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 $\beta\preceq\kappa\preceq\delta\approx\beta$, 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 $\lambda\leq\kappa$ 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 $\beta\preceq\kappa\preceq\delta\approx\beta$, 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 $w(X)\le\aleph_0$ and $d(X)\le\aleph_0$ say exactly that such a basis and such a dense subset are at most countable.

By [L1], $L(X)\le\aleph_0$ says that every open cover has a subcover of at most countable cardinality, and $c(X)\le\aleph_0$ 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 $\chi(X)\le\aleph_0$ exactly when every point has a local base of cardinality at most $\aleph_0$.

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 $U_n=\operatorname{int}(B_n)$. 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 $D=\varnothing$, then $X=\varnothing$ and the empty family is a basis. Otherwise the family $\mathcal B=\{B(d,1/n):d\in D,\ n\ge1\}$ is at most countable. It is a basis: if $x\in U$ with $U$ open, choose $\varepsilon>0$ with $B(x,\varepsilon)\subseteq U$, then choose $n$ with $2/n<\varepsilon$ and $d\in D\cap B(x,1/n)$. Now $x\in B(d,1/n)\subseteq B(x,2/n)\subseteq U$. Thus separability implies second countability.

[L1] gives second countable implies Lindelöf.

Suppose $X$ is Lindelöf. For each $n\ge1$, the radius-$1/n$ balls cover $X$. Using [A1], choose an at most countable set $D_n$ of centres whose radius-$1/n$ balls cover $X$. Then $D=\bigcup_{n\ge1}D_n$ is at most countable by [L2]. It is dense: for $x\in U$ open, choose $\varepsilon>0$ with $B(x,\varepsilon)\subseteq U$ and $n$ with $1/n<\varepsilon$; some $d\in D_n$ has $x\in B(d,1/n)$, so $d\in B(x,\varepsilon)\subseteq U$. 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 $\mathcal U=\varnothing$, 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 $|\mathcal B|$.

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 $|\mathcal B|$, 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 $|\mathcal B|$ 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 $|\mathcal U|\le d(X)$ 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 $|\mathcal B|=w(X)$.

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 $|\mathcal N_y|=\chi(y,X)$.

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 $|D|=d(X)$, 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 $\kappa\cdot\aleph_0=\kappa$ by [L2] and [L3]. It is a basis: if $x\in U$ with $U$ open, choose $\varepsilon>0$ with $B(x,\varepsilon)\subseteq U$, choose $d\in D$ with $d(x,d)<\varepsilon/3$, and then by [L2] choose a positive rational $q$ with $d(x,d)<q<\varepsilon-d(x,d)$. Thus $x\in B(d,q)\subseteq B(x,\varepsilon)\subseteq U$. Hence $w(X)\le\kappa=d(X)$.

Suppose $\kappa$ is finite. If $D=\varnothing$, density forces $X=\varnothing$ and both raw invariants are $0$. Otherwise $X=D$: if $x\notin D$, 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 $|X|$, and every basis of a discrete space must contain each singleton; also every dense set must contain every point. Therefore $w(X)=|X|=d(X)=\kappa$.

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 $\mathcal F_n=\{A\in\mathcal F:|A|=n\}$ 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 $\{A\setminus\{x\}:A\in\mathcal F_n,\ x\in A\}.$ An uncountable $\Delta$-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 $\mathcal F_n=\bigcup_{x\in M}\mathcal F_n(x).$ 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 $\Delta$-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 $\Delta$-system with finite root $R$. There are only finitely many functions $R\to2$, so one further uncountable subfamily has the same restriction $p_U\restriction R$.

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 $D\subseteq2^I$ 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 $c_i\in2^{\mathbb N}$ by $c_i(n)=s(n)(i)$.

By [L2] there are only $2^{\aleph_0}$ possible columns, whereas $|I|>2^{\aleph_0}$. 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 $\{x\in2^I:x(i)=0,\ x(j)=1\}$ 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.