Suppose is monotonic increasing and bounded. The set is non-empty and bounded above, so by the least upper bound axiom it has a supremum . Given , the number is not an upper bound, so some ; since the sequence is increasing, for all , giving and hence . The decreasing case is similar.

For an increasing sequence, boundedness above alone suffices. Boundedness as such cannot be dropped: increases without bound and diverges. The theorem is in fact equivalent to the least upper bound axiom, and every sequence has a monotonic subsequence.