Easton's Theorem and Cardinal Invariants of the Continuum: Examples and Counterexamples

For all $n$ and all $i\le n$ the difference $A_n\setminus A_i$ is finite: iterating the hypothesis and [F1] gives $A_n{\subseteq }^{\ast }{A}_i$ whenever $i\le n$.

For every $n$ the finite intersection $B_n:={\bigcap }_{i\le n}{A}_i$ is infinite: $A_n\setminus B_n\subseteq {\bigcup }_{i\le n}(A_n\setminus A_i)$ is a finite union of finite sets by step 1.1, hence finite by [F4], and $A_n=B_n\cup (A_n\setminus B_n)$ is infinite, so $B_n$ cannot be finite.

The recursion of the display is legitimate: assume $x_0,\dots ,x_{n-1}$ have been chosen; the set $B_n\setminus \{x_0,\dots ,x_{n-1}\}$ is infinite minus a finite set, hence nonempty, so it has a least element $x_n$ by [F3], and $x_n\mathrm{\notin }\{x_0,\dots ,x_{n-1}\}$; recursion gives the whole sequence.

The sequence is strictly increasing: $B_{n+1}\subseteq B_n$, and $x_n$ is the least member of $B_n$ outside $\{x_0,\dots ,x_{n-1}\}$. Since $x_{n+1}$ belongs to this latter set but differs from $x_n$, minimality gives $x_n<x_{n+1}$. By step 3.1 the $x_n$ are pairwise distinct, so $X$ is infinite; also $x_n\in B_n\subseteq A_i$ for every $i\le n$. Thus $X\in [\omega {]}^{\omega }$ by [F4].

For fixed $k$, every $x_n$ with $n\ge k$ lies in $B_n\subseteq A_k$ by step 4.1, so $X\setminus A_k\subseteq \{x_0,\dots ,x_{k-1}\}$ is finite; hence $X{\subseteq }^{\ast }{A}_k$ for every $k$, and $X$ is a pseudointersection of $\{A_k:k\in \mathbb{N}\}$, the countable case recorded in [F2]. ∎

By [F1], ${\mathrm{\aleph }}_1\le p\le t\le \mathfrak{c}={\mathrm{\aleph }}_1$, so $p$ and $t$ are pinched between ${\mathrm{\aleph }}_1$ and ${\mathrm{\aleph }}_1$.

By [F2], ${\mathrm{\aleph }}_1\le b\le d\le \mathfrak{c}={\mathrm{\aleph }}_1$, so $b$ and $d$ are pinched between ${\mathrm{\aleph }}_1$ and ${\mathrm{\aleph }}_1$.

By [F4], for $\mathcal{I}=\mathcal{N}$ and $\mathcal{I}=\mathcal{M}$ each of the four numbers lies between ${\mathrm{\aleph }}_1$ and $\mathfrak{c}={\mathrm{\aleph }}_1$.

By [F3] together with step 1.2, ${\mathrm{\aleph }}_1\le s\le d\le {\mathrm{\aleph }}_1$ and ${\mathrm{\aleph }}_1\le b\le r\le \mathfrak{c}={\mathrm{\aleph }}_1$, so $s$ and $r$ are pinched between ${\mathrm{\aleph }}_1$ and ${\mathrm{\aleph }}_1$.

Every one of the fourteen numbers is an ordinal $x$ with ${\mathrm{\aleph }}_1\le x\le {\mathrm{\aleph }}_1$, so $x={\mathrm{\aleph }}_1$ by the antisymmetry clause of [F5]; this gives $p=t=b=d=s=r={\mathrm{\aleph }}_1$ and $\mathrm{add}(\mathcal{I})=\mathrm{cov}(\mathcal{I})=\mathrm{non}(\mathcal{I})=\mathrm{cof}(\mathcal{I})={\mathrm{\aleph }}_1$ for both ideals $\mathcal{I}=\mathcal{N},\mathcal{M}$, as claimed. ∎

The domain $\{{\mathrm{\aleph }}_0,{\mathrm{\aleph }}_1\}$ is a set of infinite regular cardinals by [F2], the values $F({\mathrm{\aleph }}_0)=F({\mathrm{\aleph }}_1)={\mathrm{\aleph }}_3$ are cardinals by [F4], and $F$ is nondecreasing because its two values are equal.

$\mathrm{cf}(F(\kappa ))>\kappa$ holds at both domain points: $\mathrm{cf}({\mathrm{\aleph }}_3)={\mathrm{\aleph }}_3$, since ${\mathrm{\aleph }}_3$ is regular by [F2] and [F4], and ${\mathrm{\aleph }}_3>{\mathrm{\aleph }}_1>{\mathrm{\aleph }}_0$ by [F4].

Steps 1.1 and 1.2 verify all four clauses of [F1], so $F$ is a set-sized Easton function; the hypothesis of [F3] is met by the given ground model $M$ of ZFC+GCH and by the $M$-generic $G$.

Applying [F3] at the two domain points gives $(2^{{\mathrm{\aleph }}_0}{)}^{M[G]}=F({\mathrm{\aleph }}_0)={\mathrm{\aleph }}_3$ and $(2^{{\mathrm{\aleph }}_1}{)}^{M[G]}=F({\mathrm{\aleph }}_1)={\mathrm{\aleph }}_3$, while clause (a) of [F3] gives that $M[G]$ has the same ordinals, cofinalities and cardinals as $M$.

The Axiom of Choice is used in the regularity of ${\mathrm{\aleph }}_1$ and ${\mathrm{\aleph }}_3$ of steps 1.1 and 1.2 and is part of the ZFC ground model on which [F3] is stated; no value at a singular cardinal is asserted, since $F$ is only defined on $\{{\mathrm{\aleph }}_0,{\mathrm{\aleph }}_1\}$. The displayed equalities are exactly the statement. ∎

In the model of the given iteration, MA holds and $\mathfrak{c}=2^{{\mathrm{\aleph }}_0}={\mathrm{\aleph }}_2$ by [F1]; in particular the continuum is the cardinal ${\mathrm{\aleph }}_2>0$, so the phrase "fewer than $2^{{\mathrm{\aleph }}_0}$" in [F2] and [F3] means "of cardinality below ${\mathrm{\aleph }}_2$".

$\mathrm{add}(\mathcal{N})\le {\mathrm{\aleph }}_2$ and $\mathrm{add}(\mathcal{M})\le {\mathrm{\aleph }}_2$: by [F4] the additivity is at most the continuum, which is ${\mathrm{\aleph }}_2$ by step 1.1.

$\mathrm{add}(\mathcal{N})\ge {\mathrm{\aleph }}_2$: let $\mathcal{A}\subseteq \mathcal{N}$ with $\mid \mathcal{A}\mid <{\mathrm{\aleph }}_2$; by step 1.1 the family has fewer than $2^{{\mathrm{\aleph }}_0}$ members, so [F2] makes $\bigcup \mathcal{A}$ null, that is, $\bigcup \mathcal{A}\in \mathcal{N}$. Hence no subfamily of $\mathcal{N}$ of size below ${\mathrm{\aleph }}_2$ witnesses the additivity, and the minimum clause of [F4] gives $\mathrm{add}(\mathcal{N})\ge {\mathrm{\aleph }}_2$.

$\mathrm{add}(\mathcal{M})\ge {\mathrm{\aleph }}_2$: the same argument with [F3] in place of [F2] gives that every subfamily of $\mathcal{M}$ of size below ${\mathrm{\aleph }}_2$ has its union in $\mathcal{M}$, hence $\mathrm{add}(\mathcal{M})\ge {\mathrm{\aleph }}_2$ by the minimum clause of [F4].

Combining step 2.1 with steps 2.2 and 2.3 gives $\mathrm{add}(\mathcal{N})={\mathrm{\aleph }}_2=\mathrm{add}(\mathcal{M})$, and ${\mathrm{\aleph }}_2=\mathfrak{c}$ by step 1.1; the Axiom of Choice is used in the iteration and in reading the cardinalities of [F4] as cardinals. ∎