I wanted to prove that the closure of a set is a closed set. The book I use suggested that it would be easier to prove it by first proving that the set (L) of all limit points of a set A is a closed set. Starting the proof I wrote: Let x be a limit point of L such that x it's not in L. This implies that for every epsilon&gt;0, there exist y different from x, such that y lies both in L and in the open interval (x-eps,x+eps). But since y Is in L, for every epsilon&gt;0, there exist z different from y such that z lies in the set A and in the open interval (y-eps,y+eps). I then used the number y lies in (x-eps,x+eps) can be written as |x-y|&lt;eps/2 (since it's true for every eps. it's true for eps/2) and using the same fact i wrote |y-z|&lt;eps/2, adding these two and using the triangular inequality i was able to write that |x-z|&lt;eps Since z is already in A, the only thing left to prove in order for me to prove that x is a limit point of A (and thus produce the contradiction that x is in L), is prove that z is not equal to x, but I'm stuck. Is there a way to prove it? Or maybe is this proof wrong?
From Rudin's Principles of Mathematical Analysis (Chapter 2, Exercise 6)  Let $E'$ be the set of all limit points of a set $E$. Prove that $E'$ is closed.  I think I got it but my argument is a bit hand wavy:  If $x$ is a limit point of $E'$, then every neighborhood of $x$ contains some $y\in E'$, and every neighborhood of $y$ contains some $z\in E$. Therefore every neighborhood of $x$ contains some $z\in E$, and so $x$ is a limit point of $E$. Then $x\in E'$, so $E'$ is closed.  The thing that's bugging me is the leap from one neighborhood to another. Is this formally correct?
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