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An uncountable cardinal number $\kappa$ is weakly compact when every $\kappa$-satisfiable theory in an infinitary language $L_{\kappa ,\kappa }$ with at most $\kappa$ nonlogical symbols is satisfiable. A cardinal is inaccessible when it is uncountable, regular, and a strong limit cardinal.

Two standard results supply the proof. First, every weakly compact cardinal is inaccessible. Second, every weakly compact $\kappa$ has the Keisler extension property: there is a transitive set $X⊋V_{\kappa }$ such thatand $\kappa \in X$. Since $\kappa$ is inaccessible, the relevant downward absoluteness makes $X\models$ “$\kappa$ is inaccessible”. Hence $X$ satisfies “there is an inaccessible cardinal”. By elementarity, $V_{\kappa }$ satisfies the same sentence, so it contains some inaccessible $\lambda$. Every ordinal in $V_{\kappa }$ is below $\kappa$, and inaccessibility is absolute here, giving

Fix $i\in \{0,1\}$ and write $\kappa ={\kappa }_i$, $j=j_i$, and $M=M_i$. Every ordinal below $j(\kappa )$ is represented in the ultrapower by a function $\kappa \to \kappa$. Consequently, in $V_{\lambda }$,where the last equality uses the Generalized continuum hypothesis.

By elementarity, $M$ regards $j(\kappa )$ as measurable and hence as a strong limit cardinal. Moreover $V_{\kappa +1}\subseteq M$, so $M$ and $V_{\lambda }$ have the same subsets of $\kappa$ and the same ${\kappa }^{+}$. It follows inside $M$ thatThus, in the ambient $V_{\lambda }$, the ordinal $j(\kappa )$ is strictly larger than ${\kappa }^{+}$ but has cardinality at most ${\kappa }^{+}$. It cannot be a cardinal number. Applying this argument to both $i=0$ and $i=1$ proves that neither $j_0({\kappa }_0)$ nor $j_1({\kappa }_1)$ is a cardinal in $V_{\lambda }$, exactly as in moved critical point is not an ambient cardinal under GCH.

Since the strongly inaccessible cardinal $\lambda$ is inaccessible, $(V_{\lambda },\in )$ is a model of ZFC. The Downward Lowenheim-Skolem theorem gives an elementary substructureof cardinality $\kappa$ such thatOne may obtain $X$ concretely as the Skolem hull of this set; its cardinality remains $\kappa$ because the language of set theory is countable and $|V_{\kappa }|=\kappa$.

Apply the Mostowski collapse theorem to $X$ and write $\pi :X\to M$ for the collapse. Then $M$ is a transitive set, $|M|=\kappa$, and $\pi$ fixes $V_{\kappa }$ pointwise. It also fixes $\kappa$, because it fixes every ordinal below $\kappa$. By elementarity, $X$ satisfies ZFC and regards $U$ as a kappa-complete filter that is a nonprincipal ultrafilter on $\kappa$. Therefore, with $\bar{U}=\pi (U)$,The internal ultrafilter $\bar{U}$ need not equal the original $U$.

No. Letthe successor cardinal of $\kappa$ computed by $M$. Then $\alpha \in M$, and $M$ regards $\alpha$ as a cardinal number. Because $M$ is transitive, $\alpha \subseteq M$, so externallyOn the other hand $\kappa <\alpha$, hence $|\alpha |=\kappa$ in the ambient universe. A corresponding bijection belongs to $V_{\lambda }$ because its rank is below the inaccessible limit $\lambda$. Thus $V_{\lambda }$ regards $\alpha$ as equinumerous with $\kappa$ and therefore not as a cardinal. This is an instance of cardinal nonabsoluteness in a small transitive model.

An infinite cardinal number $\kappa$ is a regular cardinal if every function from an ordinal $\delta <\kappa$ into $\kappa$ has bounded range. Write this as $\forall f\forall \delta \in \kappa (\mathrm{Map}(f,\delta ,\kappa )\Rightarrow \exists \beta \in \kappa \forall \xi \in \delta f(\xi )<\beta )$. The graph predicates are bounded formulas in set theory, and cardinalhood is Pi-one definable, so adding infinite cardinalhood yields a Pi-one formula modulo ZF.

The second uncountable ordinal ${\omega }_2$ is the least ordinal whose cardinal number is strictly larger than that of the first uncountable ordinal ${\omega }_1$. Equivalently, it is the initial ordinal of the successor cardinal ${\aleph }_2$.

An infinite cardinal number $\kappa$ is singular when $\mathrm{cf}(\kappa )<\kappa$. Thus a short cofinal function reaches arbitrarily high ordinals below $\kappa$. For example, ${\aleph }_{\omega }$ has cofinality $\omega$. The complementary case is a regular cardinal.

For a regular infinite $\theta <\kappa$, with $\kappa$ a regular uncountable cardinal number, $S_{\theta }^{\kappa }=\{\delta <\kappa :\mathrm{cf}(\delta )=\theta \}$ is stationary. Build a strictly increasing continuous $\theta$-sequence in any club set, and take its supremum. Closure places it in that club and the cofinality of an increasing ordinal supremum gives cofinality $\theta$. In particular the two disjoint sets $S_{\omega }^{{\omega }_2}$ and $S_{{\omega }_1}^{{\omega }_2}$ show that the club filter on ${\omega }_2$ is not an ultrafilter.

For any regular uncountable cardinal number $\kappa$, the set $\kappa \setminus \{\omega \}$ contains the final-segment club set $[\omega +1,\kappa )$, so belongs to the club filter and is a stationary set. It is not closed, since its finite ordinals have supremum $\omega$, which is missing.

A subset of a regular uncountable cardinal number $\kappa$ is stationary if it meets every club set in $\kappa$. Intersecting a stationary set with a club set preserves stationarity. Every member of the club filter is stationary. At a regular infinite $\theta <\kappa$, the stationarity of ordinals of prescribed cofinality supplies important examples.

If transitive models of enough set theory contain the same $V_{\alpha +1}$, then they agree on whether $\alpha$ is a strongly inaccessible cardinal. They have the same subsets and functions on every ordinal below $\alpha$, so they agree on cardinality, regularity, and the strong limit cardinal property.

The successor cardinal ${\kappa }^{+}$ is the least cardinal number strictly larger than $\kappa$.

For an infinite cardinal number $\kappa$, every ${\kappa }^{+}$-Suslin set is a union of ${\kappa }^{+}$ many $\kappa$-Suslin sets. A countable sequence of ordinals below the successor cardinal ${\kappa }^{+}$ is bounded there, so restrict the representing tree successively to labels below each $\alpha <{\kappa }^{+}$.

Below any uncountable strongly inaccessible cardinal $\kappa$, the worldly cardinals contain a closed unbounded subset of $\kappa$ and hence have cardinality $\kappa$. The structure $V_{\kappa }$ models ZFC. Reflect each finite group from an enumeration of all first-order formulas to a closed unbounded subset of $\kappa$, using $|V_{\alpha }|<\kappa$ and regularity for the witness bounds. The countable intersection remains closed unbounded, so its ranks are elementary substructures of $V_{\kappa }$. Intersect with the closed unbounded subset of infinite cardinal numbers below $\kappa$. Uncountability is essential: the regular strong limit $\omega$ has no worldly cardinal below it.