If K is a subset of all real numbers then prove that every infinite subset of K has a limit point in K?

Answer 1

We prove it for the interval (0,1) and the proof is easily extended to any subset of real numbers. An alternative way to state this is every infinite set of points in (0,1) contains a sequence converging to a point not in the sequence.

So proof is by contradiction. We want to show that some subset (arbitrary) of (0,1) has a limit point. So let's assume this is NOT true.

Let K be the subset of (0,1) consisting of all reals in the interval, of course K is infinite, Now let's start by forming a sequence and let A0 be the first term and let A0 =K
An+1=An - lub(An) where lub means the least upper bound

1. Now we note that the lub of A1 exists because A1 is bounded by 1 above.
2. A1 is non-empty sine A0 is infinite and one point is removed at each step.
3. lub(Ai) ∈ Vi. Otherwise, it would be a limit point of Ai which is a contradiction.
4. lub(Ai+1) < lub(Ai). Since Ai+1 ⊂ Ai, ∀a∈Ai+1, a < lub(Ai )

Now let's look at two mutually exclusive possibilities:
Case 1: We know there is some k such that Ak =Ak+1
This can't be because it violates #3 since since Ak = Ak+1 = Ak - lub(Aj), lub(Ak) ∉ Aj
So let's say for all k Ak ≠ Ak+1 which we assume since case 1 is not possible.
Then
Form the set S={lub(Ak)}. A is itself a subset of (0,1) bounded below by 0, so it has a greatest lower bound. Let s=glb(S). If s ∉ S, s is a limit point of S, a contradiction. If s ∈ S, s=lub(Ak) for some k. However, by (4), Ak+1 < s, a contradiction.

So we assume our subset does have a limit point and the proof is complete.