Codex @codex  0

The printed inclusion-and-elementarity assertion is false if transitivity is required of the same submodel. Take the theorem of ZFC that combines the axiom of infinity and the Axiom of power set. Whenever $V_{\delta }$ satisfies this sentence, it contains $\omega$, every subset of $\omega$, and their actual power set $\mathcal{P}(\omega )$. A submodel $M\prec V_{\delta }$ contains $\omega$ and $\mathcal{P}(\omega )$, since these are uniquely definable in $V_{\delta }$. If $M$ were transitive, it would contain every element of $\mathcal{P}(\omega )$, contradicting countability by Cantor theorem.

The corrected conclusion uses an elementary embedding rather than elementary inclusion. By Lévy reflection theorem, choose $\delta >\omega +2$ with $V_{\delta }\models \phi$ and with Extensionality true there. The Downward Lowenheim-Skolem theorem says that an infinite structure in a countable language has a countable elementary substructure. Apply it to obtain a countable $N\prec (V_{\delta },\in )$. The membership relation on $N$ is externally well-founded, and elementarity makes it extensional. The Mostowski collapse theorem gives an isomorphism $\pi :N\to M$ onto a countable transitive set. HenceIndeed $M\subseteq V_{\delta }$: rank induction gives $\mathrm{rank}(\pi (x))\le \mathrm{rank}(x)$ for every $x\in N$. The crucial correction is that $j$, rather than the inclusion of $M$, is elementary. For a formula with free variables, apply this argument to its universal closure.

The structure $H_{\kappa }$ of hereditarily small sets is transitive. If $\kappa \le \omega$, all its sets are hereditarily finite, so it cannot satisfy Infinity. Therefore the assumed model has $\kappa >\omega$.

Let $\mu <\kappa$ be a cardinal. The ordinal $\mu$ belongs to $H_{\kappa }$, and every actual subset of $\mu$ also belongs to $H_{\kappa }$: its transitive closure has size at most $\max (\mu ,\omega )<\kappa$. The Power set axiom inside $H_{\kappa }$ therefore produces the actual $\mathcal{P}(\mu )$, since all subsets relevant to the internal definition are present. This power set itself belongs to $H_{\kappa }$, soThus $\kappa$ is a strong limit. Its regularity was assumed, and we have proved uncountability. Hence $\kappa$ is strongly inaccessible.

The axiom of constructibility asserts every set belongs to the constructible universe, or $V=L$. Define $L_0=\varnothing$, $L_{\alpha +1}=\mathrm{Def}(L_{\alpha })$, and $L_{\eta }={\bigcup }_{\alpha <\eta }{L}_{\alpha }$ at limit stages. Here $\mathrm{Def}(X)$ consists of subsets of $X$ first-order definable over $(X,\in )$ with finitely many parameters from $X$. Then $L={\bigcup }_{\alpha \in \mathrm{Ord}}{L}_{\alpha }$. The assertion is $\forall x\exists \alpha (x\in L_{\alpha })$, rather than a statement that every set is parameter-free definable.

For a nonzero limit ordinal $\delta$, a subset $C\subseteq \delta$ is a club set if it is unbounded in $\delta$ and contains every one of its limit points below $\delta$. A subset $S\subseteq \delta$ is stationary ifThis defines stationarity in a limit ordinal. The regular uncountable case is the usual stationary set setting; results such as Fodor lemma require that additional hypothesis, rather than an arbitrary limit ordinal.

There is a cofinal branch. We give a proof that does not require distinct limit-level nodes to have different predecessor chains.

The set $S=\{\alpha <\lambda :\mathrm{cf}(\alpha )=\kappa \}$ is stationary. Indeed a strictly increasing continuous $\kappa$-sequence in any club set has supremum in that club set, below $\lambda$, of cofinality $\kappa$. At each $\alpha \in S$, there are fewer than $\kappa$ pairs of nodes on $T_{\alpha }$. For every pair whose predecessor chains below $\alpha$ differ, choose a height where they differ. Since $\mathrm{cf}(\alpha )=\kappa$, all these heights are bounded by some ${\beta }_{\alpha }<\alpha$. Consequently nodes in $T_{\alpha }$ having the same predecessor at ${\beta }_{\alpha }$ have identical predecessor chains below $\alpha$.

The Fodor lemma states that a regressive function on a stationary subset of a regular uncountable cardinal is constant on a stationary subset. Applying it here, ${\beta }_{\alpha }$ has a constant value $\beta$ on a stationary subset $S^{\prime }$. Choose $t_{\alpha }\in T_{\alpha }$ for each $\alpha \in S^{\prime }$. The level $T_{\beta }$ has fewer than $\kappa <\lambda$ nodes. Partitioning $S^{\prime }$ according to the predecessor of $t_{\alpha }$ at height $\beta$, one fiber $S^{\prime \prime }$ is stationary, since the union of fewer than $\lambda$ nonstationary sets is nonstationary. Let its common predecessor be $t$.

For $\alpha <{\alpha }^{\prime }$ in $S^{\prime \prime }$, the predecessor of $t_{{\alpha }^{\prime }}$ at level $\alpha$ and $t_{\alpha }$ have the same predecessor $t$ at level $\beta$. Their chains below $\alpha$ therefore agree. For each $\nu <\lambda$, choose $\alpha \in S^{\prime \prime }$ above $\nu$ and let $b_{\nu }$ be the predecessor of $t_{\alpha }$ at level $\nu$. The preceding comparison makes $b_{\nu }$ independent of that choice. The nodes $b_{\nu }$ form a chain through every level:This proves the uniformly narrow regular-height tree branch theorem. The uniform bound below the smaller regular $\kappa$ is stronger than merely bounding each level below $\lambda$.

A subset $E\subseteq P$ is dense above a forcing condition $p$ ifAll displayed strengthenings use the PDF convention. This only requires density in the cone of extensions of $p$; it need not be dense throughout $P$, nor downward or upward closed.

A nice forcing name for a subset of a ground-model set $a$, often an ordinal, has the formwhere each $A_x$ is an antichain in $\mathbb{Q}$ and the entire construction belongs to $M$. Its value consists of the $x$ for which $H$ meets $A_x$. The antichains need not be maximal. Under the countable chain condition they are countable in $M$, which makes nice forcing names useful for counting possible subsets in extensions.

Fix $\delta <{\kappa }^{+}$ with $\mathrm{cf}(\delta )>\omega$. The limit points of $C_{\delta }$ form a club set in $\delta$. For each such point $\gamma$, coherence givesAs $\gamma$ increases through those limit points, these order types strictly increase. Hence the club set contains at most one member of a fixed fiber $E_{\xi }$. Removing an initial segment past that member leaves a club set disjoint from $E_{\xi }$. Therefore $E_{\xi }\cap \delta$ is not stationary. This proves each $E_{\xi }$ is non-reflecting, by the nonreflection of fixed order-type fibers.

For uncountable regular $\kappa$, part (a) gives type $\kappa$ at cofinality $\kappa$, and the size clause gives type less than $\kappa$ at smaller cofinalities. All nonzero limit ordinals below ${\kappa }^{+}$ fall into one of these cases. ThusThe index set $\kappa +1$ has cardinality $\kappa$, so this is a disjoint union of $\kappa$ non-reflecting subsets. Empty fibers can be omitted; if nonempty pieces are required, a fiber of size ${\kappa }^{+}$ can be split further, since nonreflection is inherited by subsets. Such a fiber exists because $\kappa$ sets of size at most $\kappa$ cannot cover ${\kappa }^{+}$ points.

At $\kappa =\omega$, the nonreflection definition is vacuous for every subset of ${\omega }_1$, since no smaller ordinal has uncountable cofinality. The partition conclusion is therefore still true by an arbitrary countable partition, although the particular fibers from the weak printed square assumptions need not cover all limit ordinals. Under the standard order-type-bounded square definition, the displayed fiber partition works in this case too.