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I am trying to show that the rational interval $ S= \{x \in \mathbb Q : a \leq x \leq b\} $ is disconnected in the metric space $(X,d)$ where $X=\mathbb R$ and $d$ is the standard metric ( $ d(x,y)=|x-y| \; \forall \; x,y \in \mathbb R$ )

The definition of disconnected I would like to use is: a metric space $(X,d)$ is disconnected if there are non-empty, open, disjoint sets $A, B$ such that $X=A \cup B$.

If $a,b \notin \mathbb Q$ I think I have shown the above by what follows.

Any rational interval contains an irrational number (assume known). Let $c$ be an irrational number and $a. Let $A=\{x\in \mathbb Q : a and $B=\{x\in \mathbb Q : c. $A$ and $B$ are disjoint and open. $A$ and $B$ are non-empty as every interval contains a rational number (assume known). As $c \notin \mathbb Q$ $S=A \cup B$ and so S is disconnected.

If either $a$ or $b$ do belong to $\mathbb Q$ then I don't know how to construct the sets $A$ and $B$ so that they are both open, disjoint and their union is $S$. As if for example if $A=\{x\in \mathbb Q : a\leq x then this is neither open nor closed I think.

How could this problem be solved?

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    It doesn't matter if $a,b\in\mathbb{Q}$ or not, since you're looking at the set $S$ with subspace topology from $\mathbb{R}$.2012-12-17
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    If you consider space $[a,b] \cap \mathbb{Q}$ then $[a,x)$ is open for any $x$, because it is a complement of closed set $[x,b]$.2012-12-17

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