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" have different choice-theoretic statuses over ZF. The compact Hausdorff version is equivalent to dependent multiple choice; DC implies DMC in ZF, the reversal remains open in ZF, and DMC does not imply DC in ZFA. That account 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).

For every real $\epsilon >0$, some natural $n\ge 1$ satisfies $1\mathrm{/}n<\epsilon$ (For every $\epsilon >0$ in a complete ordered field there is a natural $n\ge 1$ with $1\mathrm{/}n<\epsilon$).

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 $C([0,1])$).

The pairs of natural numbers have a specified countable enumeration ($\mathbb{N}\times \mathbb{N}\approx \mathbb{N}$).

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 $F_{\sigma }$).