Subsequence

Definition 2.3.1. (Subsequence) Suppone that {a_n} is any sequence and let

be infinite subset of ℕ i.e. a strictly increasing sequence of natural numbers. Then {a_nk} is called the subsequence of the squence {a_n}. For every intege k ∈ ℕ, by induction on k, n_k is at least as large as k  □

If {a_n} is a sequence, if for k ∈ ℕ we set n_k = 2k, we get the subsequence {a_nk} of even-indexed term. Similarly, if by letting n_k = 2k − 1, we obtain the subsequence {a_2k−1} of odd-indexed terms.

For example consider the sequence {1/n}. Then, the sequence {1/2, 1/4,1/6, ..., 1/2n, ...} is the subsequence of even-indexed terms of {1/n}.

Theorem 2.3.2. If a sequence {a_n} of real numbers converges to a limit L, then every subsequence {a_nk} of {a_n} converges to the same limit.

Proof. Since {a_n} converges to L, for each ε > 0, there exists a positive integer n₀ such that

Then taking n ≥ n_k ≥ n₀ we have

Hence {a_nk} converges to L.  □

The converge of the last theorem is not generally true. For example consider {(−1)ⁿ}. This sequence is not convergent. But the subsequence {a_nk} = {1,1,1,..} whose all terms are 1 converges to 1 as n ⟶ ∞

Theorem (Nested intervals property) Consider a decreasing sequnce of nested closed intervals in ℝ:

If the lenghts of these intervals vanish, i.e,

then there is exactly one real number ℓ that is common to all of these intervals, i.e.

$$\bigcap _{n=1}^{\mathrm{\infty }}[a_n,b_n]=\{\ell \}$$

Proof. The nondecreasing sequence of the left end points a₁, a₂, a₃, ... is bounded by any one of the right endpoints b_n; so by the Complete Axiom, it has a least upper bound, say α. Similarly, the nonincreasing sequence b₁, b₂, b₃,.. of right end points has a greatest lower bound β. Therefore

In particular, 0 ≤ β − α b_n − a_n for every index n. Since lim_{n ⟶ ∞} (b_n − a_n) = 0, the squeeze theorem implies that β − α = 0. So α = β, and this common value is the number r.  □

The nested intervla property holds because the completeness of real numbers enusres that ther are no gaps in ℝ.

We have proved that a convergent sequence is bounded and that the converse of this theorem is false take {(−1)ⁿ} for example. The Bolzano–Weierstrass theorem states that every bounded sequence has a convergent subsequence. Considering a_n = {(−1)ⁿ}, note that the subsequence a_2n is the constant sequence 1 and so converges to 1. On the ohter hand a_{2n + 1} is the constant sequence −1 and so congerges to −1.

Consider the following sequences:

Neither {a_n} nor {b_n} converges. Nervertheless, these two instances of nonconvergent sequences don't feel quite the same: the sequence {a_n} has a convergent sequence "inside it" while {b_n} does not have. Indeed {a_n} is a bounded sequence.

Theorem (The Bolzano–Weierstrass theorem) Every bounded sequence of real numbers has a convergent subsequence (hence, an accumulation point).

Proof. Let {a_n} be a bounded sequence in ℝ so that there exists b > 0 satisfying |a_n| ≤ b and let S = {a_n : n ∈ ℕ} be the range of {a_n}. If S is a finite set, then at least one point ℓ in it must repeat infinitely ofter; i.e., there is an increasing sequence of indicides n₁ < n₂ < ... such that a_nj for all j = 1, 2, 3,.. . It follows that the subsequence a_nj converges (trivially) to a, making ℓ an accumulation point.

If S is infinite, then given the boundedness of {a_n}, there are real numbers a,b such that S ⊂ [a,b]. If we partition the closed interval [a,b] into halves

then at least one on the half interval must contain infinitely many disctinct elements of S. Denote that half interval by [a₁,b₁] where either a₁ or b₁ is (a+b)/2 and note that the lenght of [a₁,b₁] is half the lenght of [a,b], i.e. (b − a)/2.
If we further partition [a₁,b₁] into half intervals, then one of the new hald intervals, call it [a₂,b₂], contains infinitely many (distinct) points of S, and its lenght is (b − a)/4. Repeating this partitioning process generates a nested sequence of hald intervals [a_k, b_k] such that

1. [a₁, b₁] ⊃ [a₂, b₂] ⊃ [a₃, b₃] ⊃ ...

2. the lenght of [a_k, b_k] is (b − a)/2^k for every k = 1, 2, 3, ...

3. [a_k, b_k] contains infinitely many ponts of S.

By the nested interval property

$$\bigcap _{n=1}^{\mathrm{\infty }}[a_k,b_k]=\{\ell \}$$

This number r is in [a_k, b_k] for every k, which by (iii) above must also contain at least one term a_n, say a_nk, where a_nk ≠ ℓ. It follows that the distance between a_nk and ℓ cannot exceed the lenght of the interval [a_k, b_k], i.e.

We conclude that the subsequence a_nk converges to ℓ and therefore, ℓ is an accumulation point.  ■

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