Question

Find the limiting points of the set {1, 2, 3, 4, … }.

Answer 1

Answer:

the set A below is the set of all limit points.

A={0}∪{1a∣a∈N}

I came up with a proof inspired by that of Gregory Grant. In fact mine is another way to put his. To show that A contains every limit point of S, let's assume real number x and x∉A. We have to show that x is not a limit point for S.

Since x∉A, there exists an ϵ>0 for which the interval I=(x−ϵ,x+ϵ) has no real number of the form 1a where a∈N. Also, ϵ is such that 0∉I, and hence, x−ϵ>0. These are hold for the interval I′=(x−ϵ2,x+ϵ2), too. Our goal is to show that there are only finite number of elements both in S and I′, if any. To do so, assume y=1m+1n (n,m∈N) to be an element of both S and I′, and without loss of generality, 1n≤1m. Since I has no real number of the form 1a (a∈N), then:

1n≤1m≤x−ϵ(1)

For y∈I′ it is necessary that:

x−ϵ2<1n+1m(2)

and since 1n≤1m, we have x−ϵ2<2m, or:

0<12(x−ϵ2)<1m

So there are only finite natural numbers for m. On the other hand, using (1) and (2):

x−ϵ2<1n+x−ϵ

or:

0<ϵ2<1n

which shows there are finite natural numbers for n. Thus, there are only finite elements like y which belong to both S and I′, and hence, x cannot be a limit point for S.

Step-by-step explanation:

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