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  • 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 $\mathbb{Q}$?

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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 $A=A^\ast\cap\mathbb Q$.

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  • $\begingroup$ Thanks forn answer $\endgroup$ – James Ensor Nov 17 '19 at 9:39
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$\mathbb Q$ 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.

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  • $\begingroup$ Thanks for answer $\endgroup$ – James Ensor Nov 17 '19 at 9:39

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