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

(mathematics, topology)

**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 V_{0}=V and form a sequence according to the following rule:

V_{n+1}=V_{n} - lub(V_{n}) (lub indicates the least upper bound)

Observe that for all i:

- lub(V
_{i}) exists because V_{i} is bounded above by 1.
- V
_{i} is non-empty because V_{0} is infinite and only a single point is removed at each step.
- lub(V
_{i}) ∈ V_{i}. Otherwise, it would be a limit point of V_{i}, a contradiction.
- lub(V
_{i+1}) < lub(V_{i}). Since V_{i+1} ⊂ V_{i}, ∀v∈V_{i+1}, v < lub(V_{i}). (The inequality is strict because V_{i+1}=V_{i}-lub(V_{i})). Therefore, by (3), lub(V_{i+1}) < lub(V_{i}).

Now, consider the following two cases:

- There exists a j such that V
_{j}=V_{j+1}
- Since V
_{j} = V_{j+1} = V_{j} - lub(V_{j}), lub(V_{j}) ∉ V_{j}, contradicting (3).
- For all j, V
_{j} ≠ V_{j+1}
- Form the set A={lub(V
_{j})}. 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(V_{k}) for some k. However, by (4), V_{k+1} < a, a contradiction. **QED**