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### Every infinite subset of (0,1) has a limit point

Theorem:Every infinite subset V of (0,1) has a limit point. (Equivalently, every infinite set of points in (0,1) contains a sequence converging to a point not in the sequence, though perhaps in the set )

Proof: It is sufficient to show that some subset of V has a limit point. Assume the contrary. Let V0=V and form a sequence according to the following rule:
Vn+1=Vn - lub(Vn) (lub indicates the least upper bound)

Observe that for all i:

1. lub(Vi) exists because Vi is bounded above by 1.
2. Vi is non-empty because V0 is infinite and only a single point is removed at each step.
3. lub(Vi) ∈ Vi. Otherwise, it would be a limit point of Vi, a contradiction.
4. lub(Vi+1) < lub(Vi). Since Vi+1 ⊂ Vi, ∀v∈Vi+1, v < lub(Vi). (The inequality is strict because Vi+1=Vi-lub(Vi)). Therefore, by (3), lub(Vi+1) < lub(Vi).

Now, consider the following two cases:

There exists a j such that Vj=Vj+1
Since Vj = Vj+1 = Vj - lub(Vj), lub(Vj) ∉ Vj, contradicting (3).
For all j, Vj ≠ Vj+1
Form the set A={lub(Vj)}. A is itself a subset of (0,1) bounded below by 0, and thus has a greatest lower bound. Let a=glb(A). If a ∉ A, a is a limit point of A, a contradiction. If a ∈ A, a=lub(Vk) for some k. However, by (4), Vk+1 < a, a contradiction. QED