The answer are the singletons of $\mathbb{Q}$. I can show that the open intervals of $\mathbb{Q}$ are disconnected by choosing some irrational in the open set and using it to form a separation. But strangely enough, I am having a hard time seeing why the subset $\{p,q \}$ of $\mathbb{Q}$ is disconnected. If say $p < q$ and we choose some irrational $p < x , the separation of $\{p,q \}$ would, I think, be given by $ \{p\}$ and $\{q\}$. But these aren't open in $\mathbb{Q}$ ...?
What are all of the connected subsets of $\mathbb{Q}$?
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real-analysis
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0I do believe $\Bbb Q$ is what you would call totally disconnected. – 2012-03-16
1 Answers
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Let $A\subseteq\mathbb Q$ with at least two elements, thus $x\in A$, $y\in A$, and $x\ne y$. Let $z$ be an irrational such that $x\lt z\lt y$. Then $A=A_+\cup A_-$ with $A_+=A\cap(z,+\infty)$ and $A_-=A\cap(-\infty,z)$. Since $A_+$ and $A_-$ are two disjoint nonempty sets and are open in $A$, $A$ is not connected.
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1@palio: Indeed $(z,+\infty)$ is not contained in $A$... and this fact is irrelevant. To prove that $A_+$ is open in (the relative topology on) $A$, choose any point $x$ in $A_+$ and find a positive $r$ such that $B(x,r)\cap A\subset A_+$. – 2012-03-16