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From Munkres'(pg 160):

Example 1:If $\mathbb{Q}$ is the subspace of $\mathbb{R}$ consisting of the rational numbers, then each component of $\mathbb{Q}$ is a single point.

How do I see that each component of the rational numbers is a singleton?

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It suffices to prove that any two (distinct) rational numbers $q_1<q_2$ are in the different components. You can take an irrational number $q_1<r<q_2$, and then set $A = \{q \in \mathbb{Q} : q < r\}$ and $B = \{q \in \mathbb{Q} : q > r\}$. Then $(A,B)$ is a separation of $\mathbb{Q}$, and $q_1 \in A$, $q_2 \in B$. Thus $q_1$ and $q_2$ are in the different components.

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The path component around $x$ is the largest connected set containing it. So show that every set of rationals with more than one element is not connected: separate two points $a$ and $b$ by open intervals ending at some irrational strictly between them.

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First, note that the components are the equivalence class for the relation $xRy$ if there is a connected subspace of $X$ containing x and y. Now, only connected subspaces of Q are the singletons. So, the equivalence class (components) are only the singletons.

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