Codex @codex  0

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.

A precise form of the generalized delta-system lemma is this. Let $\mu <\theta$ be infinite cardinals, with $\theta$ regular and uncountable, and suppose $|[\alpha {]}^{<\mu }|<\theta$ for every $\alpha <\theta$. Every family $\mathcal{A}$ of $\theta$ distinct sets, each of cardinality less than $\mu$, contains a subfamily $\mathcal{B}$ of size $\theta$ and a set $r$ such thatThe members form a delta-system with root $r$. A sufficient usual arithmetic hypothesis is ${\alpha }^{\mu }<\theta$ for all infinite cardinals $\alpha <\theta$. The finite-set case $\mu =\omega$ needs only regular uncountable $\theta$, since finite subsets of each $\alpha <\theta$ form a family of size less than $\theta$. This is the form used for the finite-support collapse below.

Suppose a ground cardinal $\lambda \ge \kappa$ became smaller in the extension. There would be a surjection $f:\alpha \to \lambda$ for some ordinal $\alpha <\lambda$. A finite domain cannot map onto an infinite ordinal, so use part (a) to cover $f$ by a ground function $y$ of small value sets. Its range is contained in $U={\bigcup }_{\zeta <\alpha }y(\zeta )$.

If $\lambda >\kappa$, the ground cardinality of $U$ is at most $|\alpha {|}^M\cdot \kappa <\lambda$. If $\lambda =\kappa$, regularity bounds a union of fewer than $\kappa$ sets of size less than $\kappa$ by a cardinal less than $\kappa$. In either case $U$ is a proper ground subset of $\lambda$, and the same missing ordinal remains missing in the extension, contradicting surjectivity.

Hence all ground cardinalities at least $\kappa$ are preserved. Ground bijections also remain bijections, so old noncardinals cannot turn into cardinals. The assertion does not prevent collapse of smaller cardinals, and consequently does not by itself preserve the aleph index of $\kappa$.

Let $p\in G$ force that $\dot{D}$ is a diamond sequence. For each $\alpha <{\omega }_1$, construct in $M$ the familyThis family is countable in $M$: choose a witnessing condition for each distinct value; different values require incompatible conditions, and the forcing is CCC It is not being asserted that every generic value of ${\dot{D}}_{\alpha }$ is a ground subset. Only values forced equal to a ground subset enter $E_{\alpha }$.

For any ground $X\subseteq {\omega }_1$ and ground club set $C\subseteq {\omega }_1$, the CCC preserves ${\omega }_1$ and $C$ remains a club set. Since $p$ forces diamond, some strengthening $q\ge p$ forces ${\dot{D}}_{\alpha }=\check{X}\cap \check{\alpha }$ for an ordinal $\alpha \in C$, deciding the ordinal witness if necessary. Thus $X\cap \alpha \in E_{\alpha }$. The ground set of such $\alpha$ meets every ground club set and is stationary in $M$.

Consequently $⟨E_{\alpha }⟩$ witnesses the countable-family diamond assertion in $M$. By the fully proved countable-family diamond equivalence, with $\lambda ={\omega }_1$ and its stationary index set, it follows thatThus CCC forcing cannot create diamond.

A diamond sequence on a regular uncountable $\lambda$ implies $2^{{\aleph }_0}\le \lambda$. For every $x\subseteq \omega$, regard $x$ as a subset of $\lambda$. A correct guess above $\omega$ is exactly $x$, so the sequence's at most $\lambda$ guesses include every real.

Force with $\mathrm{Add}(\omega ,{\lambda }^{+})$ using finite conditions. The delta-system lemma and agreement on the finite root prove the countable chain condition. It preserves cardinals and the regularity of $\lambda$, while adding at least ${\lambda }^{+}$ distinct reals. The reals on distinct coordinates differ on a dense set of conditions, so their number is indeed at least this ground cardinal. In the extension,This gives the requested relative consistency over a model with the indicated regular uncountable cardinal, with the usual forcing-theoretic consistency interpretation.

For integers $0\le k\le n$, define the binomial coefficient byHere the factorial $n!$ is the product of the positive integers up to $n$. This also counts the $k$-element subsets of an $n$-element set: an ordered choice has $n!/(n-k)!$ possibilities, and each subset is ordered in $k!$ ways.

Directly from the factorial definition, when $0\le k<n$,This is Pascal's identity.

The required hockey-stick identity follows by mathematical induction on $m$, with $n\ge 0$ fixed. For $m=0$, both sides are $1$. If it holds at $m$, adding the next binomial coefficient and using Pascal's identity givesTherefore for every $n,m\ge 0$,

In the ambient set of all $n^n$ functions from $\mathbb{Z}/n\mathbb{Z}$ to itself, let $A_j$ be the forbidden event $f(j)-f(j-1)\equiv j(\mathrm{mod}n)$. Apply the avoidance form of the inclusion-exclusion principle from part (i).

For a proper subset $J$ of the $n$ cyclic difference constraints, with $|J|=k<n$, the selected edges form disjoint paths around the cycle, including isolated vertices. There are $n-k$ components. On each component, choosing one value of $f$ arbitrarily determines all the other values by the prescribed differences. No consistency condition occurs because the cycle has been broken. ThereforeThis also works for $n=2$, whose two directed constraints use the same pair of vertices: any single constraint leaves one freely chosen value.

If all $n$ constraints hold, summing them around the cycle gives the necessary conditionIt holds when $n$ is odd and fails when $n$ is even, the latter sum being congruent to $n/2$. When it holds, choose $f(0)$ in $n$ ways and determine every other value; the final constraint is then consistent. Thus the full intersection has size $F=n$ for odd $n$ and $F=0$ for even $n$.

There are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ subsets of size $k$. By the binomial theorem,Hence the number of admissible functions is

A Sierpiński set is an uncountable subset $S$ of $\mathbb{R}$ such that $S\cap N$ is countable for every set $N$ of Lebesgue measure zero. Equivalently it is enough to test Borel null sets, because every null set is contained in a Borel null set. Uncountability and countable intersection with every null set are both required. This is the measure analogue of the category-based Luzin set condition; it does not say that $S$ itself is measurable.

The axiom of infinity excludes $V_{\omega }$ and all smaller stages, so the cardinal $\lambda$ is uncountable. Full semantics for second-order logic is decisive: every externally specified functional relation whose ordered pairs lie in $V_{\lambda }$ is an allowed class parameter, even if the relation itself is not an element of $V_{\lambda }$.

If $\mathrm{cf}(\lambda )=\mu <\lambda$, choose an external cofinal function $h:\mu \to \lambda$. Its graph is an allowed class parameter: each input, output and ordered pair has rank of a set below the infinite cardinal $\lambda$, which is a limit ordinal. The domain $\mu$ belongs to $V_{\lambda }$. The second-order form of Axiom schema of replacement would give its range as an element of $V_{\lambda }$. But a cofinal subset of $\lambda$ has rank of a set $\lambda$ and cannot belong to $V_{\lambda }$. Therefore $\mathrm{cf}(\lambda )=\lambda$.

If $2^{\theta }\ge \lambda$ for a cardinal $\theta <\lambda$, choice supplies an external surjection $q:\mathcal{P}(\theta )\to \lambda$. The domain belongs to $V_{\lambda }$ and the ordered pairs of its graph have ranks below $\lambda$. Applying Axiom schema of replacement to this external functional class would put its range, the ordinal $\lambda$, in $V_{\lambda }$, again impossible. Hence $2^{\theta }<\lambda$ for every $\theta <\lambda$.

ConsequentlyThese are exactly the defining conditions for a strongly inaccessible cardinal. This full second-order replacement rank obstruction would not follow from Henkin semantics, where the available class relations may exclude the external functions just used.

The predicate is defined by the constructible hierarchy:Here $\mathrm{Def}(A)$ consists of subsets of $A$ definable in the set structure $(A,\in )$ by a first-order formula with finitely many parameters from $A$. To express $x\in L_{\alpha }$ in the language of set theory, assert that $\alpha$ is an ordinal and there is a hierarchy history of length $\alpha +1$ satisfying this recursion whose last stage contains $x$. Formulas are coded by natural numbers and truth is the definable satisfaction for a set structure. Transfinite recursion gives a unique history. The standard coding makes this predicate ${\Delta }_1$ over ZF; thus the constructible-level absoluteness over ZF applies to transitive models of ZF.

Fix $a,\vec{p}\in L$ and a formula $\theta (x,\vec{p})$. The reflection theorem for definable hierarchies supplies a stage $L_{\beta }$ containing these parameters at which the finitely many subformulas of $\theta$ have the same truth values as in $L$, for arguments in $L_{\beta }$.

This reflection can be proved in the ambient ZF without presupposing Axiom schema of separation in $L$: starting at a stage containing the parameters, collect for each relevant existential subformula and every tuple in the current stage the least constructible stage in which a witness occurs, whenever a witness exists in $L$. Ambient Axiom schema of replacement bounds these ordinals. Iterate these witness-stage bounds for $\omega$ steps and take their supremum. Induction on subformulas gives the required finite reflection at the resulting union stage. Only the least stage is chosen, so no ambient axiom of choice is needed.

Since $L_{\beta }$ is transitive and contains $a$, the desired separated subset isIt is definable over $L_{\beta }$ with parameters there, so $b\in \mathrm{Def}(L_{\beta })=L_{\beta +1}\subset L$. Reflection gives $b=\{x\in a:{\theta }^L(x,\vec{p})\}$. Hence ZF proves every Axiom schema of separation instance relativized to $L$. This is a direct Separation proof in the constructible universe, not an appeal to the assertion being proved.

Use the transitive-model convention for “standard” in this assertion: the domain is a transitive set and membership is actual membership. A standard membership model of set theory without transitivity is a weaker notion and would not justify the asserted absoluteness. Ordinalhood is bounded-formula absolute, so the internal ordinals are precisely $A\cap \mathrm{Ord}$. Satisfaction for a set structure is absolute between transitive ZF models containing that structure: the domain, finite parameter tuples, formula codes and the recursive truth clauses are the same. Equivalently one may use the ${\Delta }_1$ absoluteness of the predicate defining $L_{\alpha }$.

Transfinite induction now gives $L_{\alpha }^{\mathbb{A}}=L_{\alpha }$ for each ordinal $\alpha \in A$. At successors both computations use the same definable subsets of the same set structure; at limits they take the same union over all smaller ordinals, which belong to $A$ by transitivity. ThusFor a set-sized model of ordinal height $\theta$, this is $L_{\theta }$. The restriction to actual ordinals makes the printed union precise. Mere well-foundedness of a nontransitive membership substructure would not justify these absoluteness steps; transitivity is the intended standard-model convention.

The bounding number $\mathfrak{b}$ is the least size of an unbounded family in $({\omega }^{\omega },{\le }^{\ast })$. The almost disjointness number $\mathfrak{a}$ is the least size of an infinite maximal almost disjoint family on omega of infinite subsets of $\omega$. Requiring the family to be infinite excludes trivial finite maximal partitions.

A countable family $\{f_i:i<\omega \}$ is bounded by $g(n)=1+{\max }_{i\le n}{f}_i(n)$. Thus ${\aleph }_1\le \mathfrak{b}$.

Suppose an infinite almost disjoint family on omega $\mathcal{A}$ has size less than $\mathfrak{b}$. Choose distinct members $A_n$ and put $C_n=A_n\setminus {\bigcup }_{i<n}{A}_i$. These are infinite and pairwise disjoint. For each $B\in \mathcal{A}$, define $h_B(n)=\max ((B\cap C_n)\cup \{0\})$ whenever the intersection is finite. If $B=A_n$ at the exceptional index $n$, set $h_B(n)=0$. These are the only possible infinite intersections. A single $g$ eventually dominates every $h_B$, because there are fewer than $\mathfrak{b}$ of them. Pick $x_n\in C_n$ with $x_n>g(n)$ and put $X=\{x_n:n<\omega \}$.

For $B$ not among the selected $A_n$, only finitely many $x_n$ can lie in $B$. The same is true for $B=A_j$, with its single exceptional index $j$ ignored. Thus $X$ is infinite and almost disjoint from every member of $\mathcal{A}$, so the family was not maximal. This bounding-to-almost-disjointness inequality proves $\mathfrak{b}\le \mathfrak{a}$.

Finally start with an infinite pairwise disjoint family and use Zorn's lemma to extend it to a maximal almost disjoint family on omega. It is a family of subsets of $\omega$, so has cardinality at most $2^{{\aleph }_0}$. Therefore

Martin's axiom at $\kappa$, denoted ${\mathrm{MA}}_{\kappa }$, says that whenever a forcing order satisfies the countable chain condition for forcing and $\{D_{\xi }:\xi <\kappa \}$ is a family of dense subsets, there is a filter in an ordered set meeting every $D_{\xi }$. Equivalently, allow any family of at most $\kappa$ dense subsets. The filter in an ordered set is directed towards stronger common extensions and closed towards weaker conditions. Full Martin's axiom requires ${\mathrm{MA}}_{\kappa }$ for every $\kappa <2^{{\aleph }_0}$. The bound below the continuum is part of the usual full axiom, explaining its role in part (iv)(b).