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 $\aleph_0$.

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 $[x,x+1/n)$ 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 $\chi(S)=\aleph_0$.

The rationals are countable and meet every nonempty half-open interval, so $d(S)\le\aleph_0$; 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 $\{[-n,n):n\ge1\}$ has no finite subcover, so $d(S)=L(S)=\aleph_0$. Density bounds cellularity above, and the disjoint family $\{[n,n+1):n\in\mathbb Z\}$ bounds it below, giving $c(S)=\aleph_0$.

All half-open intervals form a basis of cardinality at most $|\mathbb R|^2=|\mathbb R|$. 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 $B_x\ne B_y$. Thus $w(S)=|\mathbb R|=2^{\aleph_0}$.

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 $D^*=D\cup\{\infty\}$. The one-point compactification theorem makes $D^*$ compact, hence Lindelöf, and every point of $D$ remains isolated.

A neighbourhood of $\infty$ has finite complement in $D$, since compact subsets of a discrete space are finite. If $(N_n)$ were a countable local base at $\infty$, put $F_n=D\setminus N_n$. For every $x\in D$, the neighbourhood $D^*\setminus\{x\}$ would contain some $N_n$, so $D=\bigcup_nF_n$. 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 $D^*$ 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 $D^*$ 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 $|I|>2^{\aleph_0}$. Under choice, Under choice, every Cantor cube $2^I$ satisfies ccc makes $2^I$ ccc and Under choice, if $|I|>2^{\aleph_0}$, then the Cantor cube $2^I$ is not separable makes it nonseparable.

Separability is hereditary.

Assuming countable choice, Lindelöfness is productive.