Countability Axioms and Cardinal Functions

By [A1], every basis has a cardinality. The topology of $X$ is itself a basis, so the set of cardinalities of bases is nonempty.

By [L1], its nonempty collection of cardinal values has a least member; that member is exactly the minimum in the definition of $w(X)$.

By [A1], every dense subset has a cardinality. The subset $X$ is dense in itself, so cardinalities of dense subsets form a nonempty collection.

[L1] gives its least member, which is the value defined as $d(X)$.

The neighbourhood filter at $x$ is a local base, so local-base cardinalities form a nonempty set.

The candidate cardinalities are ordinals, so their nonempty set has a least member, namely $\chi (x,X)$.

Let $K=\{\chi (x,X):x\in X\}$ and $\delta =\bigcup K$. This is an ordinal and the least ordinal upper bound of $K$ by [L3]. It is a cardinal: if $\beta <\delta$ and $\beta \approx \delta$, choose $\kappa \in K$ with $\beta <\kappa \le \delta$. Then $\beta ⪯\kappa ⪯\delta \approx \beta$, so [L3] gives $\beta \approx \kappa$, contradicting that $\kappa$ is a cardinal. Thus $\delta$ is the cardinal supremum $\chi (X)$.

By [A1], the topology $\tau$ has a cardinality $\kappa$. Every open cover is a subcover of itself and has cardinality at most $\kappa$, so $\kappa$ bounds every cover's subcover size.

Let $S$ be the set of cardinals $\lambda \le \kappa$ such that every open cover of $X$ has a subcover of cardinality at most $\lambda$. By Step 1.1, $\kappa \in S$, so [L1] supplies a least member of $S$. Any bounding cardinal larger than $\kappa$ cannot be smaller than that member, and hence this least member is exactly $L(X)$.

Each cellular family is a subfamily of the topology, so its cardinality is bounded by the cardinality of the topology.

Let $K$ be the set of cardinalities of cellular families and put $\delta =\bigcup K$. By [L2], $\delta$ is the least ordinal upper bound of $K$. It is a cardinal: if $\beta <\delta$ and $\beta \approx \delta$, choose $\kappa \in K$ with $\beta <\kappa \le \delta$; then $\beta ⪯\kappa ⪯\delta \approx \beta$, so [L2] yields $\beta \approx \kappa$, contrary to $\kappa$ being a cardinal. Hence $\delta =\sup K=c(X)$.

By [L1], $w(X)$ and $d(X)$ are the least cardinalities of a basis and a dense subset, respectively; hence $w(X)\le {\mathrm{\aleph }}_0$ and $d(X)\le {\mathrm{\aleph }}_0$ say exactly that such a basis and such a dense subset are at most countable.

By [L1], $L(X)\le {\mathrm{\aleph }}_0$ says that every open cover has a subcover of at most countable cardinality, and $c(X)\le {\mathrm{\aleph }}_0$ says that every pairwise-disjoint family of nonempty open sets is at most countable.

Since $\chi (X)=\sup \{\chi (x,X):x\in X\}$, one has $\chi (X)\le {\mathrm{\aleph }}_0$ exactly when every point has a local base of cardinality at most ${\mathrm{\aleph }}_0$.

The descriptions in steps 1.1, 1.2 and 1.3 are precisely the definitions in [L2], so they yield the five asserted equivalences.

A local base is nonempty, so enumerate it as $(B_n)$ by [L2], with repetitions allowed. Put $U_n=\mathrm{int}(B_n)$. Since $B_n$ is a neighbourhood of $x$, its interior is open, contains $x$, and is contained in $B_n$. This definition is canonical and uses no countable choice.

Put $V_n=U_0\cap \cdots \cap U_n$; each $V_n$ is open, contains $x$, and $V_{n+1}\subseteq V_n$.

Since $V_n\subseteq B_n$, every neighbourhood contains some $V_n$, so $(V_n)$ is the required decreasing local base.

For $x\in X$, the subfamily $\{B\in \mathcal{B}:x\in B\}$ is countable and refines every neighbourhood of $x$ because $\mathcal{B}$ is a basis.

Thus it is a countable neighbourhood base at every $x$, which is first countability.

Apply [A1] to the nonempty members of $\mathcal{B}$, and let $D$ be the selected points.

The set $D$ is countable and meets every nonempty basic open set, hence every nonempty open set, so it is dense.

Therefore $X$ is separable.

For each $B\in \mathcal{B}$ that lies in some $U\in \mathcal{U}$, use [A1] to select one such $U_B$.

The selected family is countable and covers $X$: a point lies in a cover member, and a basis member containing it lies inside that member.

Thus every open cover has a countable subcover, so $X$ is Lindelöf.

Suppose $D$ is at most countable and dense. If $D=\mathrm{\varnothing }$, then $X=\mathrm{\varnothing }$ and the empty family is a basis. Otherwise the family $\mathcal{B}=\{B(d,1\mathrm{/}n):d\in D,n\ge 1\}$ is at most countable. It is a basis: if $x\in U$ with $U$ open, choose $\epsilon >0$ with $B(x,\epsilon )\subseteq U$, then choose $n$ with $2\mathrm{/}n<\epsilon$ and $d\in D\cap B(x,1\mathrm{/}n)$. Now $x\in B(d,1\mathrm{/}n)\subseteq B(x,2\mathrm{/}n)\subseteq U$. Thus separability implies second countability.

[L1] gives second countable implies Lindelöf.

Suppose $X$ is Lindelöf. For each $n\ge 1$, the radius-$1\mathrm{/}n$ balls cover $X$. Using [A1], choose an at most countable set $D_n$ of centres whose radius-$1\mathrm{/}n$ balls cover $X$. Then $D={\bigcup }_{n\ge 1}{D}_n$ is at most countable by [L2]. It is dense: for $x\in U$ open, choose $\epsilon >0$ with $B(x,\epsilon )\subseteq U$ and $n$ with $1\mathrm{/}n<\epsilon$; some $d\in D_n$ has $x\in B(d,1\mathrm{/}n)$, so $d\in B(x,\epsilon )\subseteq U$. Thus Lindelöf implies separable.

The three implications prove the equivalence.

The nonempty traces $B\cap Y$ for $B\in \mathcal{B}$ form a countable basis for the subspace topology.

Hence every subspace is second countable.

Use [A1] to choose countable factor bases.

Finite-support boxes with selected basic coordinates form a basis for the product by [F1].

The finite subsets of a countable index set form an at most countable family: after enumerating the index set, subsets of size at most $n$ are coded by $n$-tuples of natural numbers, which are countable by finite induction using the product theorem in [L1], and their union over $n$ is countable by the union theorem in [L1]. For each fixed finite support $F$, the choices of one member of the selected basis in every coordinate of $F$ form a finite product of countable sets and are countable by the same induction. A final application of the countable-union theorem shows that all finite-support boxes form an at most countable family.

Thus the product is second countable.

Choose the coordinate local bases by [A1].

Finite-support products of their members form a local base at the given point: refine each of the finitely many restricted coordinates of a basic product neighbourhood by a member of its selected local base.

Finite subsets of the countable index set are countable in total: code the subsets of size at most $n$ by $n$-tuples and use finite induction on the product theorem in [L1], followed by the countable-union theorem. For each fixed finite support, the possible coordinate choices are a finite product of countable local bases and hence countable. The union over all finite supports is countable by [L1].

The product is first countable.

If $\mathcal{U}=\mathrm{\varnothing }$, it is already at most countable. Otherwise $X$ is nonempty, so the dense set $D$ is nonempty; enumerate $D$ and assign to each $U\in \mathcal{U}$ the first enumerated point of $D\cap U$, which is nonempty by density.

Disjointness makes this assignment injective into a countable set.

Hence $\mathcal{U}$ is countable and $X$ is ccc.

Choose one point from each nonempty $B\in \mathcal{B}$; the chosen set meets every nonempty open set because $\mathcal{B}$ is a basis, so it is dense and has cardinality at most $\mathrm{\mid }\mathcal{B}\mathrm{\mid }$.

For each $x\in X$, the subfamily $\{B\in \mathcal{B}:x\in B\}$ is a local base at $x$ and has cardinality at most $\mathrm{\mid }\mathcal{B}\mathrm{\mid }$, so every local character, and therefore its supremum $\chi (X)$, is at most $w(X)$.

Given an open cover, choose for each $B\in \mathcal{B}$ that lies in a cover member one such member; these at most $\mathrm{\mid }\mathcal{B}\mathrm{\mid }$ chosen sets still cover $X$, so $L(X)\le w(X)$.

For a pairwise-disjoint family $\mathcal{U}$ of nonempty open sets, choose a point of $D\cap U$ for each $U\in \mathcal{U}$; disjointness makes this assignment injective into $D$, so $\mathrm{\mid }\mathcal{U}\mathrm{\mid }\le d(X)$ and $c(X)\le d(X)$.

Steps 1.1, 1.2, 1.3 and 1.4 give $c(X)\le d(X)\le w(X)$ and $\chi (X),L(X)\le w(X)$.

The family $\{B\cap Y:B\in \mathcal{B}\}$ is a basis of $Y$ by [L1] and has cardinality at most $\mathrm{\mid }\mathcal{B}\mathrm{\mid }=w(X)$.

The family $\{N\cap Y:N\in {\mathcal{N}}_y\}$ is a local base at $y$ in $Y$ by [L1] and has cardinality at most $\mathrm{\mid }{\mathcal{N}}_y\mathrm{\mid }=\chi (y,X)$.

Applying the two minima in [L2] to the families of steps 1.1 and 1.2 yields $w(Y)\le w(X)$ and $\chi (y,Y)\le \chi (y,X)$.

If $D\subseteq X$ is dense, then $f[D]$ is dense in $Y$: a nonempty open $V\subseteq Y$ has nonempty open preimage by surjectivity and [L1], so that preimage meets $D$ and $V$ meets $f[D]$.

For an open cover $\mathcal{U}$ of $Y$, the family $\{f^{-1}[U]:U\in \mathcal{U}\}$ is an open cover of $X$; a subfamily indexed by at most $L(X)$ members covers $X$, and the corresponding members of $\mathcal{U}$ cover $Y$ by surjectivity.

Taking $D$ with $\mathrm{\mid }D\mathrm{\mid }=d(X)$, step 1.1 gives a dense subset of $Y$ of cardinality at most $d(X)$; hence $d(Y)\le d(X)$ by [L2].

Step 1.2 gives $L(Y)\le L(X)$ by [L2], and together with step 2.1 this proves both inequalities.

Suppose first that $\kappa$ is infinite. The family $\mathcal{B}=\{B(d,q):d\in D,q\in \mathbb{Q},q>0\}$ has cardinality at most $\kappa \cdot {\mathrm{\aleph }}_0=\kappa$ by [L2] and [L3]. It is a basis: if $x\in U$ with $U$ open, choose $\epsilon >0$ with $B(x,\epsilon )\subseteq U$, choose $d\in D$ with $d(x,d)<\epsilon \mathrm{/}3$, and then by [L2] choose a positive rational $q$ with $d(x,d)<q<\epsilon -d(x,d)$. Thus $x\in B(d,q)\subseteq B(x,\epsilon )\subseteq U$. Hence $w(X)\le \kappa =d(X)$.

Suppose $\kappa$ is finite. If $D=\mathrm{\varnothing }$, density forces $X=\mathrm{\varnothing }$ and both raw invariants are $0$. Otherwise $X=D$: if $x\mathrm{\notin }D$, the finitely many positive distances $d(x,a)$ for $a\in D$ have a positive minimum, and a smaller ball about $x$ misses $D$, contradicting density. A finite metric space is discrete, since at each point a ball smaller than all distances to the other finitely many points is a singleton. The singleton family is a basis of size $\mathrm{\mid }X\mathrm{\mid }$, and every basis of a discrete space must contain each singleton; also every dense set must contain every point. Therefore $w(X)=\mathrm{\mid }X\mathrm{\mid }=d(X)=\kappa$.

Step 1.1 handles infinite density and step 1.2 handles finite density; combining the resulting upper bound with [L1] gives $w(X)=d(X)$ in every case.

Partition the family $\mathcal{F}$ by finite cardinality. Some layer ${\mathcal{F}}_n=\{A\in \mathcal{F}:\mathrm{\mid }A\mathrm{\mid }=n\}$ is uncountable; otherwise [L1] would make their countable union $\mathcal{F}$ countable. It therefore suffices to prove the assertion by induction on the common size $n$.

The case $n=0$ is vacuous, since there is only one empty set.

Suppose the result holds for $(n-1)$-element sets and ${\mathcal{F}}_n$ is uncountable. If some point $x$ belongs to uncountably many members, apply the induction hypothesis to $\{A\setminus \{x\}:A\in {\mathcal{F}}_n,x\in A\}\mathrm{.}$ An uncountable $\mathrm{\Delta }$-subfamily with root $R$ then restores to one with root $R\cup \{x\}$.

It remains to suppose that ${\mathcal{F}}_n(x)=\{A\in {\mathcal{F}}_n:x\in A\}$ is at most countable for every $x$. Order the pairwise-disjoint subfamilies of ${\mathcal{F}}_n$ by inclusion. The union of a chain is again pairwise disjoint, so Zorn's lemma in [A1] gives a maximal such family $\mathcal{G}$.

If $\mathcal{G}$ were at most countable, then $M=\bigcup \mathcal{G}$ would be at most countable by [L1], because its members are finite. Maximality says every $A\in {\mathcal{F}}_n$ meets $M$, so ${\mathcal{F}}_n={\bigcup }_{x\in M}{\mathcal{F}}_n(x)\mathrm{.}$ The right side is a countable union of at most countable families and is at most countable by [L1], a contradiction. Hence $\mathcal{G}$ is uncountable.

The family $\mathcal{G}$ is pairwise disjoint, hence is an uncountable $\mathrm{\Delta }$-system with empty root. Together with step 1.3 this completes the induction and proves the lemma.

Suppose $\mathcal{U}$ is an uncountable pairwise-disjoint family of nonempty open sets. By [A1] and [F1], choose for every $U\in \mathcal{U}$ a nonempty basic cylinder $[p_U]\subseteq U$, where $p_U$ is a function from a finite support $F_U\subseteq I$ to $2$. Distinct $U$ give distinct cylinders.

By [L1], after passing to an uncountable subfamily the supports form a $\mathrm{\Delta }$-system with finite root $R$. There are only finitely many functions $R\to 2$, so one further uncountable subfamily has the same restriction $p_U\upharpoonright R$.

Choose two members of that subfamily. Their supports meet exactly in $R$ and their partial functions agree there, so the union of the two partial functions extends—by assigning $0$ elsewhere—to a point of $2^I$ lying in both cylinders. The corresponding members of $\mathcal{U}$ intersect, a contradiction. Thus every such family is at most countable and $2^I$ is ccc.

Suppose $D\subseteq 2^I$ is at most countable and dense. It is nonempty because $2^I$ is nonempty, so choose a surjection $s:\mathbb{N}\to D$ by [L1]. For each $i\in I$ define its column $c_i\in 2^{\mathbb{N}}$ by $c_i(n)=s(n)(i)$.

By [L2] there are only $2^{{\mathrm{\aleph }}_0}$ possible columns, whereas $\mathrm{\mid }I\mathrm{\mid }>2^{{\mathrm{\aleph }}_0}$. Thus distinct $i,j\in I$ have $c_i=c_j$, which says $d(i)=d(j)$ for every $d\in D$ because $s$ is onto.

The cylinder $\{x\in 2^I:x(i)=0,x(j)=1\}$ is nonempty and open by [F1], but step 2.1 makes it disjoint from $D$, contradicting density. Hence $2^I$ is not separable.

Let $X=2^I$ with its product topology.

The space $X$ satisfies the hypothesis of the proposed implication because it is ccc by [L1].

The same space fails the proposed conclusion because it is not separable by [L2].

Thus $X$ is a ccc nonseparable space, which refutes the statement.

For each $x\in D$, the one-member family $\{\{x\}\}$ is a local base, because every neighbourhood of $x$ contains the open singleton $\{x\}$ by [L1].

If $\mathcal{B}$ is any basis of $D$, then for each $x\in D$ some $B_x\in \mathcal{B}$ satisfies $x\in B_x\subseteq \{x\}$, so $B_x=\{x\}$ and $\mathcal{B}$ contains every singleton.

Step 1.1 makes $D$ first countable, while steps 1.2 and [L3] make every basis uncountable; hence $D$ is not second countable.

Every nonempty basic interval $[a,b)$ meets $\mathbb{Q}$, so $\mathbb{Q}$ is a countable dense subset of the lower-limit line.

If an at most countable basis $\mathcal{B}$ existed, it would be nonempty, so enumerate it as $(B_n)$ by [L4]. For $x\in \mathbb{R}$, the set $E_x=\{n:x\in B_n\subseteq [x,x+1)\}$ is nonempty by [L2]; let $n(x)$ be its least member. If $x<y$ and $n(x)=n(y)$, then the common basis member contains $x$ but is contained in $[y,y+1)$, impossible. Thus $x\mapsto n(x)$ would inject $\mathbb{R}$ into $\mathbb{N}$, contradicting [L1].

Step 1.1 gives separability, whereas step 1.2 rules out an at most countable basis; thus this separable space is not second countable.

The half-open intervals cover $\mathbb{R}$, and if two contain $x$, then $[x,c)$ lies in their intersection for some $c>x$; hence [F1] makes them a basis. The rational grid $\mathbb{Q}\times \mathbb{Q}$ is at most countable by [L1] and [L2], and density of $\mathbb{Q}$ makes it meet every nonempty basic lower-limit rectangle, so it is dense in $P$.

For each $x\in \mathbb{R}$, the basic rectangle $[x,x+1)\times [-x,-x+1)$ meets $A$ only in $(x,-x)$; hence $A$ is discrete in its subspace topology.

The map $x\mapsto (x,-x)$ is a bijection from the uncountable set $\mathbb{R}$ onto $A$, so a dense subset of the discrete space $A$ must be all of $A$ and cannot be at most countable.

Thus $P$ is separable by step 1.1 but has the nonseparable subspace $A$ by step 2.1, refuting heredity of separability.

The discrete space $D$ is Hausdorff because distinct singleton neighbourhoods are disjoint, and locally compact because each point has the compact singleton neighbourhood; it is not compact because its singleton cover has no finite subcover. Thus its one-point compactification has the usual compact Hausdorff behavior, and in any case [L1] makes $D^{\ast }$ compact with $D$ as an open subspace. By [L2], $D^{\ast }$ is Lindelöf.

The subspace $D$ is discrete and has the open cover $\{\{x\}:x\in D\}$; any subcover must contain every singleton, so no at most countable subfamily covers the uncountable set $D$.

Thus the Lindelöf space $D^{\ast }$ has the non-Lindelöf subspace $D$, so Lindelöfness is not hereditary.

For an open cover $\mathcal{U}$ of $S$, let $\mathcal{D}$ be all basic intervals $[a,b)$ lying in members of $\mathcal{U}$, and put $C=\bigcup \{(a,b):[a,b)\in \mathcal{D}\}$. For every rational pair $p<q$ for which $(p,q)\subseteq (a,b)$ for some $[a,b)\in \mathcal{D}$, use [A1] to select one such member of $\mathcal{D}$. The selected family is at most countable and covers $C$: every point of an interval $(a,b)$ lies in some rational interval $(p,q)\subseteq (a,b)$.

The antidiagonal $A=\{(x,-x):x\in \mathbb{R}\}$ is uncountable and discrete in $S^2$, because $[x,x+1)\times [-x,-x+1)$ meets $A$ only in $(x,-x)$.

The antidiagonal is closed: a point $(u,v)$ with $u+v\mathrm{\ne }0$ has a basic rectangle avoiding $A$, using $[u,u+1)\times [v,v+1)$ when $u+v>0$ and sufficiently short intervals ending before the sum reaches $0$ when $u+v<0$.

Fix an enumeration of $\mathbb{Q}$. If $x\mathrm{\notin }C$, some member of $\mathcal{D}$ containing $x$ must have left endpoint exactly $x$; otherwise $x$ would lie in its ordinary interior and hence in $C$. Let $r_x$ be the first rational in the fixed enumeration satisfying $x<r_x$ and $[x,r_x)\in \mathcal{D}$; such a rational exists by [L2]. If $x<y$ and $r_x=r_y$, then $y\in (x,r_x)\subseteq C$, a contradiction. Thus $x\mapsto r_x$ injects $\mathbb{R}\setminus C$ into $\mathbb{Q}$, so $\mathbb{R}\setminus C$ is at most countable.