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

- We have seen some bounded sequences which do not converge. We can, however, say something about such sequences.
- A
**subsequence**is an infinite ordered subset of a sequence. **Examples**- (
*a*_{2},*a*_{4},*a*_{6}, ... ) is a subsequence of (*a*_{1},*a*_{2},*a*_{3},*a*_{4}, ... ). So is (*a*_{1},*a*_{10},*a*_{100},*a*_{1000}, ... ). **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 (
*a*_{1},*a*_{2},*a*_{3},*a*_{4}, ... ) 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*x*_{1}to be one of these terms.

Then split this interval in half again and repeat the process choosing*x*_{2}to be further down the sequence than*x*_{1}. Continuing in this way, at the*n*th stage we will choose a term*x*_{n}lying in an interval [*a*_{n},*b*_{n}].**Claim:**The subsequence (*x*_{n}) is convergent.**Proof of claim**

The sequence (*l*_{n}) of "Left-hand ends" of intervals is monotonic increasing, bounded above by 1 and hence has a limit*α*.

The sequence (*r*_{n}) 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 [*l*_{n},*r*_{n}] has length (^{1}/_{2})^{n}, we must have*α*=*β*and since the sequence (*x*_{n}) is trapped between (*l*_{n}) and (*r*_{n}), it converges to the same limit.

**Definition**

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