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True. A useful fact is that squaring is strictly increasing on ordinals. If $A<B$, monotonicity in both arguments gives $A^2\le BA$, and strict increase in the right argument for the positive left factor $B$ gives $BA<B^2$. Hence $A^2<B^2$.

Suppose, for a contradiction, that $\alpha \beta <\beta \alpha$. Again using monotonicity and associativity of ordinal multiplication,This contradicts the hypothesized equality of the outer terms. Reversing the roles excludes $\beta \alpha <\alpha \beta$. If either factor is zero, they already commute. Therefore the commuting squares of ordinals property isThe proof uses monotonicity, which follows directly from the ordered-copy definition; it does not assume a cancellation law for ordinals.

Zorn lemma states that a partially ordered set in which every chain in a partially ordered set has an upper bound has a maximal element. Include the empty chain in a partially ordered set, so the hypotheses imply that the set $P$ is nonempty. Suppose there were no maximal element. Every chain in a partially ordered set $C$ has an upper bound $u$, and some $v>u$; therefore the set of elements strictly above every member of $C$ is nonempty. The axiom of choice selects one such element $f(C)$ for each chain in a partially ordered set (and an element of $P$ when $C$ is empty).

By Hartogs theorem, there is an ordinal $\kappa$ which does not inject into $P$. Transfinite recursion defines $x_{\alpha }=f(\{x_{\beta }:\beta <\alpha \})$ for all $\alpha <\kappa$: previous elements form a strictly increasing chain in a partially ordered set, including at limit ordinal stages. All $x_{\alpha }$ are distinct, giving the forbidden injection. This proves Zorn lemma; the choice of $f$ is the use of the axiom of choice.

Apply Zorn lemma to linearly independent subsets of $\mathbb{R}$ over $\mathbb{Q}$, ordered by inclusion. A union along a chain in a partially ordered set is linearly independent because each finite linear relation occurs in one member of that chain in a partially ordered set. A maximal independent set spans $\mathbb{R}$, since a vector outside its span could be adjoined. Thus $\mathbb{R}$ has a Hamel basis over $\mathbb{Q}$.

For a finite basis, the Steinitz exchange lemma proves equality of basis sizes. This also rules out an infinite basis when a finite spanning set exists: write each member of the finite spanning set using finitely many elements of another basis; their finite union spans, and independence leaves no other basis elements. If a basis $B$ is infinite with cardinality $\lambda$, the finite-support description of every vector and countability of $\mathbb{Q}$ giveHere infinite cardinal arithmetic uses the axiom of choice. Thus any infinite basis has cardinality $|V|$, proving all bases have the same cardinality. The zero space has its unique empty basis.

A cardinal number is identified with an initial ordinal, namely an ordinal not in bijection with any smaller ordinal. An inaccessible cardinal, in the strong sense used here, is an uncountable regular cardinal that is also a strong limit cardinal. ThusA worldly cardinal is a cardinal number whose level of the Von Neumann hierarchy satisfies all of ZFC:Worldliness is this model-theoretic condition; regularity is not part of its definition. The original PDF asks for these two notions; the extra comma in the local TeX makes “cardinal” look like a separate requested item.

Induct on $\alpha$ using the transitivity and inclusion already established. The zero stage is immediate. Suppose $L_{\alpha }\cap \mathrm{Ord}=\alpha$. Being an ordinal is expressible by a bounded formula in set theory, for example transitivity, transitivity of all members, and membership comparability. It is therefore absolute for the transitive set $L_{\alpha }$. The subset of its elements that are ordinals is internally definable, soBy inclusion, all smaller ordinals are present as well. Conversely, if an ordinal $\gamma$ belongs to $L_{\alpha +1}$, then $\gamma \subseteq L_{\alpha }$, since every member of a definable power set is a subset of the preceding domain. All members of $\gamma$ are ordinals, so $\gamma \subseteq L_{\alpha }\cap \mathrm{Ord}=\alpha$ and $\gamma \le \alpha$. Hence the ordinal part of the successor level is exactly $\alpha +1$.

At a nonzero limit ordinal $\lambda$,Thus

D. Downward absoluteness of cardinalhood holds because, if the larger transitive model sees no bijection with a smaller ordinal, neither can the smaller model, whose functions form only a subset of those in the larger model. Cardinalhood is not described by an upward absolute formula, because the larger model can contain a new bijection collapsing an ordinal that the smaller model regards as a cardinal number.

Let $N$ be a well-founded model of set theory of the complete theory $T$, and suppose that a Paris model $M\models T$ were ill-founded. Its internal ordinals would then contain an external descending sequenceFor every $n$, choose a first-order formula ${\phi }_n$ that uniquely defines ${\alpha }_n$ in $M$. The sentences asserting that ${\phi }_n$ uniquely defines an ordinal and that the object defined by ${\phi }_{n+1}$ belongs to the object defined by ${\phi }_n$ are true in $M$. Since $T$ is complete, all its models satisfy the same first-order sentences, so the corresponding uniquely defined ordinals in $N$ form an external descending membership sequence. This contradicts the well-foundedness of $N$. Thus Paris models are well-founded when their complete theory has a well-founded model proves that every Paris model of $T$ is well-founded.

Fix $a\in L$. Inside the ambient transitive model $M$, the Axiom of power set makes the collection of constructible subsets of $a$ a set. For each such subset $x$, choose the least stage of the constructible hierarchy at which $x$ appears. The Axiom schema of replacement and the supremum of a set of ordinals give an ordinal $\beta$ bounding all these stages; enlarge $\beta$ so that $a\in L_{\beta }$.

Nowis definable over $L_{\beta }$ with parameter $a$. It therefore belongs to the definable power set $\mathcal{D}(L_{\beta })=L_{\beta +1}\subseteq L$. This set contains exactly the subsets of $a$ that belong to the constructible universe, so it witnesses the Axiom of power set in $L$. Therefore $L\models$ Power Set.

False. Take $x=\mathcal{P}(\omega )$, the power set of the first infinite ordinal $\omega$. Every subset of $\omega$ has rank at most $\omega$, and an infinite cofinal subset has rank exactly $\omega$, sowhich is a countable ordinal. But Cantor theorem shows that $\mathcal{P}(\omega )$ is an uncountable set.

True. Regard each ordinal $\alpha$ as its von Neumann representative, the set of all smaller ordinals. Transfinite induction givesThus the set $\alpha$ itself has rank $\alpha$.

The Von Neumann hierarchy is defined by transfinite recursion:for every limit ordinal $\lambda$.

We prove simultaneously by transfinite induction that every $V_{\alpha }$ is a transitive set and that $V_{\alpha }\subseteq V_{\alpha +1}$. The claims are immediate for $V_0$. Suppose first that $V_{\alpha }$ is transitive. If $x\in y\in V_{\alpha +1}=\mathcal{P}(V_{\alpha })$, then $y\subseteq V_{\alpha }$, so $x\in V_{\alpha }$. Moreover, transitivity says that every $x\in V_{\alpha }$ satisfies $x\subseteq V_{\alpha }$, and henceThus $V_{\alpha +1}$ is transitive and $V_{\alpha }\subseteq V_{\alpha +1}$. At a limit $\lambda$, if $x\in y\in V_{\lambda }$, then $y\in V_{\beta }$ for some $\beta <\lambda$; transitivity of $V_{\beta }$ gives $x\in V_{\beta }\subseteq V_{\lambda }$. The inclusions at successor stages also imply

Finally, the rank of a set $x$ is the ordinalEvery $y\in x$ belongs to $V_{\mathrm{rank}(y)+1}\subseteq V_{\mathrm{rank}(x)}$, so $x\subseteq V_{\mathrm{rank}(x)}$. Thereforewhich proves that every set occurs in the hierarchy.

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

The critical point of an elementary embedding $j$ is the least ordinal $\alpha$ for which $j(\alpha )\ne \alpha$.

We first prove by transfinite induction that $j(\alpha )=\alpha$ for every $\alpha <\kappa$. Suppose this is known below $\alpha <\kappa$. If $[f]E[{\mathrm{const}}_{\alpha }]$, thenThe sets $A_{\beta }=\{\xi \in A:f(\xi )=\beta \}$ for $\beta <\alpha$ partition $A$ into fewer than $\kappa$ pieces. A kappa-complete filter that is an ultrafilter must contain one cell $A_{\beta }$: otherwise all their complements would belong to $U$, and their intersection would contradict $A\in U$. Hence $[f]=[{\mathrm{const}}_{\beta }]$. The predecessors of $j(\alpha )$ are consequently exactly the already-fixed ordinals below $\alpha$, so $j(\alpha )=\alpha$.

Now let $\mathrm{id}(\xi )=\xi$. An identical argument shows that the predecessors of $[\mathrm{id}]$ in the ultrapower are exactly $[{\mathrm{const}}_{\beta }]$ for $\beta <\kappa$, soSince $\{\xi <\kappa :\xi <\kappa \}=\kappa \in U$, one has $[\mathrm{id}]E[{\mathrm{const}}_{\kappa }]$. After collapsing, $\kappa \in j(\kappa )$, and therefore $j(\kappa )>\kappa$. Thus $\mathrm{crit}(j)=\kappa$.

For a nonzero ordinal $\alpha$, letThe set is nonempty because $0\in S$, and it is an initial segment because ordinal exponentiation by $\omega$ is strictly increasing. Put $\beta =\sup S$.

If $\beta$ is a successor, the definition of the supremum forces $\beta \in S$. If $\beta$ is a limit ordinal, then continuity of ordinal exponentiation givesso again $\beta \in S$. Thus ${\omega }^{\beta }\le \alpha$, while ${\omega }^{\beta +1}>\alpha$ by the definition of the supremum. Hence $\beta$ is the greatest required exponent.

Apply division by an additively indecomposable ordinal with $\lambda ={\omega }^{\beta }$:Since $\alpha \ge {\omega }^{\beta }$, the quotient $q$ is nonzero. It must be finite: if $q\ge \omega$, thencontradicting the maximality of $\beta$. Writing $q=n$ givesThis is the leading-term decomposition.

Assume the axiom of choice. By the well-ordering theorem, every set is bijective with an ordinal, and hence with an initial ordinal. Any two ordinals are comparable, so for any sets $K,L$ either $K$ injects into $L$ or $L$ injects into $K$. Transitivity is immediate, and antisymmetry on cardinalities follows from the Cantor-Schröder-Bernstein theorem. Thus cardinal comparison is a total order.

Conversely, assume the cardinal comparability principle. Given any set $X$, let $h(X)$ be its Hartogs ordinal. By Hartogs theorem, there is no injection $h(X)\to X$. Comparability therefore supplies an injection $X\to h(X)$. Pulling the ordinal order back to $X$ gives a well-order of $X$. Hence every set can be well-ordered, and the well-ordering theorem implies the axiom of choice.

The Lévy reflection theorem says that for every finite collection $\Phi$ of first-order formulas there are arbitrarily large ordinals $\alpha$ such that, for every $\phi \in \Phi$ and every tuple of parameters $a_1,\dots ,a_n\in V_{\alpha }$,Equivalently, the ordinals simultaneously reflecting all formulas in $\Phi$ form a closed unbounded class.

Take $\lambda =\omega \cdot 2$. This is a limit ordinal greater than $\omega$. There is a recursive well-order code $R$ for ${\omega }^2$: for example, use a recursive pairing of the natural numbers with $\omega \times \omega$ and the lexicographic order consisting of $\omega$ successive blocks of order type $\omega$. Since $R$ is definable over $L_{\omega }$, it belongs to $L_{\omega +1}$ and hence to $L_{\omega \cdot 2}$.

The representation of $R$ is ${\omega }^2$, butand the ordinals belonging to $L_{\lambda }$ are exactly those below $\lambda$. Thus $R\in L_{\lambda }$ while its representation is not in $L_{\lambda }$, violating the second requirement for a coding level of the constructible hierarchy.

Let $a,p\in M[G]$, and suppose a first-order formula $\psi (x,y,p)$ defines exactly one $y$ for every $x\in a$. Apply the Lévy reflection theorem to the formulas needed to express this assertion, choosing an ordinal $\alpha$ with $a,p\in V_{\alpha }^{M[G]}$ such thatfor every $x\in a$. Therefore every required value lies in the set $V_{\alpha }^{M[G]}$.

The already established axiom schema of separation forms the setFunctionality makes $b$ exactly the range of the definable function on $a$. This proves every instance of the Axiom schema of replacement in the generic extension $M[G]$.

A Scott formula describes a finite tuple at a transfinite stage of the back-and-forth method. Stage zero specifies the complete atomic description; successor stages specify matching one-element extensions in both directions; limit stages conjoin the earlier descriptions. For a countable structure the equivalences on its finite tuples stabilize at a countable ordinal.

Let $\mathcal{C}$ be the ordered category of all ordinals with an added greatest object $\infty$. The categorical presheaf with $X(\alpha )=\{\ast \}$ at ordinals and $X(\infty )=\varnothing$ preserves every small categorical limit: a small supremum is an ordinal unless its diagram contains $\infty$. Its category of elements consists of all ordinals, and their projection has the large colimit $\infty$. Nevertheless $X$ is not representable: an ordinal representer has bounded support, while the representer $\infty$ has a nonempty value at $\infty$. Thus preservation must include the possibly large opposite elements diagram. The dual functor $X^{\mathrm{op}}:\mathcal{C}\to {\mathbf{Set}}^{\mathrm{op}}$ also disproves the comma-projection criterion for a right adjoint under small-colimit preservation alone.

For an uncountable regular cardinal $\kappa$, choose cofinal sequences $c_{\alpha }:\omega \to \alpha$ for the stationary set of ordinals of cofinality $\omega$. Above any bound, some fixed coordinate exceeds the bound on a stationary subset; Fodor lemma makes that coordinate constant on a stationary subset. Thus the stationary constant fibers, over all coordinates, have unboundedly many values. Regularity makes one coordinate have $\kappa$ such values. Its fibers are disjoint stationary sets; adding all leftover ordinals to one piece partitions $\kappa$.