a1,a2,a3,...
b1,b2,b3,...
c1,c2,c3,...
证明存在自然数m,n,使得
a_m >= a_n
b_m >= b_n
c_m >= c_n
Rearrange the sequence such that a1≤a2≤... now we only need to consider sequences b and c. If bn is the minimum of sequence b, we have two cases: (i) There exists an index m, m>n, such that cn≤cm; (ii) For all m, m>n, cn>cm. For case (i), we already find the desired m, n. For case (ii), we can simply remove index n and do the above analysis again using the next minimum in sequence b. Since natural numbers are bounded from below, after a finite number of steps, we will stop at case (i). QED. |
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.] |
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