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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>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>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|<eps/2 (since it's true for every eps. it's true for eps/2) and using the same fact i wrote |y-z|<eps/2, adding these two and using the triangular inequality i was able to write that |x-z|<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?

Evan
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    What definition of a "closed set" are you using? - there are different ways of approaching the topic and several equivalent characterisations. The proof will depend on where you begin. A clear statement of the definition at the start also helps to focus the mind on the goal – Mark Bennet Apr 16 '21 at 09:15
  • https://math.stackexchange.com/questions/1986774/prove-that-a-set-is-closed-iff-it-contains-all-its-accumulation-points – C Squared Apr 16 '21 at 09:28

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