How to prove this language is not regular (via Fooling Set)
Let $s_1 &lt; s_2 &lt; \cdots$ be a subset of $\mathbb{N}$ with the following property:

$$ \limsup_{n\to\infty} s_{n+1}-s_n = \infty. $$

I will show that the language $L = \{ 0^{s_n} : n \geq 0 \}$ isn't regular using the fooling set method.

The first step is to find indices $i_1 &lt; i_2 &lt; \cdots$ such that
$$
s_{i_1+1}-s_{i_1} &lt; s_{i_2+1}-s_{i_2} &lt; \cdots.
$$
This can easily be done using the given property.

Our fooling set is $\{ 0^{s_{i_n}} : n \geq 0 \}$. To show that this is a fooling set, let $j &lt; k$, and consider the words $x = 0^{s_{i_j}}$ and $y = 0^{s_{i_k}}$. Let $z = 0^{s_{i_j+1}-s_{i_j}}$. Then $xz = 0^{s_{i_j+1}} \in L$ but $yz = 0^{s_{i_k}+s_{i_j+1}-s_{i_j}} \notin L$ since by construction,
$$ s_{i_k} &lt; s_{i_k}+s_{i_j+1}-s_{i_j} &lt; s_{i_k+1}. $$



More generally, a unary language $\{0^s : s \in S\}$ is regular if and only if there exist $N,M$ such that $n \in S$ iff $n+M \in S$ for all $n \geq N$. There are many ways to show it, the most elementary being that a graph of unary DFA consists of a path leading to a cycle. This result implies yours.