Complete Metrizability, Čech-Completeness, and Baire Category: Examples and Counterexamples

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

Assume the Axiom of Dependent Choice (def-dependent-choice). Let $(X,\mathcal{T})$ be a locally compact Hausdorff space (def-locally-compact-space, def-hausdorff-space, def-topological-space). Then $X$ is a Baire space (def-baire-space): for every sequence $(U_n{)}_{n\in \mathbb{N}}$ of dense open subsets of $X$ (def-dense-top, def-sequence-convergence-top), the intersection ${\bigcap }_{n\in \mathbb{N}}{U}_n$ is dense in $X$. Dependent choice is sufficient here and no claim of necessity is made. The several statements that go by the name "Baire category theorem" are inequivalent over ZF, and the choice principles they correspond to differ; that account, including the fact that the compact Hausdorff version is equivalent to a principle strictly weaker than dependent choice, is rem-baire-category-choice-strength, which this library states and does not prove. Nothing below asserts that dependent choice is needed for the statement above. (Assuming dependent choice, every locally compact Hausdorff space is a Baire space).

Assume the Axiom of Choice (def-axiom-of-choice). Let $I$ be a set and let $(X_i,{\mathcal{T}}_i{)}_{i\in I}$ be a family of compact topological spaces (def-compact-space, def-topological-space). Then the product $P:={\prod }_{i\in I}{X}_i$ with the product topology (def-product-topology) is compact. The Axiom of Choice is spent twice, and both uses are flagged below. Once inside thm-alexander-subbase-lemma, through Zorn's lemma (thm-zorn), and once directly at step 2.1, to produce a point of a product of nonempty sets. (Tychonoff's theorem: an arbitrary product of compact spaces is compact in the product topology, assuming the Axiom of Choice).

A topological space $(X,\mathcal{T})$ (def-topological-space) is first countable if every point of $X$ has an at most countable neighbourhood base: for each $x\in X$ there is a family ${\mathcal{B}}_x\subseteq \mathcal{N}(x)$ that is at most countable (def-countable, def-equinumerous) and such that every neighbourhood of $x$ contains a member of ${\mathcal{B}}_x$ (def-neighbourhood-top). (First countable space: a countable neighbourhood base at every point).

Assume the Axiom of Countable Choice. Let $(A_n{)}_{n\in \mathbb{N}}$ be a family of at most countable sets indexed by $\mathbb{N}$. Then $U={\bigcup }_{n\in \mathbb{N}}{A}_n$ is at most countable (Countable unions of at most countable sets, assuming ${\mathrm{A}\mathrm{C}}_{\omega }$).

Let $(X,d)$ be a metric space (def-metric-space) and let ${\mathcal{T}}_d$ be its metric topology (def-metric-topology). Call ${\mathcal{T}}_d$ completely metrizable if some metric $\rho$ on $X$ is topologically equivalent to $d$, that is ${\mathcal{T}}_{\rho }={\mathcal{T}}_d$ (def-equivalent-metrics), and makes $(X,\rho )$ complete (def-complete-metric-space). Then: 1. Homeomorphism invariance. Let $(Y,e)$ be a metric space and let $h:X\to Y$ be a bijection (def-injection-surjection-bijection) such that $h$ and $h^{-1}$ are continuous (def-metric-continuity). If ${\mathcal{T}}_d$ is completely metrizable then so is ${\mathcal{T}}_e$. 2. Closed subspaces. If ${\mathcal{T}}_d$ is completely metrizable and $A\subseteq X$ is closed in $(X,d)$, then ${\mathcal{T}}_{d_A}$ is completely metrizable, $d_A$ being the subspace metric (def-isometry-and-metric-embedding). 3. The property is strictly weaker than completeness. Let $P:=(0,\mathrm{\infty })\subseteq \mathbb{R}$ (def-interval) carry $d(x,y):=\mathrm{\mid }x-y\mathrm{\mid }$ (lem-real-line-is-a-metric-space). Then $(P,d)$ is not complete, while ${\rho }_P(x,y):=\mathrm{\mid }x-y\mathrm{\mid }+\mid \frac{1}{x}-\frac{1}{y}\mid$ is a complete metric on $P$ with ${\mathcal{T}}_{{\rho }_P}={\mathcal{T}}_d$. So ${\mathcal{T}}_d$ is completely metrizable although no completeness assumption holds for $d$ itself. Complete metrizability is a condition on the collection of open sets alone: the metric is quantified over and does not survive into the statement. That is exactly what completeness fails to be, and claim 3 shows the two conditions are genuinely different rather than merely stated differently. (Complete metrizability: admitting a topologically equivalent complete metric is preserved by homeomorphism and by closed subspaces, and $(0,\mathrm{\infty })$ has it without being complete).

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).

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

Baire sequence space ${\mathbb{N}}^{\mathbb{N}}$ is homeomorphic to the irrational subspace $\mathbb{R}\setminus \mathbb{Q}$. (Baire sequence space is homeomorphic to the irrational real numbers).

For simple continued fractions with $a_0\in \mathbb{Z}$ and $a_n\ge 1$ for $n\ge 1$, the convergents satisfy $p_n=a_n{p}_{n-1}+p_{n-2}$ and $q_n=a_n{q}_{n-1}+q_{n-2}$, with $p_n{q}_{n-1}-p_{n-1}{q}_n=(-1{)}^{n-1}$. A code cylinder $C(a_0,\dots ,a_n)\subseteq {\mathbb{N}}^{\mathbb{N}}$ and the real interval $J(a_0,\dots ,a_n)$ with endpoints $p_n\mathrm{/}{q}_n$ and $(p_n+p_{n-1})\mathrm{/}(q_n+q_{n-1})$ are different objects and are not identified; it is the intervals $J$ that are nested as the prefix is extended, with $\mathrm{diam}J(a_0,\dots ,a_n)=1\mathrm{/}(q_n(q_n+q_{n-1}))\to 0$. (Continued-fraction convergents, determinant identities, and nested irrational cylinders).

The continued-fraction coding of def-simple-continued-fraction-coding is a bijection from the sequences $(a_0,a_1,\dots )$ with $a_0\in \mathbb{Z}$ and $a_n\ge 1$ for $n\ge 1$ onto $\mathbb{R}\setminus \mathbb{Q}$, and both the coding map and its inverse are continuous for the cylinder and subspace topologies (Infinite simple continued fractions parametrise the irrational real numbers).

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

Let $((X_n,d_n){)}_{n\in \mathbb{N}}$ be complete metric spaces with $d_n\le 1$. On ${\prod }_n{X}_n$, the formula $D(x,y)={\sum }_{n=0}^{\mathrm{\infty }}{2}^{-(n+1)}{d}_n(x_n,y_n)$ defines a complete metric inducing the product topology. The empty product is the one-point space. (The standard weighted metric on a countable product of bounded complete metric spaces is complete).

Every separable metrizable space is homeomorphic to a subspace of the Hilbert cube $[0,1{]}^{\mathbb{N}}$. (Every separable metrizable space embeds in the Hilbert cube $[0,1{]}^{\mathbb{N}}$).

Assume the Axiom of Choice (def-axiom-of-choice). Let $I$ be a set and let $(X_i,{\mathcal{T}}_i{)}_{i\in I}$ be a family of compact topological spaces (def-compact-space, def-topological-space). Then the product $P:={\prod }_{i\in I}{X}_i$ with the product topology (def-product-topology) is compact. The Axiom of Choice is spent twice, and both uses are flagged below. Once inside thm-alexander-subbase-lemma, through Zorn's lemma (thm-zorn), and once directly at step 2.1, to produce a point of a product of nonempty sets. (Tychonoff's theorem: an arbitrary product of compact spaces is compact in the product topology, assuming the Axiom of Choice).

A topological space is Polish when it is separable (def-separable-space) and completely metrizable: its topology is induced by some complete metric (lem-complete-remetrisation). No particular compatible complete metric or countable dense subset is part of the structure. (Polish spaces are separable completely metrizable spaces).

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

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).

For $p,q\in {\mathbb{N}}_{>0}$, let $E_{p,q}$ be the functions $f\in C([0,1],\mathbb{R})$ for which some $a\in [0,1]$ satisfies $\mathrm{\mid }f(t)-f(a)\mathrm{\mid }\le p\mathrm{\mid }t-a\mathrm{\mid }$ whenever $t\in [0,1]$ and $\mathrm{\mid }t-a\mathrm{\mid }<1\mathrm{/}q$. Then $E_{p,q}$ is closed in the supremum metric. (Functions satisfying a fixed local Lipschitz bound somewhere form a closed subset of $C([0,1])$).

For every $f\in C([0,1],\mathbb{R})$, every $\epsilon >0$, and every $M>0$, there is a piecewise-affine $h$ with finitely many vertices such that $\parallel f-h{\parallel }_{\mathrm{\infty }}<\epsilon$ and every slope on a nonvertex affine piece has absolute value greater than $M$. (Polygonal functions with sufficiently steep nonvertex slopes are dense in $C([0,1])$).

Assume the Axiom of Dependent Choice ($\mathrm{D}\mathrm{C}$). Then the set of continuous functions $[0,1]\to \mathbb{R}$ having no finite two-sided derivative at an interior point and no finite one-sided derivative at either endpoint is dense in $C([0,1],\mathbb{R})$ for the supremum metric. (Under Dependent Choice, continuous nowhere differentiable functions form a dense subset of $C([0,1],\mathbb{R})$).

For every topological space $X$, the meagre subsets of $X$ contain $\mathrm{\varnothing }$, are closed under taking subsets, and are closed under countable unions (The meagre subsets of a topological space form a sigma-ideal).

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

$\mathbb{Q}$, with its usual subspace topology from $\mathbb{R}$, is not a Baire space. The cited item is a false-statement item: the sentence displayed under its Statement heading is the claim it refutes, and what it establishes is that negation (FALSE: the rational numbers form a Baire space).

Assume Dependent Choice. Every Čech-complete space is a Baire space. (Under Dependent Choice, every Čech-complete space is Baire).

Assume the ultrafilter lemma and the Axiom of Choice. A metrizable space is Čech-complete if and only if it is completely metrizable. (Under the ultrafilter lemma and the Axiom of Choice, a metrizable space is Čech-complete exactly when it is completely metrizable).

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 $F_{\sigma }$).