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The condition of 'X is a Baire space': (cite from Topology by Munkres)

Given any countable collection $\{ A_n \}$ of closed sets of X each of which has empty interior in X, their union $\bigcup A_n$ also has empty interior in X.

So for rational numbers $\mathbb{Q}$ with the usual topology of $\mathbb{R}$:

(1) countable collection $\{ Q_i \}$ : one can index all rational numbers with positive integers, so let $Q_i = \{q_i\}$ be one-point set.

(2) $Q_i$ is closed, since $\mathbb{R}- Q_i= (-\infty, q_i) \cup (q_i, \infty)$ is open in $\mathbb{R}$.

(3) $Q_i$ has empty interior, since $Q_i$ contains no open set of $\mathbb{R}$.

(4) $\mathbb{Q}=\bigcup Q_i$ has empty interior, since $\mathbb{Q}$ contains no open set of $\mathbb{R}$.

Thus, $Q_i$ is a Baire space. What's wrong here?

whitegreen
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2 Answers2

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$\Bbb Q$ doesn't have an empty interior. The relevant topology is the topology on $\Bbb Q$. That topology is the topology inherited from the usual topology on $\Bbb R$, so a basis for the open sets of the topology are the rational points of any open interval.

Robert Shore
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You have to consider the topology of $\Bbb Q$ in its own right: it has the order topology, or equivalently the metric topology induced by $d(x,y)= \max(y-x,x-y)$ for $x,y \in \Bbb Q$. There is no need to refer to the topology on $\Bbb R$ to define it.

Indeed $\{q\}$ is closed (as in any ordered or metric space), and your proof explicitly shows this for the order topology: $\Bbb Q\setminus \{q\} = \{r\in \Bbb Q\mid r < q\} \cup \{r \in \Bbb Q\mid r > q\}$ and both right hand sets are subbasic order-open sets.

$\{q\}$ has empty interior because if $q$ were an interior point there must be rationals $r_1, r_2$ such that $r_1 < q < r_2$ and $\langle r_1,r_2\rangle \subseteq \{q\}$, as $q$ is not the max or min and so has a local base of open intervals $\langle r_1,r_2\rangle = \{r \in \Bbb Q: r_1 < r < r_2\}$ in the order topology. But inbetween any two rationals there is a third ($\Bbb Q$ is order dense) so any $r_3$ with $r_1 < r_3 < q$ contradicts the inclusion $\langle r_1,r_2\rangle \subseteq \{q\}$, contradiction. Or simply note that all open sets of $\Bbb Q$ are infinite.

And the interior of $\Bbb Q$ in $\Bbb Q$ is $\Bbb Q$ itself, of course. Any space is open in itself. We don't need to refer to points outside $\Bbb Q$; ignore them.

Henno Brandsma
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