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## Homework Statement

Let ##E'## be the set of all limit points of a set ##E##. Prove that ##E'## is closed. Prove that ##E## and ##\bar E = E \cup E'## have the same limit points. Do ##E## and ##E'## always have the same limit points?

## Homework Equations

Theorem:

(i) ##\bar E## is closed

(ii) ##E=\bar E## if and only ##E## is closed.

Theorem:

For any collection of ##\{G_\alpha\}## of open sets, ##\bigcup_\alpha G_\alpha## is open.

## The Attempt at a Solution

I tried to prove the above but since my solution seems quite different from the one in the solution manual I suspect I may have done something wrong. I Would appreciate if someone could take a look and see if it's wrong/incomplete or if I done something unnecessarily complicated.

We split the first part of proof into two cases:

Case 1: Assume ##E## is open. Then ##E'## has to be closed since the union of two open sets is open and ##\bar E## is closed (contradiction otherwise).

Case 2: Assume ##E## is closed. Then ##E= \bar E##. Then it follows that ##(\bar E)' = E'## but then ##(E')' \subseteq E'##. Hence every limit point of ##E'## is in ##E'## so ##E'## is closed.

Take any ##x\in E'## then ##x \in \bar E##. Since ##x## is a limit point of ##E## and ##E \subseteq \bar E##, ##x## is also a limit point of ##\bar E##. We then have ##E' \subseteq (\bar E)'##.

Take ##x \in (\bar E)'##. Since ##(\bar E)'## is closed (as we showed before) this means ##x \in \bar E##. Then at least one of the following is true ##x \in E## or ##x \in E'##. If the second case is true we're done.

Otherwise if ##x \in E##. We can further assume ##E## is open (If ##E## is closed we're done since then ##E = \bar E##). This means that there is an open ball around ##x## ##B(x,r) = \{y;d(x,y)<r\}## that is a subset of ##E## and hence ##\bar E##. Then since ##x## is a limit point of ##\bar x## that open ball must contain another point ##q \in E## such that ##q \ne p##.

But then every open ball with a larger radius must also contain this point. If instead the radius is smaller that open ball is contained in ##B## and hence in ##\bar E## and hence that ball must also contain a point ##q_2 \in E## such that ##q_2 \ne x##. Hence ##x \in E'## and ##E' = (\bar E)'##.