FALSE: the rational numbers form a Baire space

Given: The objects, hypotheses, and choice principles stated above.

For a topological space $X$, the following are equivalent: every countable intersection of dense open sets is dense; every countable union of closed sets with empty interior has empty interior; no nonempty open subset is meagre in $X$; and every residual subset meets every nonempty open set. The equivalence includes the empty space. (Equivalent forms of the Baire property).

Let $X$ be a topological space and let $A\subseteq X$. The set $A$ is nowhere dense when $\mathrm{int}(\bar{A})=\mathrm{\varnothing }$ (def-interior-closure-boundary-top). It is meagre when there is a sequence $(N_n{)}_{n\in \mathbb{N}}$ of nowhere dense subsets of $X$ with $A\subseteq {\bigcup }_n{N}_n$. It is residual, or comeagre, when $X\setminus A$ is meagre. The empty union shows that $\mathrm{\varnothing }$ is meagre, including when $X=\mathrm{\varnothing }$. (Nowhere dense, meagre, residual, and comeagre subsets of a topological space).

Write ${\mathbb{Q}}_{\mathbb{R}}$ for the image of $\mathbb{Q}$ in $\mathbb{R}$ under the canonical embedding $q\mapsto \widehat{q}$ (lem-rat-embeds-dense), the set usually written $\mathbb{Q}$ once the identification is made, and put $X:=\mathbb{R}\setminus {\mathbb{Q}}_{\mathbb{R}}$ for the irrationals. Then: 1. ${\mathbb{Q}}_{\mathbb{R}}$ is an $F_{\sigma }$ set (def-f-sigma-g-delta) and is meager (def-nowhere-dense-meager); 2. $X$ is a $G_{\delta }$ set and is residual; 3. ${\mathbb{Q}}_{\mathbb{R}}$ is not a $G_{\delta }$ set, and $X$ is not an $F_{\sigma }$ set. Claims 1 and 2 are bookkeeping. Claim 3 is the substance and is exactly where thm-baire-category-r is spent: no argument from the algebra of open and closed sets alone can reach it, since ${\mathbb{Q}}_{\mathbb{R}}$ and $X$ are interchanged by complementation while $F_{\sigma }$ and $G_{\delta }$ are, so any such argument would prove the same thing about both sets and about neither. ($\mathbb{Q}$ is $F_{\sigma }$, meager and not $G_{\delta }$, while the irrationals are $G_{\delta }$, residual and not $F_{\sigma }$).

$\mathbb{Q}\approx \mathbb{N}$ (def-equinumerous): the rationals are countably infinite (def-countable). No choice principle is used. The one place where a reader expects a choice, "pick a representative $a\mathrm{/}b$ of each rational", is exactly where lem-rat-positive-denominator applies: every rational has a representative with positive denominator, so the map $(a,b)\mapsto [(a,b)]$ defined on $\mathbb{Z}\times {\mathbb{Z}}_{>0}$ is already surjective onto $\mathbb{Q}$, and countability follows from a surjection without ever selecting a representative. The same device handles $\mathbb{Z}$, which is a surjective image of $\mathbb{N}\times \mathbb{N}$ by construction (def-integers). ($\mathbb{Q}$ is countably infinite).

Let $(X,d)$ be a metric space (def-metric-space) and let $p,q\in X$ with $p\mathrm{\ne }q$. Put $r:=d(p,q)\mathrm{/}2$. Then $r>0$ and $B(p,r)\cap B(q,r)=\mathrm{\varnothing }\mathrm{.}$ Both sets are open (thm-metric-open-set-algebra) and contain $p$ respectively $q$ (def-metric-ball), so every metric space is Hausdorff: distinct points are separated by disjoint open sets (def-metric-topology). (Distinct points of a metric space have disjoint balls around them).

Let $(X,\mathcal{T})$ be a topological space (def-topological-space). The following implications hold, and each is proved by an earlier item of this page. 1. Perfectly normal implies completely normal, assuming the Axiom of Countable Choice (def-countable-choice). 2. Completely normal implies normal, and perfectly normal implies normal. 3. Normal together with $T_1$ implies $T_3$, that is regular together with $T_1$. 4. Completely regular implies regular, and Tychonoff implies $T_3$. 5. Regular together with $T_1$ implies Urysohn, which implies Hausdorff, which implies $T_1$, which implies $T_0$. 6. Metrizable implies every property named above: a metrizable space is perfectly normal, completely normal, normal, Tychonoff, completely regular, $T_3$, regular, Urysohn, Hausdorff, $T_1$ and $T_0$, with no choice principle used. Reading the numbered axioms in order, clauses 1 to 5 give $T_6\Rightarrow T_5\Rightarrow T_4\Rightarrow T_3\Rightarrow T_{2\frac{1}{2}}\Rightarrow T_2\Rightarrow T_1\Rightarrow T_0,$ the first arrow under ${\mathrm{A}\mathrm{C}}_{\omega }$, together with $T_{3\frac{1}{2}}\Rightarrow T_3$. This is the whole of the classical chain that this page proves, and it is one arrow short of the classical chain. The implication $T_4\Rightarrow T_{3\frac{1}{2}}$ — a normal $T_1$ space is completely regular — is Urysohn's lemma and is not available at this point in the reading order. Its absence is recorded, with what would license it, in this page's conventions remark; it is deliberately not asserted here, and no clause above may be read as giving it. (The implications proved on this page: perfectly normal gives completely normal under countable choice, and completely normal gives normal; normal with $T_1$ gives $T_3$; completely regular gives regular; regular with $T_1$ gives Urysohn, hence Hausdorff, hence $T_1$, hence $T_0$; and metrizable gives every one of them).

Let $(X,\mathcal{T})$ be a topological space (def-topological-space) and let ${\mathcal{T}}_{\mathrm{c}\mathrm{o}\mathrm{f}}$ be the cofinite topology on the set $X$ (def-standard-topologies). The following four conditions are equivalent. - (a) $X$ is $T_1$ (def-t0-and-t1-spaces). - (b) $\{x\}$ is closed for every $x\in X$. - (c) $F$ is closed for every finite $F\subseteq X$ (def-countable). - (d) ${\mathcal{T}}_{\mathrm{c}\mathrm{o}\mathrm{f}}\subseteq \mathcal{T}$, that is, the topology of $X$ is finer than the cofinite topology on the same set. Condition (d) says that the cofinite topology is the coarsest $T_1$ topology on any set: it is $T_1$ by the equivalence, and every $T_1$ topology on that set contains it. (A space is $T_1$ if and only if every singleton is closed, if and only if every finite subset is closed, if and only if its topology contains the cofinite topology).