $E$ complete.

Let's consider an arbitrary CS $(p_n)$ in $E$. Since $M$ is complete it converges to a point $p\in M$.
This means that $\forall \epsilon > 0$ $\exists N$ so that $d(p_n,p)< \epsilon$ $\forall n \ge N$.
If $p_n = p$ for some $n$ we have $p\in E$. If $p_n \ne p$ it means that $p$ is a limit point of $E$ since $p_n \in E$ and $\epsilon$ is arbitrary small. And since $E' \subseteq E$ since $E$ is closed we have that $E$ is complete.

Seems ok!