Bolzano-Weierstrass Theorem

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 (a_n) be a sequence of real numbers. Say that k is a maximal index if a_k>=a_l for all l>=k. Then there are two possibilities:

• There are infinitely many maximal indices k₁<k₂<k₃<...
  Then a_{k_1}>=a_{k_2}>=a_{k_3}>... so (a_n) 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 (a_{k_j}). Let k₁=b+1. k₁ isn't a maximal index, so there exists k₂>k₁ with a_{k_2}>a_{k_1}. Now, suppose we have found k_j. k_j isn't a maximal index, so there exists k_j+1>k_j with a_{k_j+1}>a_{k_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 {a_n} be a bounded real sequence. Then there is a real M such that for each n,a_n is in [-M,M ], which is Heine-Borel compact. Let E be the set of points in {a_n}. Then E is either finite or infinite.

Suppose E is finite. Label the points p₁,...,p_k and associate with each p_j the set E_j={n: a_n=p_j }. The E_j form a partition of the positive integers. So one of the E_j must be countable. Label its elements as a strictly increasing sequence {n_k}. Thus {a_{n_k} } is a convergent subsequence of {a_n}.

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 {a_n}. Thus there is a convergent subsequence of {a_n}.

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 V₀=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:

1. lub(V_i) exists because V_i is bounded above by 1.
2. V_i is non-empty because V₀ is infinite and only a single point is removed at each step.
3. lub(V_i) ∈ V_i. Otherwise, it would be a limit point of V_i, a contradiction.
4. 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