Is every subset of the natural numbers that is closed under successor also closed under addition?

Question

Suppose that $S$ is a subset of $\mathbb{N}$ such that $S$ is closed under the successor operation. Does it follow that $S$ is closed under addition?

Answer 1

Any non-empty subset of $\mathbb{N}$ which is closed under the successor operator is of the form $$S=\{n, n+1, n+2, \ldots\}$$ for some $n\in\mathbb{N}$. This is because as soon as you have an element $n$ in your (non-empty) subset its successor must also be in it, and by iteration all numbers greater than $n$ will be in it as well.

Clearly, any set of this form is closed by addition.