|
6#
楼主 |
发表于 2006-2-1 19:49:35
|
只看该作者
Another proof
Given any infinite sequence of natural numbers we can find a non-decreasing subsequence (proof below). So suppose the three sequences are ai, bi, and ci. Take a non-decreasing subsequence of ai. Suppose it is ai1, ai2, ai3, ... . Now consider the infinite sequence bi1, bi2, ... . It must have a non-decreasing subsequence. Suppose it is bj1, bj2, ... . Now consider the infinite sequence cj1, cj2, ... . It must have a non-decreasing subsequence ck1, ck2, ... . Each of the three sub-sequences ak1, ak2, ... , bk1, bk2, ... , ck1, ck2, ... is non-decreasing. So we may take, for example, m=k2 and n=k1. [Proof that any infinite sequence of natural numbers has a non-decreasing subsequence: if the original sequence is unbounded, then we can take a strictly increasing subsequence. If not, then since there are only finitely many possible numbers not exceeding the bound, at least one of them must occur infinitely often.]
www.ddhw.com
|
|
|