Easton's Theorem and Cardinal Invariants of the Continuum

Assume [F4]. Let $\kappa$ be an infinite cardinal. By [F1], $2^{\kappa }=\mid \mathcal{P}(\kappa )\mid$ is a cardinal and $\kappa <2^{\kappa }$. Restricting to infinite regular $\kappa$ gives clause (a).

Let $\kappa \le \lambda$ be infinite regular cardinals. Every subset of $\kappa$ is a subset of $\lambda$, so the inclusion $\mathcal{P}(\kappa )\subseteq \mathcal{P}(\lambda )$ is an injection; by the injection criterion of [F3], $\mid \mathcal{P}(\kappa )\mid \le \mid \mathcal{P}(\lambda )\mid$. Applying [F1] at $\kappa$ and at $\lambda$ turns this into $2^{\kappa }\le 2^{\lambda }$, which is clause (b).

Clause (c) is [F2] at $\kappa$: $\mathrm{cf}(2^{\kappa })>\kappa$ for every infinite cardinal $\kappa$, in particular for every infinite regular one.

Clauses (a), (b) and (c) hold for all infinite regular cardinals, so in ZFC the continuum function on infinite regular cardinals satisfies exactly the displayed constraints. The Axiom of Choice enters only through [F1] and [F2] -- through the identification of $2^{\kappa }$ with $\mid \mathcal{P}(\kappa )\mid$ and through König's theorem -- and no further selection is made in steps 1.2 and 1.3. ∎

Under GCH every infinite cardinal $\mu$ satisfies $2^{\mu }={\mu }^{+}$, by [F4]. If $\lambda ={\mathrm{\aleph }}_0$, then $2^{<\lambda }={\sup }_{n<\omega }{2}^n={\mathrm{\aleph }}_0=\lambda$. If $\lambda ={\nu }^{+}$ for an infinite cardinal $\nu$, then GCH gives $2^{\nu }=\lambda$ and monotonicity gives $2^{\mu }\le \lambda$ for every $\mu <\lambda$, so again $2^{<\lambda }=\lambda$. If $\lambda$ is a limit cardinal above ${\mathrm{\aleph }}_0$, the successor cardinals ${\mu }^{+}$ for infinite $\mu <\lambda$ are cofinal in $\lambda$, while each is at most $\lambda$; thus $2^{<\lambda }={\sup }_{\mu <\lambda }{\mu }^{+}=\lambda$. These cases prove clause (a) for every infinite regular $\lambda$.

Now let $⟨p_{\xi }:\xi <\delta ⟩$ be a descending sequence in $P^{>\lambda }$ with $\delta <{\lambda }^{+}$, and put $p={\bigcup }_{\xi <\delta }{p}_{\xi }$. The conditions form a $\subseteq$-chain of functions, so $p$ is a function with values in $\{0,1\}$, and every triple in its domain has first coordinate $>\lambda$. Fix an infinite regular $\gamma >\lambda$ and let $A_{\xi }=\{(\kappa ,\alpha ,\beta )\in \mathrm{dom}(p_{\xi }):\kappa \le \gamma \}$. Each $\mathrm{\mid }{A}_{\xi }\mathrm{\mid }<\gamma$ by [F8], and $\mathrm{\mid }\delta \mathrm{\mid }\le \lambda <\gamma$. Regularity of $\gamma$ gives $\sigma :={\sup }_{\xi <\delta }\mathrm{\mid }{A}_{\xi }\mathrm{\mid }<\gamma$, so $\mathrm{\mid }{\bigcup }_{\xi <\delta }{A}_{\xi }\mathrm{\mid }\le \mathrm{\mid }\delta \mathrm{\mid }\cdot \sigma <\gamma$ (with the finite or empty cases immediate). For $\gamma \le \lambda$ the set in question is empty. Hence $p\in P^{>\lambda }$ and $p$ extends every $p_{\xi }$, so $P^{>\lambda }$ is ${\lambda }^{+}$-closed, clause (c), in the sense of [F3]; the same cardinal bound applies to a pairwise compatible family of at most $\lambda$ conditions, whose union is a function because the members agree on overlaps.

Consequently $2^{\mathrm{\mid }r\mathrm{\mid }}\le 2^{<\lambda }=\lambda$ for every set $r$ with $\mathrm{\mid }r\mathrm{\mid }<\lambda$, and $\lambda \otimes \lambda =\lambda$ by [F6]; with Choice [F9], this bounds a union of $\lambda$ many sets each of size at most $\lambda$ by $\lambda$.

We prove clause (b) by contraposition. Suppose $W\subseteq P^{\le \lambda }$ is an antichain with $\mathrm{\mid }W\mathrm{\mid }={\lambda }^{+}$. By the support condition of [F8] at the regular cardinal $\lambda$, every $p\in W$ has $\mathrm{\mid }\mathrm{dom}(p)\mathrm{\mid }<\lambda$. For any fixed domain there are at most $2^{<\lambda }=\lambda$ bit assignments. Thus there are ${\lambda }^{+}$ distinct domains, since otherwise Choice [F9] and $\lambda \otimes \lambda =\lambda$ would bound $\mathrm{\mid }W\mathrm{\mid }$ by $\lambda$. Select one condition for each distinct domain. Now [F1] applies with $\kappa =\lambda$, $\rho =2^{<\lambda }=\lambda$ of step 1.1 and yields $W^{\mathrm{\prime }}\subseteq W$ of size ${\lambda }^{+}$ and a root $r$ with $\mathrm{dom}(p)\cap \mathrm{dom}(q)=r$ for distinct $p,q\in W^{\mathrm{\prime }}$, in the sense of [F2].

The map $p\mapsto p\upharpoonright r$ takes at most $2^{\mathrm{\mid }r\mathrm{\mid }}\le \lambda$ values on $W^{\mathrm{\prime }}$ by step 2.1, and a union of $\lambda$ many classes each of size at most $\lambda$ has size at most $\lambda$; since $\mathrm{\mid }{W}^{\mathrm{\prime }}\mathrm{\mid }={\lambda }^{+}>\lambda$, two distinct $p,q\in W^{\mathrm{\prime }}$ have $p\upharpoonright r=q\upharpoonright r$.

For such $p,q$ the union $p\cup q$ is a function, because the domains meet exactly in $r$ and the two agree on $r$; all its triples still have first coordinate $\le \lambda$. Let $\gamma$ be an infinite regular cardinal with $\gamma \le \lambda$ and let $A_p=\{(\kappa ,\alpha ,\beta )\in \mathrm{dom}(p):\kappa \le \gamma \}$ and similarly $A_q$. By [F8] both have cardinality below $\gamma$, and $\mathrm{\mid }{A}_p\cup A_q\mathrm{\mid }\le \mathrm{\mid }{A}_p\mathrm{\mid }\oplus \mathrm{\mid }{A}_q\mathrm{\mid }<\gamma$ by [F6], the case of finite cardinalities being immediate since $\gamma$ is infinite. Hence $p\cup q$ satisfies the support condition at every regular $\gamma \le \lambda$, and the bound at $\lambda$ also implies the bound at every regular $\gamma >\lambda$. Thus it is a condition of $P^{\le \lambda }$ extending both $p$ and $q$. This contradicts the antichain property of $W$, so no antichain of $P^{\le \lambda }$ has cardinality ${\lambda }^{+}$: every antichain has cardinality below ${\lambda }^{+}$, which is clause (b) by [F3].

Every condition of $P(F)$ splits uniquely as $p=p^{\le \lambda }\cup p^{>\lambda }$ with each restriction supported on its respective side of $\lambda$; since the support condition at a regular $\gamma$ refers only to triples with $\kappa \le \gamma$, the two parts are conditions of $P^{\le \lambda }$ and $P^{>\lambda }$ respectively, every head-tail pair has disjoint domains and its union satisfies each support bound because the union of two sets of size below an infinite regular $\gamma$ still has size below $\gamma$, and the extension order is preserved in both directions. Hence $p\mapsto (p^{\le \lambda },p^{>\lambda })$ is an isomorphism $P(F)\cong P^{\le \lambda }\times P^{>\lambda }$ when $\mathrm{dom}(F)$ is a set, and for a class Easton function the same computation applies to each set-sized condition; $P^{\le \lambda }$ is then a set, since its conditions are partial functions on the set of triples with $\kappa \le \lambda$ of size below $\lambda$. This is clause (d) and completes the proof. [F8, given] ∎

Choose a $P\times Q$-name $\dot{f}\in M$ with $f={\mathrm{val}}_{G\times H}(\dot{f})$, and put $\rho ={\mathrm{rk}}_{P\times Q}(\dot{f})$. By [F4], $\mathrm{rank}(f)\le \rho$. Every value $f(\alpha )$ belongs to $M$ and has rank below $\rho +1$; transitivity of $M$ therefore puts it in the ground set $A:=V_{\rho +1}^M$. The product extension satisfies ZFC by [F4], so the truth lemma supplies $r=(p_r,q_r)\in G\times H$ forcing that $\dot{f}$ is a function from $\check{\lambda }$ into $\check{A}$. We will only use forcing below $r$.

Form the cones $P_r=\{p\in P:p\le p_r\}$ and $Q_r=\{q\in Q:q\le q_r\}$ in $M$. They are nonempty. Every descending sequence in $P_r$ of length below ${\lambda }^{+}$ has a lower bound still in $P_r$ (include $p_r$ when the sequence is empty); every pairwise-incompatible subset of $Q_r$ is one of $Q$, so $Q_r$ is ${\lambda }^{+}$-cc. Here a maximal antichain of $Q_r$ means a pairwise-incompatible subset $W$ such that every member of $Q_r$ is compatible with some member of $W$. This agrees with maximality by inclusion: a condition incompatible with all of $W$ could be added, and conversely.

For each $\alpha <\lambda$ let $D_{\alpha }\subseteq P_r$ be the set of $p$ for which there are a maximal antichain $W\subseteq Q_r$ and $a:W\to A$ such that $(p,q)⊩\dot{f}(\check{\alpha })=\check{a(q)}$ for every $q\in W$. The coordinate notation abbreviates the fixed first-order formula saying that $\dot{f}$ has value $\check{a(q)}$ at $\check{\alpha }$; no function-value term is added to the forcing language. This forcing relation is definable in $M$, and $W$ and $a$ range over ground sets, so Separation gives $D_{\alpha }\in M$. Monotonicity makes $D_{\alpha }$ downward closed in $P_r$.

Fix $p_0\in P_r$. Recursively, for $\gamma <{\lambda }^{+}$, suppose $⟨(p_{\xi },q_{\xi },a_{\xi }):\xi <\gamma ⟩$ has been chosen with the $p_{\xi }$ descending and the $q_{\xi }$ pairwise incompatible. If $W_{\gamma }=\{q_{\xi }:\xi <\gamma \}$ is maximal in $Q_r$, stop. Otherwise choose $q^{\mathrm{\prime }}\in Q_r$ incompatible with every member of $W_{\gamma }$. Closure gives $\bar{p}\in P_r$ below $p_0$ and every preceding $p_{\xi }$, since their number is below ${\lambda }^{+}$. Because $(\bar{p},q^{\mathrm{\prime }})\le r$ forces that the value of $\dot{f}$ at $\check{\alpha }$ lies in $\check{A}$, the existential-name clause first gives, densely below $(\bar{p},q^{\mathrm{\prime }})$, a name $\sigma$ forced to be that value and to belong to $\check{A}$. The atomic membership clause for $\check{A}$ then gives a further pair $(p_{\gamma },q_{\gamma })\le (\bar{p},q^{\mathrm{\prime }})$ and $a_{\gamma }\in A$ forcing $\sigma =\check{a_{\gamma }}$, hence $\dot{f}(\check{\alpha })=\check{a_{\gamma }}$. In particular $q_{\gamma }$ remains incompatible with all earlier $q_{\xi }$. The selected triples lie in the ground set $P_r\times Q_r\times A$; Choice in $M$ well-orders this set and makes the recursion deterministic. No choice from the proper class of all names is needed.

The recursion stops at some $\beta <{\lambda }^{+}$: otherwise the distinct $q_{\gamma }$, $\gamma <{\lambda }^{+}$, form a pairwise-incompatible subset of $Q_r$ of size ${\lambda }^{+}$. At the stopping stage $W_{\beta }$ is maximal. Closure supplies $p^{†}\in P_r$ below $p_0$ and every $p_{\xi }$ for $\xi <\beta$. Monotonicity then gives $(p^{†},q_{\xi })⊩\dot{f}(\check{\alpha })=\check{a_{\xi }}$ for all $\xi <\beta$, so $W_{\beta }$ and $a(q_{\xi })=a_{\xi }$ witness $p^{†}\in D_{\alpha }$. Thus every $D_{\alpha }$ is dense open in $P_r$.

The intersection $D={\bigcap }_{\alpha <\lambda }{D}_{\alpha }$ belongs to $M$ and is dense open in $P_r$. Indeed, starting below any $p\in P_r$, use [F5] to meet $D_{\alpha }$ successively for $\alpha <\lambda$, taking lower bounds at limits and at the end. All these sequences have length at most $\lambda <{\lambda }^{+}$; downward closure keeps the final condition in every $D_{\alpha }$.

The projection $G$ is $M$-generic for $P$: if $E\in M$ is dense in $P$, then $E\times Q$ is dense in $P\times Q$ and the product generic meets it. To meet the cone-dense set $D$, use the global dense set $E_D=D\cup \{p\in P:p\perp p_r\}$: if $p$ is compatible with $p_r$, first strengthen into $P_r$ and then into $D$. Since $p_r\in G$ and a filter cannot contain incompatible conditions, $G\cap E_D\subseteq D$. Fix $p^{\ast }\in G\cap D$. Choice in $M$ selects, for all $\alpha <\lambda$, witnesses $(W_{\alpha },a_{\alpha })$ to $p^{\ast }\in D_{\alpha }$; their full sequence belongs to $M$.

Similarly $H$ is $M$-generic for $Q$. For each $\alpha$, the downward closure of $W_{\alpha }$ is dense in $Q_r$: compatibility with a member of the maximal antichain yields a common stronger condition. Lifting this cone-dense set to $Q$ by adjoining conditions incompatible with $q_r$ shows that $H$ meets it. Upward closure then gives some $w_{\alpha }\in H\cap W_{\alpha }$, and downward directedness makes $w_{\alpha }$ unique because distinct members of $W_{\alpha }$ are incompatible.

For each $\alpha <\lambda$, $(p^{\ast },w_{\alpha })\in G\times H$ forces $\dot{f}(\check{\alpha })=\check{a_{\alpha }(w_{\alpha })}$, so the truth lemma gives $f(\alpha )=a_{\alpha }(w_{\alpha })$. The ground sequence $⟨(W_{\alpha },a_{\alpha }):\alpha <\lambda ⟩$ and $H$ belong to $M[H]$, which satisfies ZFC by [F4]. Its Separation and Replacement therefore construct $\alpha \mapsto a_{\alpha }(w_{\alpha })$ there. This function is $f$, hence $f\in M[H]$. Characteristic functions and cofinal maps are special cases of such sequences, proving the stated relative conclusions. ∎

Fix the data of the statement, so that $M\subseteq M[G]$ are transitive models with the same ordinals and $M[G]\models \mathrm{Z}\mathrm{F}$, and let $\delta \in M$ be an ordinal.

For every infinite ground regular cardinal $\lambda$ the factorization $P(F)\cong P^{\le \lambda }\times P^{>\lambda }$ holds in $M$. The coordinate projections $G^{>\lambda }$ and $G^{\le \lambda }$ of $G$ are $M$-generic: if $D\subseteq P^{\le \lambda }$ is dense and $D\in M$, then the set of conditions $p\in P(F)$ with $p^{\le \lambda }\in D$ is dense in $P(F)$ and lies in $M$, so $G$ meets it and $G^{\le \lambda }$ meets $D$; the same computation with a dense $D\subseteq P^{>\lambda }$ handles the tail. Hence $M[G]=M[G^{>\lambda }][G^{\le \lambda }]$.

Every ground regular cardinal remains regular in $M[G]$. Suppose $\kappa$ is infinite and regular in $M$ but not in $M[G]$, and let $\delta ={\mathrm{cf}}^{M[G]}(\kappa )<\kappa$ with a cofinal $f:\delta \to \kappa$ in $M[G]$. Then $\delta$ is an infinite regular cardinal of $M[G]$ and hence also of $M$: if ${\mathrm{cf}}^M(\delta )<\delta$ then $M[G]$ would contain a cofinal map ${\mathrm{cf}}^M(\delta )\to \delta$ by [F7], making $\delta$ singular in $M[G]$. So $\delta$ is an infinite regular cardinal of $M$ with $\delta <\kappa$, and by step 1.2 the lemma [F4] applies with the ${\delta }^{+}$-closed tail $P^{>\delta }$ and the ${\delta }^{+}$-cc head $P^{\le \delta }$, giving $f\in M[G^{\le \delta }]$. Thus ${\mathrm{cf}}^{M[G^{\le \delta }]}(\kappa )\le \delta$, while [F5] at $\theta ={\delta }^{+}\le \kappa$ gives ${\mathrm{cf}}^{M[G^{\le \delta }]}(\kappa )={\mathrm{cf}}^M(\kappa )=\kappa >\delta$, a contradiction.

Every ground cardinal remains a cardinal of $M[G]$. Suppose not, and let $\kappa$ be the least ground cardinal with $\mid \kappa {\mid }^{M[G]}=\mu <\kappa$. By step 2.1, $\kappa$ cannot be regular in $M$: an ordinal regular in the ZFC extension is a cardinal there. Thus $\kappa$ is a singular ground cardinal, hence a limit cardinal, and the ground cardinals below it are cofinal in it. Choose a ground cardinal $\nu$ with $\mu <\nu <\kappa$. Minimality of $\kappa$ makes $\nu$ a cardinal of $M[G]$. A bijection $\kappa \to \mu$ in $M[G]$ restricts to an injection $\nu \to \mu$, since $\nu \subseteq \kappa$ and $\nu$ is an ordinal of $M[G]$; hence $\mid \nu {\mid }^{M[G]}\le \mu <\nu$, a contradiction.

All ground cofinalities are preserved. Let $\kappa ={\mathrm{cf}}^M(\delta )$ and fix a strictly increasing cofinal $g:\kappa \to \delta$ in $M$, so that ${\mathrm{cf}}^{M[G]}(\delta )\le \kappa$ because $g$ still has unbounded range in $M[G]$. If $\eta ={\mathrm{cf}}^{M[G]}(\delta )<\kappa$, take a cofinal $h:\eta \to \delta$ in $M[G]$ and define $h^{\mathrm{\prime }}:\eta \to \kappa$ in $M[G]$ by letting $h^{\mathrm{\prime }}(\xi )$ be the least $\beta <\kappa$ with $h(\xi )\le g(\beta )$. Then $h^{\mathrm{\prime }}$ has cofinal range in $\kappa$: given ${\beta }_0<\kappa$, cofinality of $h$ gives $\xi$ with $h(\xi )\ge g({\beta }_0)$, hence $g(h^{\mathrm{\prime }}(\xi ))\ge h(\xi )\ge g({\beta }_0)$ and $h^{\mathrm{\prime }}(\xi )\ge {\beta }_0$ by strict increase of $g$. So ${\mathrm{cf}}^{M[G]}(\kappa )\le \eta <\kappa$ would hold, contradicting step 2.1 since $\kappa$ is ground regular; therefore ${\mathrm{cf}}^{M[G]}(\delta )=\kappa ={\mathrm{cf}}^M(\delta )$.

Steps 3.2 and 3.1 show that $M$ and $M[G]$ have the same ordinals, the same cofinality function on the ordinals of $M$, and the same cardinals, which is the statement. ∎