Determine the accumulation points of the set 2nlk n and k ar

Determine the accumulation points of the set {2^n+l/k: n and k are positive integers}.

Solution

I claim that {2^n | n is a positive integer} is the set of all accumulation points of the given set S.

(i) Fix n. Given epsilon > 0, there exists an integer N > 0 such that 1/N < epsilon. Thus, N_(epsilon)(2^n) intersect S contains at least {2^n + 1/(N+1), 2^n + 1/(N+2), ...}. This this intersection is not empty, we see that 2^n is an accumulation point of S.

(ii) It should be clear that not other points (other than those in (i) that are NOT in S can be accumulation points; just take the nbhd of such a point to have epsilon < (the distance to the nearest point in S). This will give an intersection with S being the empty set.

(iii) As for the points in S, fix n as in (i). Also fix n. By choosing
0< epsilon < min{1/k - 1/(k+1), 1/(k-1) - 1/k} [yes, this is overkill!], we see that N_(epsilon)(2^n + 1/k) intersect S is empty. Hence, no element of S is an accumulation point of S.