Let $\mathbb{N}$ denote the set of natural numbers, then a subbasis on $\mathbb{N}$ is
$$S = \{(-\infty, b), b \in \mathbb{N}\} \cup \{(a,\infty), a \in \mathbb{N}\}$$
Let $\leq$ be the relation on $\mathbb{N}$ identified with "less or equal to"
Then I saw a claim that says: (the order topology) $(\mathbb{N}, \leq)$ is a discrete space
Proof: Let $x \in \mathbb{N}, \{x\} = (x-1,x+1) = (-\infty,x+1) \cap (x-1, \infty) \quad\quad\square$
I have trouble with the last equivalent. If $(-\infty,x+1)$ and $(x-1, \infty)$ are subbasic elements i.e. open sets, then intersection of opens are open and I have no problem with that. But my confusion is whether $(-\infty,x+1)$ and $(x-1, \infty)$ are subbasic open sets to begin with.
The confusion stems from definition of the subbasis which is $$S = \{(-\infty, b), b \in \mathbb{N}\} \cup \{(a,\infty), a \in \mathbb{N}\}$$
Then every element in the subbasis has to be like $(-\infty, b)\cup (a,\infty)$. So it doesn't seem we can get rid of the $(a,\infty)$ part to get a pure subbasic element of the form $(-\infty, b)$. So $(-\infty, b)$ is not a subbasic element.
Can someone help?
The subbase is a union of two families: the family of left intervals, and the family of right intervals. So the proof is correct.
If your interpretation would hold, you would write it as
$$S = \{(-\infty,b) \cup (a, \infty): a,b \in \mathbb{N} \}$$
which is different: a family of unions, parametrised by 2 parameters, and it would be only one family.
As it stands, there are two families, both consisting of one type (left or right), and we take both of them together.
The separate families are also subbases but for different, coarser topologies. Your interpretation gives a different topology yet again. So be careful reading the notation!