# How can I show the sets are open? (The set of rational numbers $\mathbb{Q}$ is not a connected topological space)

• The set of rational numbers $$\mathbb{Q}$$ is not a connected topological space

My Attempt. Let $$\alpha \in \mathbb{R}$$ be an irrational number. By definition, $$\alpha \notin \mathbb{Q}$$. Consider the sets: $$\begin{array}{l}{S:=\mathbb{Q} \cap(-\infty, \alpha)} \\ {T:=\mathbb{Q} \cap(\alpha , \infty)}\end{array}$$

So since $$S, T$$ are open sets on $$\mathbb{Q}$$. Then

$$S \cup T=\mathbb{Q}$$ $$S \cap T=\varnothing$$ $$S, T \neq \varnothing$$

So $$S,T$$ are disjoint sets of $$Q$$. Hence, $$\mathbb{Q}$$ is not connected.

My question is: How can I show $$S,T$$ are open sets on $$\mathbb{Q}$$?

Because a subset $$A$$ of $$\mathbb Q$$ is an open subset of $$\mathbb Q$$ if and only if there is an open subset $$A^\ast$$ of $$\mathbb R$$ such that $$A=A^\ast\cap\mathbb Q$$.
$$\mathbb Q$$ is trivially open on $$\mathbb Q$$. So $$S$$ and $$T$$ are simply each the intersection of two open sets on $$\mathbb Q$$, which is open.