Past exam of the analytic topology and axiomatic set theory course of the University of Oxford / 2008 / B

In first-order set theory a class in set theory is a definable collection, written schematically $\{x:\theta (x,\vec{p})\}$ using a logical formula and set parameters. A class term is an abbreviation for such a defining condition; it does not add a new kind of object to the quantifiers of ZF. A class in set theory is proper if it is not the collection of elements of any set.

Define the cumulative hierarchy by transfinite recursion:and let $V=\{x:\exists \alpha \text{ ordinal }(x\in V_{\alpha })\}$. Without foundation this denotes the well-founded part produced by the hierarchy; it must not simply be assumed to contain every ambient set. Transfinite recursion on ordinals and the relevant set constructions remain available in ZF without foundation.

Prove transitivity by transfinite induction. The empty stage is transitive. Suppose $T=V_{\alpha }$ is transitive. Then each member of $T$ is a subset of $T$, so $T\subseteq \mathcal{P}(T)$. Also $\mathcal{P}(T)$ is transitive: if $u\in v\in \mathcal{P}(T)$, then $u\in T$ and, by transitivity of $T$, $u\subseteq T$, so $u\in \mathcal{P}(T)$. At a limit stage, a union of transitive sets is transitive. This proves transitivity at every stage.

Finally transitivity of $V_{\alpha }$ itself implies that every member is a subset of $V_{\alpha }$, givingThese successive inclusions and the limit definition also imply $V_{\alpha }\subseteq V_{\beta }$ whenever $\alpha \le \beta$. The argument has not used foundation. This leaf includes the unnumbered definitions preceding the source's three claims.

For every $x\in a$, the assumption $a\subseteq V$ supplies an ordinal stage containing $x$. Let ${\alpha }_x$ be the least such ordinal. Leastness makes this a uniquely definable assignment, so ambient replacement gives the setPut $\beta =\sup S$, taking $\beta =0$ if $a$ is empty. Monotonicity of the hierarchy gives $x\in V_{{\alpha }_x}\subseteq V_{\beta }$ for every $x\in a$. Hence $a\subseteq V_{\beta }$, and thereforeThis is a rank-bounding argument: a set-sized collection of elements from the hierarchy has its stages bounded by one ordinal. No simultaneous arbitrary choices are required, since each least stage is uniquely specified.

Fix a logical formula $\theta (x,y,\vec{p})$ with parameters in $V$, and a set $a\in V$. Suppose $⟨V,\in ⟩$ satisfies that each $x\in a$ has a unique $y$ with $\theta (x,y,\vec{p})$. Relativize every unbounded quantifier in $\theta$ to the definable class in set theory $V$, obtaining an ambient first-order formula ${\theta }^V$.

Transitivity of $V$ ensures that every member of $a$ belongs to $V$. The ambient relationtherefore defines a unique ambient set $y$ for every $x\in a$. Apply ambient replacement to obtain its range set $b$. By construction every element of $b$ belongs to $V$, so part (ii) gives $b\in V$.

Membership in this set $b$ is exactly the range relation required by replacement inside the structure $⟨V,\in ⟩$. Since the logical formula and parameters were arbitrary, every instance of the replacement scheme holds in the hierarchy class in set theory. The proof uses formula relativization to a class and the closure property from part (ii), not the unsupported assertion that $V$ is the whole ambient universe.

A bounded quantifier has the form $\exists u\in v$ or $\forall u\in v$, abbreviating $\exists u(u\in v\wedge \cdots )$ or $\forall u(u\in v\to \cdots )$. A ${\Sigma }_0$ logical formula, also called ${\Delta }_0$, is built from atomic membership and equality logical formulas using Boolean connectives and only bounded quantifiers. It has no unbounded quantifier.

For example $z\subseteq y$ abbreviates the bounded logical formula $\forall u\in z(u\in y)$. Quantifiers are bounded by an existing variable or parameter, not by an unrestricted class term such as the power set of a parameter unless that set has separately been supplied as a parameter.

A ${\Sigma }_1$ logical formula has the formup to the standard logical transformations defining the hierarchy. The initial existential quantifiers may be unbounded; all quantifiers inside the matrix are bounded. A finite tuple of witnesses can be encoded by one set if desired, and logical formulas with no initial unbounded quantifier are included as the level-zero special case.

The dual ${\Pi }_1$ class in set theory has an initial universal block with a bounded matrix. Negation interchanges these two classes, but does not in general preserve the existential class in set theory. Part (iii) supplies an explicit reason for the failure in this setting.

Use the strict-order convention appropriate to the membership relation. A bounded logical formula isThe first clause is irreflexivity, the second transitivity, and the third comparability of distinct elements. Irreflexivity and transitivity exclude $u\in v\in u$, so distinct elements cannot be ordered in both directions. These are exactly the axioms of a strict total order on the set $x$, with $\in$ as its order relation.

Every displayed quantifier is bounded by $x$, so $TO(x)$ is ${\Sigma }_0$. The logical formula does not assert that $x$ is transitive as a set, and no such additional assertion is needed to say that the restriction of membership to $x$ is a total order.

There are two ways for a set to fail to be the power set of $y$: it may contain an object which is not a subset of $y$, or omit a subset of $y$. A single existential logical formula expressing these alternatives isIts matrix is bounded, so this is ${\Sigma }_1$. Its negation is precisely the first-order power-set specification

We prove the absoluteness fact needed to exclude a logically equivalent ${\Sigma }_1$ logical formula. If $M\subseteq N$ are transitive sets and all parameters lie in $M$, bounded logical formulas have the same truth value in $⟨M,\in ⟩$ and $⟨N,\in ⟩$. Atomic equality and membership agree because both structures use actual membership. Boolean operations preserve agreement. In a bounded quantifier over $v\in M$, transitivity ensures that every element of $v$ already lies in $M$, so the two structures quantify over exactly the same elements of $v$. Induction on logical formulas proves the claim. It follows that ${\Sigma }_1$ logical formulas are upward absolute: witnesses in $M$ remain available in $N$, and their bounded matrix retains its truth value.

Now use a finite counterexample, so no existence of a transitive model of ZF is assumed. Let $M=4=\{0,1,2,3\}$, the Finite von Neumann ordinal, and let $N=V_4$. These are transitive sets with $M\subseteq N$. Set $y=2$ and $x=3$, both elements of $M$. Inside $M$, the subsets of $2$ which are available as elements of the domain are exactly $0,1,2$; the remaining domain element $3$ is not a subset of $2$. Thus $M$ satisfies the displayed power-set specification with $x=3$.

Inside $N$, however, the set $\{1\}$ is also present and is a subset of $2$, while $\{1\}\notin 3$. Thus the same specification is false in $N$. If $\neg \varphi$ were logically equivalent to a ${\Sigma }_1$ logical formula, its truth in $M$ would transfer upward to $N$, a contradiction. ThereforeThis proves the requested unrestricted logical non-equivalence. The finite structures are not being claimed to satisfy all of ZF; logical equivalence must hold in them as well as in any models of a stronger theory.

König theorem for cardinal numbers states that if ${\kappa }_i<{\lambda }_i$ for each index $i\in I$, thenRepresent each cardinal by its initial ordinal, let $A=\{(i,\alpha ):i\in I,\alpha <{\kappa }_i\}$, and let $B={\prod }_{i\in I}{\lambda }_i$.

First there is an injection $A\to B$: encode $(i,\alpha )$ by the function which is zero away from $i$ and takes value $\alpha +1$ at $i$. The nonzero coordinate identifies $i$, and its value identifies $\alpha$. It is valid because $\alpha +1\le {\kappa }_i<{\lambda }_i$; an index with ${\kappa }_i=0$ contributes no domain points.

For strictness suppose a surjection $F:A\to B$ existed. At coordinate $i$, the values $F(i,\alpha )(i)$ for $\alpha <{\kappa }_i$ occupy fewer than ${\lambda }_i$ possibilities. Choose $b_i$ as the least ordinal below ${\lambda }_i$ absent from that set. The resulting $b=(b_i)\in B$ differs from $F(i,\alpha )$ at coordinate $i$ for every $(i,\alpha )\in A$, contradicting surjectivity.

An injection $B\to A$ would produce a surjection $A\to B$ by inverting on its image and assigning the zero function elsewhere. It is therefore impossible. The injection in the first direction and impossibility of the reverse give the strict cardinal inequality. This diagonal argument uses the canonical well-orders of the specified cardinals to choose missing coordinate values, rather than an unexplained arbitrary choice at that step.

For an infinite cardinal regarded as an ordinal, its cofinality $\theta =\mathrm{cf}(\kappa )$ is the least cardinality of a cofinal subset of $\kappa$. Equivalently there is a family $({\alpha }_i{)}_{i<\theta }$ of ordinals below $\kappa$ whose supremum is $\kappa$, and no smaller indexed family has that property. Since an infinite initial ordinal is a limit ordinal, $\theta$ is infinite and in particular nonzero.

Use the diagonal mechanism of König theorem for cardinal numbers directly. Suppose $F:\kappa \to {}^{\theta }\kappa$ were surjective. For each $i<\theta$, the sethas cardinality at most $|{\alpha }_i|<\kappa$. Choose the least $b_i\in \kappa \setminus E_i$. The resulting function $b:\theta \to \kappa$ is not in the range: for any $\alpha <\kappa$, cofinality supplies $i$ with $\alpha <{\alpha }_i$, and then $b_i\ne F(\alpha )(i)$.

Thus no such surjection exists. Constant functions provide an injection $\kappa \to {}^{\theta }\kappa$, and an injection in the reverse direction would give a surjection by the same inverse-and-default construction as in part (i). ConsequentlyThis argument also handles $\kappa ={\aleph }_0$, when cofinal initial segments are finite; it does not incorrectly assume that smaller cardinals themselves must form a cofinal sequence in every infinite cardinal.

In the usual arithmetic under the axiom of choice of cardinal numbers, take the singular strong-limit cardinalCantor's theorem makes this sequence strictly increasing. It is countable and cofinal in its supremum, so $\mathrm{cf}(\kappa )\le {\aleph }_0$. No finite subset is cofinal in this limit ordinal, hence $\mathrm{cf}(\kappa )={\aleph }_0$. ThereforeThis construction needs no assumption such as the Continuum hypothesis. The conventional use of cardinals here includes well-orderability of these power sets, supplied by the axiom of choice. It is a separate setting from the explicit failure of well-orderability hypothesized in question 8; one must not impose choice on that later hypothesis.

For a set $M$, let $\mathrm{Def}(M)$ consist of the subsets of $M$ definable over the structure $⟨M,\in ⟩$ by a first-order formula with finitely many parameters from $M$. It is a set contained in $\mathcal{P}(M)$: satisfaction for a set structure and the set of logical formula codes give a uniform set construction of these definable subsets. If $M$ is transitive, every $a\in M$ is one such subset, defined over $M$ by $z\in a$, so $M\subseteq \mathrm{Def}(M)$.

Define the constructible hierarchy by transfinite recursion:Gödel's constructible universe is the definable class in set theoryAt successor stages only subsets definable over the previous stage are admitted, in contrast to the full power-set construction of the cumulative hierarchy. The stages are transitive and increasing, as permitted in the question, and their union contains every ordinal. Finite parameters are essential in the definition of $\mathrm{Def}$.

Fix $a\in L$. In the ambient universe form the setby power set and separation, using definability of $L$. This is the collection which must serve as the power set of $a$ inside $L$; it need not be the entire ambient power set.

For each $y\in S$, let ${\alpha }_y$ be the least ordinal with $y\in L_{{\alpha }_y}$. Ambient replacement is used to collect the ordinals ${\alpha }_y$ into a set. Their supremum, enlarged if necessary to include a stage containing $a$, gives an ordinal $\beta$ such that $a\in L_{\beta }$ and every $y\in S$ belongs to $L_{\beta }$.

Because $L_{\beta }$ is transitive and $a\in L_{\beta }$, the bounded property $y\subseteq a$ for $y\in L_{\beta }$ has its actual meaning when evaluated in that structure: every element of $y$ already lies in $L_{\beta }$, and membership is unchanged. It follows thatThis is a definable subset of $L_{\beta }$ using the parameter $a$, so $S\in \mathrm{Def}(L_{\beta })=L_{\beta +1}\subseteq L$. Every member of $L$ which is a subset of $a$ is in $S$, and every member of $S$ is such a subset. Thereforewhich is exactly the power-set axiom in $⟨L,\in ⟩$. The replacement step is indispensable to this proof: definability over one stage requires first bounding all the constructible subsets by one common stage.

Every constructible set is well-orderable: the constructible hierarchy has a canonical well-order, obtained by ordering the stage of first appearance, logical formula codes and the already ordered finite parameter lists. Its restriction to any set belonging to $L$ is a well-order in $L$ and remains a well-order in the ambient universe. This does not imply that arbitrary ambient sets are well-orderable.

If $a\in L$ is finite, all its subsets are finite combinations of its elements and hence constructible. Moreover the finite set of all those subsets is constructible by the finite set operations available in $L$. Thus the full ambient power set $\mathcal{P}(a)$ belongs to $L$.

Conversely, suppose $a\in L$ is infinite and that its full ambient power set belonged to $L$. A well-order of $a$ supplies an injection $j:\omega \to a$ by taking its first countably many elements. The mapis injective. But $\mathcal{P}(a)\in L$ would make that set well-orderable in the ambient universe, and pulling its well-order back along this injection would well-order $\mathcal{P}(\omega )$. This contradicts the hypothesis. HenceThe superscript emphasizes that the source's final claim concerns the ambient power set. The internal power set ${\mathcal{P}}^L(a)$ exists in $L$ for every $a\in L$ by part (ii), including infinite ones.