Complete metric space proof

  • Thread starter Incand
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  • #1
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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?
 

Answers and Replies

  • #2
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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?
 
  • #3
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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,.
 
  • #4
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3,297
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.
 
  • #5
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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, thats 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.
 
  • #6
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3,297
So choosing ##N_\epsilon = floor(1+2/\epsilon)## should do it.


Thanks, thats 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!
 

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