Countability Axioms and Cardinal Functions: Examples and Counterexamples

Specializing ${\mathbb{Q}}^n$ is a countable dense subset of ${\mathbb{R}}^n$, and rational open boxes form a countable basis to $n=1$, rational-endpoint intervals and the rational points give countable bases and dense sets for the usual real line. The inequalities $c(\mathbb{R})\le d(\mathbb{R})\le w(\mathbb{R})$ and $\chi (\mathbb{R}),L(\mathbb{R})\le w(\mathbb{R})$ then give countable upper bounds for all five functions (Under choice, $c(X)\le d(X)\le w(X)$ and $\chi (X),L(X)\le w(X)$). No finite family can be a basis or a local base, and finite covers or cellular families have arbitrarily large finite witnesses. Thus the five raw functions are all ${\mathrm{\aleph }}_0$.

For an infinite discrete $X$ of cardinality $\kappa$, singletons force every basis, dense set, singleton cover, and cellular family to have size $\kappa$. At a point, $\{\{x\}\}$ is a one-member local base. Hence $w=d=L=c=\kappa$ and $\chi =1$.

Assume Choice and let $S$ be the lower-limit line. At $x$, the intervals $[x,x+1\mathrm{/}n)$ form a countable local base. No finite family is a local base: the intersection of finitely many neighbourhoods is still a neighbourhood and contains some $y>x$, whereas $[x,y)$ cannot contain any member of a finite local base contained in that intersection. Hence $\chi (S)={\mathrm{\aleph }}_0$.

The rationals are countable and meet every nonempty half-open interval, so $d(S)\le {\mathrm{\aleph }}_0$; no finite set is dense, since a short half-open interval can avoid it. The published lower-limit-line lemma gives Lindelöfness, while the cover $\{[-n,n):n\ge 1\}$ has no finite subcover, so $d(S)=L(S)={\mathrm{\aleph }}_0$. Density bounds cellularity above, and the disjoint family $\{[n,n+1):n\in \mathbb{Z}\}$ bounds it below, giving $c(S)={\mathrm{\aleph }}_0$.

All half-open intervals form a basis of cardinality at most $\mathrm{\mid }\mathbb{R}{\mathrm{\mid }}^2=\mathrm{\mid }\mathbb{R}\mathrm{\mid }$. Conversely, well order any basis and assign to each $x$ its first member $B_x$ with $x\in B_x\subseteq [x,x+1)$; if $x<y$, then $B_y$ cannot contain $x$, so $B_x\mathrm{\ne }{B}_y$. Thus $w(S)=\mathrm{\mid }\mathbb{R}\mathrm{\mid }=2^{{\mathrm{\aleph }}_0}$.

Let $S$ be the lower-limit line. Under Choice, the countable-product theorem makes $S^2$ first countable, and the rational grid is a countable dense subset; hence $S^2$ is separable and therefore ccc. The antidiagonal $A=\{(x,-x):x\in \mathbb{R}\}$ is an uncountable closed discrete subspace, as proved in Refuted: separability is hereditary and Assuming countable choice, refuted: Lindelöfness is productive. If $S^2$ were second countable, hereditary second countability would make the discrete space $A$ second countable, which is impossible because every basis of a discrete space contains all its singletons. The explicit open cover in Assuming countable choice, refuted: Lindelöfness is productive shows directly that $S^2$ is not Lindelöf.

Give $D=\mathbb{R}$ the discrete topology and form $D^{\ast }=D\cup \{\mathrm{\infty }\}$. The one-point compactification theorem makes $D^{\ast }$ compact, hence Lindelöf, and every point of $D$ remains isolated.

A neighbourhood of $\mathrm{\infty }$ has finite complement in $D$, since compact subsets of a discrete space are finite. If $(N_n)$ were a countable local base at $\mathrm{\infty }$, put $F_n=D\setminus N_n$. For every $x\in D$, the neighbourhood $D^{\ast }\setminus \{x\}$ would contain some $N_n$, so $D={\bigcup }_n{F}_n$. Each finite subset of $\mathbb{R}$ has a canonical increasing enumeration; these enumerations and countability of $\mathbb{N}\times \mathbb{N}$ make the displayed union countable, contradicting uncountability of $\mathbb{R}$. Thus $D^{\ast }$ is not first countable. Finally every dense subset must meet the open singleton $\{x\}$ for every $x\in D$, so it contains all of $D$ and cannot be countable; hence $D^{\ast }$ is not separable.

Assume countable choice. Successor ordinals and $0$ are isolated. If $\alpha <{\omega }_1$ is a nonzero limit, enumerate the at most countable ordinal $\alpha$ and take the successive finite suprema of that enumeration; the result is a countable cofinal sequence, and the corresponding final intervals $({\beta }_n,\alpha ]$ form a local base at $\alpha$. Thus ${\omega }_1$ is first countable. The published ordinal theorem makes it countably compact.

Every at most countable subset $D\subseteq {\omega }_1$ is bounded by some $\beta <{\omega }_1$, so the nonempty open tail above $\beta$ misses $D$; hence ${\omega }_1$ is not separable. The open initial segments $\{[0,\beta ]:\beta <{\omega }_1\}$ cover ${\omega }_1$, but any at most countable subfamily has bounded union and therefore fails to cover. Thus ${\omega }_1$ is not Lindelöf.

Let $I=\mathcal{P}(\mathcal{P}(\mathbb{N}))$. Cantor's theorem gives $\mathrm{\mid }I\mathrm{\mid }>2^{{\mathrm{\aleph }}_0}$. Under choice, Under choice, every Cantor cube $2^I$ satisfies ccc makes $2^I$ ccc and Under choice, if $\mathrm{\mid }I\mathrm{\mid }>2^{{\mathrm{\aleph }}_0}$, then the Cantor cube $2^I$ is not separable makes it nonseparable.

Separability is hereditary.

Assuming countable choice, Lindelöfness is productive.