Problem 1511

1. Define Terms and Assumptions:
- We call two terms $a_i$ and $a_j$ "friends" if $a_i \mid a_j$ or $a_j \mid a_i$, and "enemies" otherwise.
- A sequence is "good" if any two terms are friends.
- Assume there is no good infinite sequence.

2. Construct the Longest Good Sequence:
- Let $a_1 = a_{1_1}, a_{1_2}, \dots, a_{1_n}$ be the longest good sequence containing $a_1$, where $1_1, 1_2, \dots, 1_n$ are indices.
- This means that for any $i$ distinct from $\{1_1, 1_2, \dots, 1_n\}$, there exists $j \leq n$ such that $a_i$ and $a_{1_j}$ are enemies.

3. Identify an Element with Infinite Enemies:
- There exists a number $a_{1_k}$ with an infinite number of enemies. Denote $a_{1_k} = b_1$.
- Therefore, there exists an infinite set $I_1$ of indices such that for any $i \in I_1$, $a_i$ and $b_1$ are enemies.

4. Inductive Construction of Subsequences:
- Assume we have constructed the numbers $b_1, b_2, \dots, b_n$ such that any two are enemies and there exists an infinite set $I_n$ of indices such that for any $b_i$ and for any $j \in I_n$, $b_j$ and $a_i$ are enemies.
- Pick a number $x$ from $I_n$ such that $a_x$ is distinct from $b_i$ for any $i \leq n$, and let $a_x = a_{n_1}, a_{n_2}, \dots, a_{n_m}$ be the longest good sequence containing $a_x$, with indices $a_{n_i} \in I_n$ for any $i \leq m$.
- Therefore, any element $y$ of $I_n$, which is not from the set $\{n_1, n_2, \dots, n_m\}$, has the property that $a_y$ is an enemy with some $a_{n_j}$.

5. Identify an Element with Infinite Enemies in the Subset:
- Since the set $I_n - \{n_1, n_2, \dots, n_m\}$ is infinite, there exists an index $n_k$ which is an enemy with an infinite number of elements with indices from $I_n$.
- Let $a_{n_k} = b_{n+1}$ and let $I_{n+1} \subset I_n$ be the set of indices with which $b_{n+1}$ is an enemy.

6. Inductive Conclusion:
- By induction, we have constructed an infinite subsequence $\{b_n\}$ of enemies.

Conclusion:
Thus, we have shown that if there is no good infinite sequence, we can construct an infinite subsequence of enemies. This completes the proof.

$\blacksquare$