Prove if x is an accumulation point of a set S, then every neighborhood of x con

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Analysis: The statement is of the form pÞq where p: x is an accumulation point of the set S q: every neighborhood of x contains infinitely many points of S. The contrapositive would be of the form ~qÞ~p. In this case, ~q: there exists a neighborhood of S which contains a finite number of elements of S. ~p: x is not an accumulation point of the set S Therefore in order to prove the original statement we can prove the contrapositive which says, “If there exists a neighborhood of S which contains a finite number of elements of S, then x is not an accumulation point of the set S.” In order to prove the contrapositive, we need to find a neighborhood of x that contains a finite number of element of the set S and show that for all e>0, N*(x;e)ÇS=Æ. That is, we will show that for a point x, there exists a deleted neighborhood of x that does not contain a point in S. Proof: Suppose SÍR and there exists a neighborhood N(x;e)={x1, x2, x3, …,xn} which contains a finite number of elements of S. Further suppose that x is an accumulation point of S. If x1 is the closest point in N(x;e) to the accumulation point, then e1=|x-x1| where e1 is the radius of the neighborhood by definition 13.1. Since x1 is the closest point of N(x;e) to x, all points in the smaller neighborhood N(x; e1) will be within the radius e1 from x. By definition 13.14, a point x in R is an accumulation point of S if every deleted neighborhood of x contains a point of S. In this case, N*(x; e1) does not contain any points of S, and therefore we conclude that N*(x;e)ÇS=Æ. Because N*(x;e)ÇS=Æ, x is not an accumulation point of the set S as required. This proves the contrapositive of the original statement and we conclude that “If x is an accumulation point of the set S, then every neighborhood of x contains infinitely many points of S”.¨ This first draft was followed by a peer critique. The student commented on Donna’s paper: Critique: You say “Suppose S?R and there exists a neighborhood N(x;e)={x1, x2, x3, …,xn} which contains a finite number of elements of S.” This is ~q. And then you say “Further suppose that x is an accumulation point of S.” This is p. In your setup, I think that you are doing (-q?p) ?c, or prove by contradiction by not prove by contrapositive.

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