Let the sequence {an} prove the existence of monotone subsequence {bn} of {an}.

1.

If {an} is unbounded, you may want to set its upper limit, and then you can construct the child columns as follows:

B 1=a 1, because {an} has no previous session, the existence of n makes an >: A 1, b2=aN.

Similarly, there is aM & gtAN, b3=aM. Continuous lines can form a monotonically increasing sequence {bn}.

If {an} has no next session, you can construct a subtraction sequence as above.

2.

If {an} is bounded, it can be assumed that {an} converges to the real number A because it must have a convergent subsequence.

Divide the number axis into three areas: less than a, greater than a and equal to a, and there is at least one of these three areas.

Contains an infinite number of points in {an}.

I If there are infinitely many points in the interval equal to a, just construct the constant sequence bn = a. 。

Ii If there are infinitely many points on the interval less than a, these items less than a can form a new series {a'n} and converge to a, then it might as well be recorded as {an}, and the series can be constructed as follows:

B 1=a 1，for(A-A 1)/2 & gt； 0, the existence of n makes (a-an); A 1，b2=aN，

for(a-an)/2 >； 0, the existence of m makes (a-am); AN, b3=aM, if we continue this line, we can get monotonically increasing sequence {bn}.

If there are infinitely many points greater than a, a subtraction sequence can be constructed as above.

In a word, this proposition has been proved.