Problem 1196

To show that there exists an infinite irregular set $A \subseteq \mathbb{Z}$, we will construct such a set step-by-step.

1. Initialization:
- Start with $a_0 = 0$.

2. Inductive Hypothesis:
- Assume that we have already defined $a_0, a_1, \ldots, a_n$ such that:
- $a_0 < a_1 < \cdots < a_n$ are integers.
- The sum $s_n = \sum_{0 \leq i < j \leq n} \frac{1}{a_j - a_i}$ is strictly less than 1.

3. **Choosing $a_{n+1}$**:
- We need to choose $a_{n+1}$ such that it is not of the form $a_i + k(a_j - a_i)$ for any $0 \leq i < j \leq n$ and for any integer $k$, and also ensure that $s_{n+1} = s_n + \sum_{0 \leq i \leq n} \frac{1}{a_{n+1} - a_i}$ remains strictly less than 1.
- For sufficiently large $M$, the number of positive integers in the interval $[0, M)$ that are of the form $a_i + k(a_j - a_i)$ is approximately $\frac{M}{a_j - a_i}$. This means that at least $(1 - s_n)M$ integers in that interval are not of the form $a_i + k(a_j - a_i)$ for any $0 \leq i < j \leq n$.

4. **Ensuring $a_{n+1}$ is sufficiently large**:
- Choose $a_{n+1}$ such that it is larger than $2a_n$ and is not of the form $a_i + k(a_j - a_i)$ for any $0 \leq i < j \leq n$.
- Additionally, ensure that $a_{n+1}$ is large enough so that $s_{n+1} = s_n + \sum_{0 \leq i \leq n} \frac{1}{a_{n+1} - a_i}$ remains strictly less than 1.

5. Verification:
- By construction, $a_{n+1}$ is not of the form $a_i + k(a_j - a_i)$ for any $0 \leq i < j \leq n$ and for any integer $k$.
- The sequence $(a_n)_n$ thus forms an infinite irregular set.

$\blacksquare$