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I know that the union of all singleton sets {q} with q a rational is countable so the issue I'm having is showing that {q} is nowhere dense in X? I know this is true when X=R but having trouble when X=Q

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Every open set in the metric space $\Bbb Q$ (with the usual Euclidean absolute value metric) is meagre, i.e., first category in $\Bbb Q$; in fact every subset of $\Bbb Q$ is meagre in $\Bbb Q$, because every subset of $\Bbb Q$ is countable, and $\{x\}$ is a closed, nowhere dense subset of $\Bbb Q$ for each $x\in\Bbb Q$. Hence $\mathbb{Q}$ is of first category in $\mathbb{Q}$.

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  • $\begingroup$ How do you know that {x} is a nowhere dense subset of Q? I understand that if {x} is closed then it is its own closure so I guess the question I'm asking is how do you know that the interior of {x} is empty. $\endgroup$
    – mckennja
    Apr 18 '17 at 19:45
  • $\begingroup$ I know it from several answers at MSE, e.g. here, or here. $\endgroup$ Apr 18 '17 at 20:36

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