For each positive integer n, prove that every sequence {a1, . . . , an2+1}
of n
2 + 1 distinct real numbers either contains an increasing subsequence of length n + 1, or a
decreasing subsequence of length n + 1. (Hint: Construct a sequence poset and use Dilworth’s
theorem.)

To prove that for each positive integer \( n \), every sequence \( \{a_1, a_2, \ldots, a_{n^2+1}\} \) of \( n^2+1 \) distinct real numbers contains either an increasing subsequence of length \( n+1 \) or a decreasing subsequence of length \( n+1 \), we can use the concept of a **partially ordered set (poset)** and **Dilworth's theorem**.

### Step 1: Understand the Problem
Given a sequence \( \{a_1, a_2, \ldots, a_{n^2+1}\} \), we need to show that it contains either an increasing subsequence of length \( n+1 \) or a decreasing subsequence of length \( n+1 \).

### Step 2: Define a Poset
Consider the sequence \( \{a_1, a_2, \ldots, a_{n^2+1}\} \) as a poset (partially ordered set) where the order relation \( \leq \) is defined by \( a_i \leq a_j \) if \( i \leq j \) and \( a_i \leq a_j \). This means a subsequence is increasing if \( a_i \leq a_j \) when \( i < j \), and decreasing if \( a_i \geq a_j \) when \( i < j \).

### Step 3: Apply Dilworth's Theorem
Dilworth's theorem states that in any finite partially ordered set, the size of the largest antichain (a set of elements that are pairwise incomparable) is equal to the minimum number of chains into which the poset can be partitioned.

In our context:
- A chain corresponds to either an increasing or a decreasing subsequence.
- An antichain corresponds to a subset of the sequence where no two elements are comparable (i.e., no subsequence is either fully increasing or fully decreasing).

### Step 4: Apply the Pigeonhole Principle
Given \( n^2 + 1 \) elements in our sequence, let's apply Dilworth's theorem:
- If the sequence does not contain an increasing subsequence of length \( n+1 \), then the length of any increasing subsequence is at most \( n \).
- Similarly, if the sequence does not contain a decreasing subsequence of length \( n+1 \), then the length of any decreasing subsequence is also at most \( n \).

If we try to partition the sequence into chains (which are either increasing or decreasing subsequences), then each chain has a maximum length of \( n \). 

Given that there are \( n^2 + 1 \) elements, and each chain can have at most \( n \) elements, by the pigeonhole principle, there must be at least \( \lceil \frac{n^2+1}{n} \rceil = n+1 \) chains.

### Step 5: Conclude the Proof
Now, consider the number of chains:
- If all these chains had fewer than \( n+1 \) elements, we would contradict the fact that the poset contains \( n^2 + 1 \) elements because we would need at least \( n+1 \) chains to accommodate all elements, each of size at most \( n \).
- Therefore, at least one of these chains must have \( n+1 \) elements, meaning there is either an increasing subsequence of length \( n+1 \) or a decreasing subsequence of length \( n+1 \).

This completes the proof. Hence, every sequence of \( n^2 + 1 \) distinct real numbers contains either an increasing subsequence of length \( n+1 \) or a decreasing subsequence of length \( n+1 \).

### Would you like more details, or do you have any questions?

#### Relative questions:
1. What are some examples of posets and their applications?
2. How does Dilworth's theorem relate to other combinatorial theorems?
3. Can you provide a specific example of applying this theorem to a sequence?
4. What is the importance of the pigeonhole principle in combinatorial proofs?
5. How can this theorem be extended or generalized for other types of sequences?

#### Tip:
When solving problems related to sequences and ordering, constructing an appropriate poset and leveraging theorems like Dilworth's can often simplify the proof.

Learn how to prove that for each positive integer n, every sequence of n^2+1 distinct real numbers either contains an increasing subsequence of length n+1 or a decreasing subsequence of length n+1 using Dilworth's theorem and the pigeonhole principle. Understand the concepts of posets, chains, and antichains in this advanced mathematical proof.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation