Order Topology on $\mathbb{Z_+}$

Question

Why is the order topology on $\mathbb{Z_+}$ a discrete one? I understand that the discrete topology will have all subsets of $\mathbb{Z_+}$ which means that all subsets of $\mathbb{Z_+}$ are open which is not necessarily true.

Answer 1

Why do you say not every subset is open? Every subset is open!

All the singletons $\{ n \}$ are basic open sets in the order topology, because $$\begin{alignat}{2} \{ 1 \} &\; = \{ x \in \mathbb{Z}_+ : x < 2 \} &\ &= (-\infty, 2)\\ \{ n+1 \} &\; = \{ x \in \mathbb{Z}_+ : n < x < n+2 \} &&= (n, n+2) \end{alignat}$$ But open sets needn't all be basic open sets: in particular, any unions of basic open sets is open.

Any subset $X \subseteq \mathbb{Z}_+$ can be written as the union of singletons: $$X = \bigcup_{n \in X} \{ n \}$$ ...and so all subsets are open.