Solutions

$$\begin{aligned} n_{0} \in A & \implies n_{0} + 1 \in A \\ & \implies (n_{0} + 1) + 1 \in A \\ & \implies (n_{0} + 2) + 1 \in A \\ & \qquad \dots \end{aligned}$$

Thus, A is defined just as the natural numbers except that its first term is $n_{0}$ instead of 0.