A monotonous subcolumn 40 will be found in any one of the sequences

Proof of Sequences to prove the existence of monotonous subcolumns.

If there is no boundary, you may wish to set it to have no upper bound, and you can construct a subcolumn as follows:

b1=a1, since there is no upper session, there is n so that an>a1, b2=an

In the same way, there is am>an, b3=amSuccessive rows can be used to construct a single increment sequence.

If there is no next session, a single-subtraction sequence can be constructed similarly to the above method.

If there is a boundary, since it must have a converging subcolumn, it can be assumed to converge to the real number a

Divide the number line into 3 regions: less than a, greater than a, equal to aAt least one of these 3 areas.

Contains an infinite number of points.

i If there are infinitely many points in the interval equal to a, then it is sufficient to construct a constant column bn=a.

ii If there are an infinite number of points in an interval less than a, then these terms less than a can form a new series and converge to a, so it may be worth still denoting as , and the series can be constructed as follows:

b1=a1, for (a-a1) 2>0, n is present so that (a-an) <(a-a1) 2, so an>a1, b2=an, for (a-an) 2>0, m is present so that (a-am) <(a-an) 2, so am>an, b3=am, and so on, a single increment series can be obtained.

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

To sum up, the proposition is proven.

I want to think....Consider the set a=. If a is an infinite set, then a decreasing subcolumn can be obtained by arranging the elements in a from small to large, if a is a finite set, that is, there is an m, so that when n >m, at least one of the terms after an is not less than an, then we can first take an1, n1>m, from the previous discussion, we know that there should be an n2, such that an2>=an1, and so on, we get the increasing subcolumn.

This is a constructive proof, if a is the minimum value of the sequence, then remove a as the first element, and then find a minimum b from the rest, b is obviously greater than or equal to a, and in the same way we have a monotonous subcolumn.

Any infinite sequence has a monotonous subcolumn.

Proof: Let the infinite sequence contain no monotonically increasing subcolumns.

Then there is n0, and when n>no there is x(n) in an infinite sequence starting from n=n0, which obviously does not contain monotonically increasing subcolumns.

Thus there is n1, and when n > n1, both have x(n) so that a monotonically decreasing subcolumn is found.

Functional comprehension of sequences:

A sequence is a special kind of function. Its particularity is mainly reflected in its definition domain and value range. A sequence can be thought of as a function that defines the domain as a set of positive integers n* or a finite subset of it, where the domain cannot be omitted.

It is an important way to understand the sequence of numbers from the point of view of functions, and in general, there are three ways to represent functions, and sequences are no exception, and there are usually three ways to represent them: alist method; b。

image method; c.Analytical. The analytic method includes giving a series of numbers with a general formula and giving a series of numbers with a recursive formula.

This is a very simple conclusion, if you don't understand it, read it several times.

This is a constructive proof, if a is the minimum value of the sequence, then remove a as the first element, and find a minimum b from the rest

b is obviously greater than or equal to a, and in the same way we have a monotonous subcolumn, which can also prove that the series has a maximum value.

Let's see that there is no maximum situation in the sequence, we take an element a arbitrarily in the sequence, because the sequence does not have a maximum value, so we can find a b greater than a, and in the same way, then we can find a sub-column that is monotonous, in summary, the theorem is that any number series has a monotonous sub-column.

The first floor is chaotic, and the law of counter-evidence.

It should be obtained that "there is a sequence of numbers without monotonous subcolumns", and the difficulty of finding contradictions in this is roughly equal to directly proving the original proposition.

Monotonic denotes increasing or decreasing, but not strictly increasing or decreasing.

So 1 < = 1 is also monotonous.

Why do I think this proposition is incorrect, 1,1,1,1,1,1,1,1,1,1,1,1,1,1 This series can find monotonous subcolumns? Hope to see the answer of the master