MT2002 Analysis

 Previous page (Monotonic sequences) Contents Next page (Cauchy sequences)

## Subsequences

We have seen some bounded sequences which do not converge. We can, however, say something about such sequences.

Definition

A subsequence is an infinite ordered subset of a sequence.

Examples

(a2 , a4 , a6 , ... ) is a subsequence of (a1 , a2 , a3 , a4 , ... ). So is (a1 , a10 , a100 , a1000 , ... ).

Theorem
Any subsequence of a convergent sequence is convergent (to the same limit).

Proof
Look at the definition! The nicest thing about these subsequences is a result attributed to the Czech mathematician and philosopher Bernard Bolzano (1781 to 1848) and the German mathematician Karl Weierstrass (1815 to 1897).

The Bolzano-Weierstrass Theorem

Every bounded sequence has a convergent subsequence.

Remark

Notice that a bounded sequence may have many convergent subsequences (for example, a sequence consisting of a counting of the rationals has subsequences converging to every real number) or rather few (for example a convergent sequence has all its subsequences having the same limit).

Proof
Suppose the sequence (a1 , a2 , a3 , a4 , ... ) is bounded and lies in (say) the interval [0, 1].
Then we construct a convergent subsequence by a bisection process.
Split the interval [0, 1] into two halves [0 , 1/2] and [1/2 , 1]. Then (at least) one of the halves will contain infinitely many terms of the sequence. Suppose it is [0 , 1/2]. Choose x1to be one of these terms.
Then split this interval in half again and repeat the process choosing x2 to be further down the sequence than x1. Continuing in this way, at the nth stage we will choose a term xn lying in an interval [an , bn].

Claim: The subsequence (xn) is convergent.

Proof of claim
The sequence (ln) of "Left-hand ends" of intervals is monotonic increasing, bounded above by 1 and hence has a limit α.
The sequence (rn) of "right-hand ends" of intervals is monotonic decreasing, bounded below by 0 and hence has a limit β.
Since the length of the interval [ln , rn] has length (1/2)n, we must have α = β and since the sequence (xn) is trapped between (ln) and (rn), it converges to the same limit. Previous page (Monotonic sequences) Contents Next page (Cauchy sequences)

JOC September 2001