- #1

Incand

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## Homework Statement

Let ##E \subseteq M##, where ##M## is a metric space.

Show that

##p\in \overline E = E\cup E' \Longleftrightarrow## there exists a sequence ##(p_n)## in ##E## that converges to ##p##.

##E'## is the set of limit points to ##E## and hence ##\overline E## is the closure of ##E##.

## Homework Equations

Definition:[/B]

A point ##p## is a limit point to ##E## if ##\forall \epsilon > 0## there is a point ##q\in B(p,\epsilon)## so that ##q\in E## and ##q \neq p##.

Definition:

##(p_n)## converges if ##\forall \epsilon > 0## exists ##N## so that

##d(p,p_n)< \epsilon## for all ##n\ge N##.

## The Attempt at a Solution

I think I managed to show the above but I would like someone to go over my proof and check if it's correct:

Starting with ##\Longleftarrow## we have ##\forall \epsilon > 0##

##d(p_n,p)<\epsilon## ##\forall n \ge N## for some ##N## where ##p_n \in E##. But this means that ##p## is a limit point of ##E##. i.e. ##p \in E' \subseteq \bar E##.For ##\Longrightarrow## we have ##p \in \overline E##.

If ##p \in E## the sequence of ##p_n = p## does it.

If ##p \in E'## we select

##p_n \in B(p,1/n)## for ##n \in Z^+## so that ##p_n \in E##.

But then ##d(p_n,p)<1/n = \epsilon## for all ##n\ge N_\epsilon = 1/\epsilon##. i.e. ##p_n \to p##.

One question I have is if the detour around ##1/n## is neccesary or if I simply could've picked ##p_n \in B(p,\epsilon)##.