Question

Find the derived set of all integer point

Answer 1

Let B=A∪{0}.Then B′=A′∪({0})′=A′.We claim that A′=[0,1].For proving this let us take an element c∈[0,1].Then either c∈B or c∈[0,1]∖B.Let us take δ>0 arbitrarily.

Case-1 :

Let c∈B. Then either c=1 or c is of the form 2m+12n where n∈N and m is any non-negative integer such that m<2n–12 or in other words m≤[2n–12].

Let c=1.Then there exists k∈N such that 0<12k<δ.Then clearly 1≠1−12k∈(1−δ,1+δ).Now 1−12k=2k–12k=2(2k–1–1)+12k.Since 2k–1–1<2k so we have 1−12k∈B.This shows that 1 is a limit point of B.Now let c∈B∖{1}.Then there exists a natural number n and a non-negative integer m satisfying m<2n–12 such that c=2m+12n.Now there exists r∈N such that 0<12r+n<δ. Then clearly c−12r+n∈(c−δ,c+δ) andc−12r+n=2m+12n−12r+n=2r+1m+2r–12r+n=2(2rm+2r–1–1)+12r+n.Now since m<2n–12 so we have 2rm+2r–1–1<2r(2n–1)2+2r–1−1=2r+n−1<2r+n.Which shows that c−12r+n∈B.This shows that cis a limit point of B.Thus every element of B is a limit point of B.