Hello. I didn't know if this is the right forum to post this (since this is the Calculus section and my question pertains to real analysis), but I would like some clarifications on a proof I am preparing for the propositon "If a set S is dense, then every point in S is an accumulation point".(adsbygoogle = window.adsbygoogle || []).push({});

What I did is assume the contrary, that is, "If a set S is dense then none of its points are accumulation points". Then for an arbitrary element x in S it is possible to create a deleted ε-neighborhood for any ε>0 such that (x-ε, x+ε) does not contain any point in S. Now, i let a and b be elements in S such that a < x < b such that x-a = b-x. If I assign ε=b-x (in turn, ε=x-a), then the deleted neighborhood for this choice of ε must be empty to guarantee that x is not an accumulation point, that is, (a,x)U(x,b) should be empty, which means (a,x) is empty and (x,b) is also empty. This means that there are no numbers between a and x and between x and b, which means that S is not dense (since a dense set is defined to be that for any elements c and e and c<e there exists an element d such that c < d < e), which leads to a contradiction since we assume S to be dense. Hence, If a set S is dense, then every real point in S is an accumulation point. ///

My questions are:

(1) can I simply assume that x is between a and b since S is dense?

(2) If I choose my ε to be b-x, are a and b in the neighborhood of x? this also leads to the question that are deleted neighborhoods open sets...

Thanks for any help.

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# Some clarifications on my proving

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