The Bolzano-Weierstrass Theorem states that any bounded sequence of real numbers has a convergent subsequence.

The proof of this theorem is as follows. The Monotone convergence axiom states that any bounded monotonic sequence of real numbers converges, so it only needs to be shown that any sequence of real numbers has a monotonic subsequence.

Let (an) be a sequence of real numbers. Say that k is a maximal index if ak>=al for all l>=k. Then there are two possibilities:

• There are infinitely many maximal indices k1<k2<k3<...
Then ak_1>=ak_2>=ak_3>... so (an) has a decreasing subsequence.
• There are finitely many maximal indices. Let b be an upper bound for the set of maximal indices. Then if n>b, n isn't a maximal index. It can be shown by induction that there exists an increasing subsequence (ak_j). Let k1=b+1. k1 isn't a maximal index, so there exists k2>k1 with ak_2>ak_1. Now, suppose we have found kj. kj isn't a maximal index, so there exists kj+1>kj with ak_j+1>ak_j. So one can find an increasing subsequence.
So in either case, there is a monotonic subsequence.

QED, as they don't say.

Theorem: Each bounded real sequence has a convergent subsequence.

Let {an} be a bounded real sequence. Then there is a real M such that for each n,an is in [-M,M ], which is Heine-Borel compact. Let E be the set of points in {an}. Then E is either finite or infinite.

Suppose E is finite. Label the points p1,...,pk and associate with each pj the set Ej={n: an=pj }. The Ej form a partition of the positive integers. So one of the Ej must be countable. Label its elements as a strictly increasing sequence {nk}. Thus {an_k } is a convergent subsequence of {an}.

Suppose E is infinite. Because it is an infinite subset of the compact set [-M,M ], E has a limit point A in [-M,M ]. So each neighborhood of A contains a countable subset of E; in other words, each neighborhood of A contains countably many terms of {an}. Thus there is a convergent subsequence of {an}.

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

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