We have seen the completeness axiom for the reals in the form of the
Monotone Convergence Theorem. This web page discusses an application
of this for bounded sequences and proves a beautiful result: that every
bounded sequence has a convergent subsequence, the
We have seen that every convergent sequence is bounded, but that not every bounded sequence is convergent. For example, defines a non-convergent bounded sequence. Thus there is no full converse to the statement that every convergent sequence is bounded. The Bolzano-Weierstrass theorem, to be presented next, gives a partial converse.
Suppose that is a bounded sequence of real numbers. Then has a convergent subsequence.
Suppose that is bounded. For the purposes of this proof, say that the th term of the sequence is dominant if for all . Our proof now splits into two cases.
Case 1. Suppose that has infinitely many dominant terms
where . Then by the definition
dominant we have
and this gives a bounded monotonic nonincreasing subsequence which
converges by the monotone convergence theorem.
Case 2. If not, then the sequence has only finitely many dominant terms. Choose beyond the last dominant term. (So, for example, we may take the last dominant term , and define .) As is not dominant there is such that , and as is not dominant there is such that , and so on. Continuing in this way we obtain a monotonic bounded increasing subsequence which once again converges by the monotone convergence theorem.
The Bolzano-Weierstrass theorem is an important and powerful result related to the so-called compactness of intervals in the real numbers, and you may well see it discussed further in a course on metric spaces or topological spaces. Note that the completeness of the reals (in the form of the monotone convergence theorem) is an essential ingredient of the proof. We shall apply it in another section of these notes to give an alternative view of the completeness of .