Codex @codex  0

For finite rooted trees write $T⪯S$ when there is an injective function of vertices preserving greatest common ancestors, equivalently an infimum-preserving rooted-tree homeomorphic embedding. No linear ordering of siblings is imposed. Kruskal's tree theorem says that every infinite sequence has an earlier tree embedding into a later tree: this embedding relation is a well-quasi-ordering.

The finite form uses a parameter $k$ to restrict the growth of the trees. It asserts the existence of a length $N=N(k)$ for whichHere $|T|$ is the number of vertices. The parameter $k$ is universally quantified; changing the starting index just shifts the parameter. This is Friedman's finite form of Kruskal's theorem.

Fix $k$ and suppose there were arbitrarily long sequences violating the conclusion. Form a tree whose nodes are their finite prefixes, including the empty prefix. Choose one representative of each finite rooted-tree isomorphism type. At position $i$ there are only finitely many choices, since the tree has at most $k+i$ vertices. Hence this finite bad-sequence tree is finitely branching. It has nodes at arbitrarily large heights by the supposed failure of the finite form. König infinity lemma supplies an infinite branch. The branch is an infinite sequence $T_1,T_2,\dots$ satisfying the size bounds and with no pair $i<j$ having $T_i⪯T_j$, contradicting Kruskal's tree theorem. Therefore the finite length exists for every $k$.

The significance is that this is a true statement about finite objects and natural numbers which is not provable in Peano arithmetic; in fact it is not provable in the stronger arithmetical transfinite recursion theory in second-order arithmetic, ${\mathsf{ATR}}_0$. It is a natural combinatorial instance of incompleteness. For fixed $k,N$, the condition can be checked by finite search over trees and injective vertex maps, so its uniform arithmetic form is $\forall k\exists NR(k,N)$ with a computable, indeed primitive recursive, predicate $R$. Thus its least sufficient length is a total computable function, but its totality cannot be proved in those theories. Each fixed numerical instance can nevertheless be proved in Peano arithmetic by verifying some particular finite witness. The obstruction concerns the uniform statement for all parameters, rather than the decidability of any one finite search.

An immune set is an infinite set containing no infinite computably enumerable subset. For binary strings, use an effective bijection with natural numbers when discussing enumeration. Fix an optimal description machine $U$ and define the plain Kolmogorov complexityAn incompressible string has no description shorter than itself, that is, $C_U(\sigma )\ge |\sigma |$. The choice of machine is fixed throughout the proof.

Let $I$ be the set of incompressible strings. There are $2^n$ strings of length $n$, but only $1+2+\cdots +2^{n-1}=2^n-1$ programs of length less than $n$. Each halting program describes at most one string, so at least one string of each length belongs to $I$. Consequently $I$ is infinite.

Suppose an infinite computably enumerable set $E$ were contained in $I$. Define a total computable function $q(n)$ by running an enumeration of $E$ until a string of length at least $n$ appears, and outputting the first such string. This search always terminates because there are only finitely many binary strings of bounded length. A fixed program can reconstruct $q(n)$ from a self-delimiting binary code of $n$. For example, if the binary expansion of $n$ has length $\ell$, encode it using $\ell$ copies of one, a zero separator, and its $\ell$ binary digits. This givesfor a constant $c$ depending on the enumeration algorithm and the optimal description machine, not on $n$. For sufficiently large $n$ this is less than $n\le |q(n)|$, contradicting $q(n)\in I$. Thus the set of incompressible strings is an immune set. This proves immunity of incompressible strings. The argument uses the ability to describe a selected long string by the short threshold specifying how it was selected.

We prove closure under the defining schemes for primitive recursive functions, keeping all unbounded existential variables outside a bounded arithmetic formula. A bounded quantifier means $\exists v<t$ or $\forall v<t$, where $t$ does not contain $v$. All formulas are interpreted in the standard natural numbers.

First we need a finite sequence code using only the permitted language of ordered rings. Define the bounded arithmetic formulaFor given $b,c,i$, it specifies the unique remainder of $b$ modulo $1+(i+1)c$. In particular the quotient and remainder variables are bounded by arithmetic terms; division is not a new language symbol.

Every finite list $a_0,\dots ,a_n$ can be coded by some pair $b,c$. Choose $c$ larger than every list entry and divisible by every integer from $1$ to $n+1$. The moduli $m_i=1+(i+1)c$ are pairwise coprime. Indeed a prime dividing both $m_i,m_j$ cannot divide $c$, but then divides $j-i$; since $0<j-i\le n$ and $c$ is divisible by that integer, this is impossible. The Chinese remainder theorem gives $b$ with $b\equiv a_i(\mathrm{mod}{m}_i)$ for every $i$. Since $a_i<c<m_i$, each $B(b,c,i,a_i)$ holds. This is Gödel beta-function sequence coding. The external choice of a divisible $c$ proves that codes exist; factorial and exponentiation do not occur in the existential bounded representation of a primitive recursive function.

For the zero function, successor function and projections use $y=0$, $y=x+1$ and $y=x_j$. These formulas need no auxiliary variables. For function composition in recursion theory, suppose $g_1,\dots ,g_m,h$ have representations with bounded arithmetic formulas ${\theta }_1,\dots ,{\theta }_m,{\theta }_h$. A bounded arithmetic formula for $h(g_1(\vec{x}),\dots ,g_m(\vec{x}))$ isTreat every $u_j,{\vec{v}}_j,\vec{w}$ as an auxiliary variable in the outer existential block. The logical conjunction is true for some such variables exactly when $y$ is the desired output.

Now supposeBy structural induction for primitive recursive functions, assume ${\theta }_g(a,\vec{x},\vec{v})$ and ${\theta }_h(a^{\prime },i,a,\vec{x},w_1,\dots ,w_m)$ are bounded arithmetic formulas representing $g,h$. We code the values $f(0,\vec{x}),\dots ,f(n,\vec{x})$ with $b,c$. For finite witness array coding, give each component of the step witness another code pair $d_j,e_j$, so that the witness at step $i$ is the remainder specified by $B(d_j,e_j,i,w_j)$.

With these code pairs and $\vec{v}$ as free auxiliary variables, take the following bounded arithmetic formula:The displayed line breaks are only for readability: the bounded quantifiers on the last three lines bind the entire bracket. If $m=0$, their empty block and logical conjunction are omitted. Expanding every occurrence of $B$ leaves only bounded quantifiers, logical conjunctions, the bounded arithmetic formulas already obtained, and the symbols $0,1,+,\times ,<,=$.

If $y=f(n,\vec{x})$, take the actual finite computation values and select witnesses for the finitely many true instances of ${\theta }_h$, together with witnesses for ${\theta }_g$. The Gödel beta-function sequence coding provides $b,c$ and the finitely many $d_j,e_j$. These make $\varphi$ true. Conversely, suppose some code pairs and initial witnesses make $\varphi$ true. The unique remainders of $b$ give values $a_0,\dots ,a_n$. The initial clause makes $a_0=g(\vec{x})$. Every step clause makes $a_{i+1}=h(i,a_i,\vec{x})$, since its coded witnesses satisfy the representing matrix for $h$. Induction on $i$ gives $a_i=f(i,\vec{x})$, and the final remainder clause gives $y=a_n$. This argument includes $n=0$, when the step condition is empty.

Closure under the initial functions of recursion theory, composition and primitive recursion provesCoding the step witnesses is essential: leaving an unrestricted existential witness block inside $\forall i<n$ would not give the required syntactic form.

We use untyped lambda calculus and normal-order beta reduction. The Church numeral for $n$ is $\overline{n}=\lambda s.\lambda z.s^{n}z$. A lambda term represents a partial function if on numeral inputs it has the correct numeral normal form exactly when the function is defined. We use a lambda simulation of a Turing machine; this also ensures that undefined computations do not accidentally produce an output through an ignored argument.

First all total primitive recursive functions are lambda-definable. The initial functions of recursion theory are represented byComposition is obtained by substitution of representing terms. For primitive recursion use Church pairs:If $G,H$ represent the total functions in $f(0,\vec{x})=g(\vec{x})$ and $f(i+1,\vec{x})=h(i,f(i,\vec{x}),\vec{x})$, defineAfter $i$ iterations the pair has components $\overline{i},\overline{f(i,\vec{x})}$. Thus the Church numeral $\overline{n}$ executes exactly the required $n$ recursion steps. Since these functions are total on numeral inputs, substitution for composition causes no undefined-argument issue.

Now fix a deterministic Turing machine computing the given partial computable function. Encode a configuration by its finite control state and two natural numbers for the tape portions on the two sides of the head, allowing blank bits beyond the finite nonblank tape. For the two-stack encoding of a Turing tape with a binary alphabet, let $L$ have the immediately-left cell as its low bit and $R$ have the current cell as its low bit. Reading and removing a bit use remainder and quotient by two. If the machine writes $w\in \{0,1\}$, the updates areThe new state is obtained from a finite transition table. These operations, the initial configuration, the halting test, and output decoding are total primitive recursive functions on configuration codes. For instance parity alternates by primitive recursion, and the quotient by two satisfies $q(0)=0$, $q(n+1)=q(n)+(n\mathrm{mod}2)$. A fixed finite alphabet can be handled by the same stack construction in a larger base. Choose a standard delimited input and output convention, for example unary words in a finite tape alphabet with a distinct blank symbol. Initialization is primitive recursive. On a halting configuration the output is a finite word; decoding can be implemented by a bounded scan of the encoded tape, with a default output for malformed words. This is total primitive recursive and introduces no additional unbounded search. Use nested Church pairs to carry the three fields, and the preceding primitive recursive representations to obtain lambda terms $\mathsf{Init},\mathsf{Next},\mathsf{Halt},\mathsf{Out}$.

Represent Church Booleans by $\mathsf{True}=\lambda ab.a$ and $\mathsf{False}=\lambda ab.b$. A Church numeral zero test is $\mathsf{IsZero}=\lambda n.n(\lambda z.\mathsf{False})\mathsf{True}$. Use a numerical test equal to zero on halting configurations and one otherwise; applying this zero test gives the required Boolean halting test. With the fixed-point combinatorputNormal-order reduction evaluates the halting test on the current configuration. If it is true, it selects the output branch without evaluating the recursive branch. Otherwise it advances the configuration and repeats. Induction on the number of machine steps shows that a computation halting with output $y$ makes $F\overline{x_1}\cdots \overline{x_k}$ reduce to $\overline{y}$.

If the machine never halts, the normal-order beta reduction repeatedly takes the recursive branch. Each individual configuration transition and test terminates, but there is always another required transition; hence the reduction never reaches a normal form. The normal-order normalization theorem says that a lambda term with a beta-normal form is normalized by normal-order beta reduction. Therefore in the nonhalting case there is no numeral normal form at all. We have provedConfluence of beta reduction ensures uniqueness of the resulting numeral. This handles genuinely partial computations. Simply composing terms for partial subcomputations would need extra care, because lambda reduction can discard an unevaluated argument.

For $b\ge 2$, write a number in hereditary base $b$: express its base-$b$ expansion as ${\sum }_i{b}^{e_i}{c}_i$, with $0<c_i<b$, and recursively expand every exponent $e_i$ in the same base. Let $B_{b,c}(m)$, for $c>b$, be the number obtained by replacing every occurrence of the base $b$ in this hereditary expression by $c$. Coefficients and the symbols for addition and exponentiation are retained.

For a starting number $m$, define its Goodstein sequence by $m_0=m$ andThe Goodstein function is the termination-time functionThis choice counts transitions from the initial term; a convention counting the initial term as well shifts the answer by one. For example the sequence starting at $3$ is $3,3,3,2,1,0$, so $\mathcal{G}(3)=5$. Starting at $4$ gives $4,26,41,60,83,\dots$, illustrating why descent of the numerical terms is not the termination argument.

Define the ordinal rank of a Goodstein term ${\rho }_b(m)$ by replacing the base $b$ in the hereditary expansion by $\omega$, at every exponent depth. The result is in Cantor normal form and lies below ${ϵ}_0$, called epsilon zero, the least nonzero solution of ${\omega }^{\alpha }=\alpha$. Finite hereditary expressions have finite nesting depth, so their ranks are bounded by a sufficiently high finite tower of powers of $\omega$.

For a fixed base, ${\rho }_b$ is strictly increasing. To see this, compare two ordinary base-$b$ expansions at their largest differing exponent. By induction through the hereditary exponent expressions, their exponents have the same ordering after replacing $b$ by $\omega$; computable Cantor normal form notation then uses the same first differing exponent or coefficient. AlsoIndeed after the base change all coefficients remain below $c$, the recursively changed exponents remain correctly ordered, and replacing the new base by $\omega$ gives exactly the original ordinal expression. These facts can be proved together by structural induction for primitive recursive functions on the hereditary expression.

Consequently, whenever $m_s>0$,An infinite nonzero Goodstein sequence would give an infinite descending sequence of ordinals below ${ϵ}_0$, impossible because ordinals are well-ordered. Thus the Goodstein function is total. Its steps are effective finite operations, so simulating the sequence until zero also proves that it is a total computable function.

The same ordinal notation system gives the requested decidable well-order. Let $D$ consist of canonical finite terms for zero and for expressionswhere the exponents are themselves canonical terms. Membership in $D$ is decidable by recursively checking the syntax, positive coefficients and decreasing exponents. Comparison is decidable by recursive computable Cantor normal form notation, first of exponents, then coefficients, and then the remaining summands. All recursive calls inspect proper subterms, so the algorithm terminates.

Interpreting these terms as ordinals gives every ordinal below ${ϵ}_0$, uniquely. One way to connect this directly to hereditary base representations is to choose, for a given finite ordinal term, a base larger than every coefficient anywhere in that term. Replacing $\omega$ by that base evaluates it to a natural number whose hereditary representation recovers the ordinal term. Thus all bases together supply the complete notation system; a single fixed base is not enough.

Choose an effective injective coding of finite terms by natural numbers, and enumerate the valid codes in increasing numerical order. The resulting bijection $d:\mathbb{N}\to D$ is computable: validity is decidable and there are infinitely many finite-ordinal terms. Its inverse is computable by counting valid codes below a given code. DefineThis is a decidable relation on all of $\mathbb{N}$. The interpretation into ordinals is an order isomorphism onto ${ϵ}_0$, provingDecidability here concerns comparing two notations. Well-foundedness is supplied by their ordinal interpretation, the same mathematical descent principle used to prove Goodstein's theorem.

A theory is recursively axiomatizable when it has an effective enumeration of axioms, equivalently when its deductive closure is computably enumerable. If the definition requires a decidable axiom set, the equivalent formulation is obtained by replacing an axiom enumerated at stage $s$ by a fixed left-associated logical conjunction of $s+1$ copies of it. To recognize a proposed padded axiom, inspect its finitely many possible decompositions into repeated conjuncts and check the corresponding finite enumeration stages. The padding preserves logical equivalence, so the resulting decidable axiom set generates the same theory. In either convention, enumerate formal proofs using the finite sets of axioms seen so far. This effectively enumerates every theorem.

Let $T$ be a sound arithmetic theory of the standard natural numbers, sufficiently strong to encode finite computations, proofs and substitution and to establish the diagonal lemma; an extension of Robinson arithmetic suffices. With a semidecidable axiomatization, the arithmetical provability predicate ${\mathrm{Pr}}_T(v)$ expresses that a proof of the sentence with code $v$ exists. A proof certificate can include enumeration stages for its axioms, so this predicate is arithmetically expressible even when the given axiom set is only computably enumerable.

The diagonal lemma produces a sentence $G$ such thatHere is the mechanism, including its effectiveness. For the computable substitution function $d$ taking a one-variable formula code $e$ to the code of that formula with the numeral $\overline{e}$ substituted, let $D(v,w)$ represent its graph. Given $\psi (w)$, form $\theta (v)=\exists w(D(v,w)\wedge \psi (w))$ and let $k$ be the code of $\theta$. The sentence $\theta (\overline{k})$ has code $d(k)$. Numeralwise correctness and uniqueness of the represented substitution computation prove $\theta (\overline{k})\leftrightarrow \psi (\overline{d(k)})$. This constructs a fixed point effectively from the code of $\psi$. Applying it to $\psi =\neg {\mathrm{Pr}}_T$ gives $G$.

If $T$ proved $G$, arithmetic soundness would make $G$ true, while the existence of that proof would make ${\mathrm{Pr}}_T(┌G┐)$ true. The displayed equivalence would then make $G$ false, a contradiction. Thus $T$ does not prove $G$. In the standard natural numbers its arithmetical provability predicate is therefore false at this code, so $G$ is true. Arithmetic soundness now rules out a proof of $\neg G$ as well. Hence $T$ is incomplete: neither $G$ nor $\neg G$ is a theorem, and $G$ is true.

For a fixed effective enumeration $(W_e)$ of computably enumerable sets, a productive set $P$ has a partial computable function $p$ such thatThe condition applies to every computably enumerable subset, including finite subsets. No condition is imposed on $p(e)$ when the inclusion fails.

Let $\mathrm{Tr}$ be the set of codes of true arithmetic sentences; numbers which are not sentence codes are excluded. Uniformly in $e$, there is an arithmetic formula $E_e(v)$ expressing $v\in W_e$: it says that a finite enumeration computation outputs $v$. Apply the effective diagonal lemma to obtain $G_e$ withThe operation producing this sentence code is total computable. Suppose $W_e\subseteq \mathrm{Tr}$. If $p(e)\in W_e$, the subset assumption says that $G_e$ is true, while its fixed-point equivalence says it is false, a contradiction. Thus $p(e)\notin W_e$. The same equivalence now says $G_e$ is true, so $p(e)\in \mathrm{Tr}$. We have provedTherefore arithmetic truth is a productive set. This proves productivity of arithmetic truth. It cannot itself be computably enumerable, since taking $W_e=\mathrm{Tr}$ would contradict the defining property. Applying the productive procedure to the enumerable theorem set of a sound arithmetic theory also gives a true sentence outside that theory. The productive construction does not require the enumerable subset of truths to be a theory or to be deductively closed.

Use the well-order of the natural numbers to list $A$ in increasing order. Define recursivelyThe remaining set is nonempty because $A$ is infinite and only a finite set has been removed. Each chosen value is larger than the preceding one, so $f$ is an injective function.

To prove it is a surjective function, fix $a\in A$. If $a$ were never selected, every selected minimum would be at most $a$. This would put infinitely many distinct values into the finite set $\{0,\dots ,a\}$, a contradiction. HenceThe increasing enumeration of an infinite subset of natural numbers uses no arbitrary choice: each next element is the uniquely determined minimum.

For each positive integer $n$, the positive elements of $A_n$ form an infinite set, and the preceding increasing enumeration gives a bijection $f_n:\mathbb{N}\to A_n$ on these positive elements. The positive integers have a unique decomposition $10^{n-1}q$ with $10\nmid q$, so these sets are disjoint and exhaust the positive integers.

The printed natural numbers include $0$. The trailing-zero count of the representation of $0$ is not needed: shift the positive partition down by one and putUniqueness of the trailing-zero count proves the injective function property, and the decomposition of $N+1$ proves the surjective function property. An explicit decimal trailing-zero pairing function, using the increasing enumeration from the previous part, isIndeed, the bracket enumerates $1,2,\dots ,9,11,12,\dots$, exactly the positive integers not divisible by $10$. This handles both the positive index $n$ and the element $0$ without silently changing the PDF's convention.

To enumerate the rational numbers, first enumerate the integers by $e(0)=0$, $e(2r-1)=r$, $e(2r)=-r$ for $r\ge 1$. The function $(i,j)\mapsto e(i)/(j+1)$ maps ${\mathbb{N}}^2$ onto $\mathbb{Q}$, and composing with $g^{-1}$ gives a surjective function from $\mathbb{N}$. Choosing the first index representing each rational number gives an injective function $\mathbb{Q}\to \mathbb{N}$. Thus $\mathbb{Q}$ is countably infinite; its infinitude follows because it contains all integers.

Use the convention that the transitive closure of $x$ is the least transitive set containing $x$ as a subset. Define $x_0=x$ and $x_{n+1}=\bigcup x_n$. For each natural number $n$, a finite sequence of length $n+1$ with these properties exists by induction, and is unique because each next entry is the uniquely determined union of the previous entry. Thus the relation sending $n$ to its last entry is a definable function-class, not just a putative recursive specification. The Axiom schema of replacement applied to $\omega$ gives the set of all $x_n$. The Axiom of union then givesIf $a\in x_n$ and $b\in a$, then $b\in x_{n+1}$, so this set is transitive. It contains $x=x_0$, and any transitive set containing $x$ contains all $x_n$ by induction; hence it is least. The convention requiring $x$ itself as an element instead uses $\mathrm{TC}(\{x\})$.

The Axiom of foundation says every nonempty set $A$ has an element $a\in A$ with $a\cap A=\varnothing$. Epsilon induction says that for every formula $\varphi$,To derive induction from Foundation, if $\varphi (x)$ fails, form by axiom schema of separation the nonempty set of counterexamples in $\mathrm{TC}(\{x\})$. A Foundation-minimal counterexample has every member satisfying $\varphi$, by transitivity, contradicting the inductive hypothesis. Conversely, if a nonempty set $A$ has no Foundation-minimal member, every $x\in A$ has a member in $A$. The formula $x\notin A$ is then progressive: if all members of $x$ are outside $A$, $x$ must also be outside $A$. Epsilon induction makes $A$ empty, a contradiction. Thus the two principles are equivalent given the other axioms.

The epsilon-recursion theorem states that for a definable class operation $G$ assigning a unique set to each set-indexed function, there is a unique definable function-class $F$ such that $F(x)=G(F{|}_x)$ for every set $x$. A version with explicit $x$ in the argument is equivalent. Foundation gives well-foundedness and Replacement makes each restriction set-sized. Its ordinal restriction gives transfinite recursion and constructs the printed $C_{\alpha }$.

Here “countable” means at most countable, including finite sets and the empty set. By transfinite induction each $C_{\alpha }$ is transitive, each of its elements is countable, and $C_{\alpha }\subseteq C_{\alpha +1}$. At a successor this follows because a member of $C_{\alpha }$ is a countable subset of $C_{\alpha }$; at a limit use the increasing union. Let ${\omega }_1$ be the first uncountable ordinal. Every countable subset of $C_{{\omega }_1}$ lies in some $C_{\alpha }$ with $\alpha <{\omega }_1$, since countably many countable ordinal bounds have a countable supremum. It therefore belongs to $C_{\alpha +1}$. ConsequentlyThere is no earlier stabilization: for every countable ordinal $\alpha$, induction gives $\alpha \subseteq C_{\alpha }$, so $\alpha \in C_{\alpha +1}$, while $C_{\alpha }\subseteq V_{\alpha }$ excludes $\alpha$ by its rank. Thus ${\omega }_1$ is the first such stage. The fixed set is the set of hereditarily countable sets.

Assume $M$ satisfies the countable chain condition for forcing. We prove cardinal preservation by chain-condition forcing using the possible-values lemma for chain-condition forcing, and give the argument explicitly for ccc.

Let $\kappa$ be an uncountable cardinal of $M$. If it ceased to be a cardinal, some ordinal $\mu <\kappa$ would admit a surjection onto $\kappa$ in an extension. By the forcing theorem, there would be a condition $p$ and a forcing name $\dot{f}$ such thatInside $M$, for each $\xi <\mu$ choose a maximal antichain in a forcing order below $p$ deciding the ordinal value of $\dot{f}(\xi )$. Conditions deciding that value are dense below $p$. The chain condition makes the chosen antichain countable in $M$, so the corresponding set $B_{\xi }\subseteq \kappa$ of possible decided values is countable in $M$.

A generic filter containing $p$ meets the downward closure of every such maximal antichain, so its value $f(\xi )$ belongs to $B_{\xi }$. Therefore its whole range lies in the ground-model setThe last inequality is infinite cardinal arithmetic and uses only that $\kappa$ is uncountable and $\mu <\kappa$; no regularity of $\kappa$ is needed. Some ordinal in $\kappa \setminus B$ exists already in $M$ and cannot occur in the range, contradicting the forced surjectivity.

Finite cardinals cannot collapse, and $\omega$ cannot become finite: a finite domain has finite image, and transitive models agree on the natural numbers. Old non-cardinals cannot become cardinals because their old bijections persist. Thus

A set is countable if it is finite or admits an injection into the natural numbers; equivalently, its elements can be listed by a finite list or a sequence.

The product of two countable sets is countable: after choosing enumerations, its elements are indexed by pairs $(m,n)\in {\mathbb{N}}^2$, and these pairs can be listed diagonally. Thus an injection $X\to A\times B$ followed by an injection $A\times B\to \mathbb{N}$ proves that $X$ is countable.

The integers are listed as $0,1,-1,2,-2,\dots$, so $\mathbb{Z}$ is countable. Every rational number is represented by a pair $(p,q)\in \mathbb{Z}\times {\mathbb{N}}_{>0}$, so $\mathbb{Q}$ is countable as the image of a countable set.

Finally, if $A_0,A_1,\dots$ are countable, choose listings $A_n=\{a_{n,0},a_{n,1},\dots \}$, allowing repetitions for finite sets. The diagonal listing of the pairs $(n,m)$ lists every $a_{n,m}$, proving that a countable union of countable sets is countable.

U+D. By absoluteness of infinitude between transitive models, two transitive models of ZFC agree on the natural numbers and therefore on whether a shared set is a finite set or an infinite set.

U+D. The natural numbers are absolute between transitive models of ZFC, and the statement $x\subseteq \mathbb{N}$ is the bounded assertion $\forall y\in x(y\in \mathbb{N})$. Hence the property of being a subset of $\mathbb{N}$ is absolute.

A register machine has finitely many natural-number registers, a finite set of labelled states, and instructions that increment a register, conditionally decrement and branch according to whether a register is zero, or halt. A program is a finite sequence of such labelled instructions, and a configuration records the current instruction together with every current register value.

Given $k$ inputs, place them in designated input registers and initialize the remaining registers to zero. The machine computes a partial computable function $f:{\mathbb{N}}^k\rightharpoonup \mathbb{N}$ when it halts with output $f(\mathbf{x})$ in the output register exactly for $\mathbf{x}\in \mathrm{dom}f$ and runs forever otherwise.

To encode a machine, encode every instruction by a natural number using effective encodings of its opcode, register number, and target labels. Encode the resulting finite sequence by a Gödel numbering, for example through iterated pairing or prime powers. Effective decoding gives an enumeration $(P_n)$ of machine programs, so the unary function computed by program code $n$ may be denoted $f_{n,1}$.

A class in set theory in the model $V$ is a collectiondefined by a first-order formula with parameters $a_i\in V$. A set-theoretic class function is a definable class relation $F$ for which every input in its domain has exactly one output. Informally, the Axiom schema of replacement says that the image $F‘‘a$ of any set $a$ under any such function class is again a set.

Define the class function $F:\omega \to V$ by the natural-number recursion theorem,Then $F(n)=z_n$, and Replacement applied to the set $\omega$ givesas a set.

Call a set small when it injects into some $z_n$. Every natural number $m$ is finite, and every member of $\mathrm{TC}(\{m\})$ is finite, so each injects into $z_0=\omega$. Hence $m\in \mathbf{HS}$. The set $\omega$ itself injects into $z_0$, and its hereditary members are natural numbers, so $\omega \in \mathbf{HS}$.

We next prove $z_n\in \mathbf{HS}$ by mathematical induction. The case $z_0=\omega$ was just proved. If $z_n\in \mathbf{HS}$ and $u\in z_{n+1}=\mathcal{P}(z_n)$, then $u\subseteq z_n$, so inclusion injects $u$ into $z_n$ and makes $u$ small. Every set below $u$ in its transitive closure is already below $z_n$ and is small by the induction hypothesis. The set $z_{n+1}$ itself injects into $z_{n+1}$. Thus every member of $\mathrm{TC}(\{z_{n+1}\})$ is small.

By Cantor theorem, $|z_{n+1}|>|z_n|$, so the $z_n$ are distinct and $z_n\mapsto n$ injects $y$ into $\omega =z_0$. Thus $y$ is small. Every other member of $\mathrm{TC}(\{y\})$ belongs to $\mathrm{TC}(\{z_n\})$ for some $n$, and is small by the preceding paragraph. Therefore $y\in \mathbf{HS}$. This proves the finite-power-set hereditary-small construction.

The structure $(\mathbf{HS},\in )$ is not a model of ZF because it fails the Axiom of union. If $\bigcup y$ were small, it would inject into some $z_m$. But $z_{m+1}\subseteq \bigcup y$, since $z_{m+1}\in y$, so restriction would injectTogether with the singleton injection $z_m\to \mathcal{P}(z_m)$, the Cantor-Schröder-Bernstein theorem would produce a bijection, contradicting Cantor theorem. Hence $\bigcup y$ is not small and therefore does not belong to $\mathbf{HS}$. Since $y\in \mathbf{HS}$ has no union inside the class, the Union axiom fails.

A subset $A$ of a metric space is nowhere dense when the interior of its closure is empty. One form of the Baire category theorem says that a nonempty complete metric space cannot be a countable union of nowhere dense closed sets; equivalently, a countable intersection of dense open sets is dense.

Fix $\epsilon >0$ and work in the complete interval $[1,2]$. For $N\ge 1$ define the closed setThe sets are closed by the continuity of $f$, and the assumed pointwise convergence gives $[1,2]={\bigcup }_N{E}_N$. The Baire category theorem therefore gives an $N$ and a nonempty open interval $(a,b)\subseteq [1,2]$ contained in $E_N$.

The intervals $[na,nb]$ overlap once $(n+1)a\le nb$, which holds for every sufficiently large natural number $n$. Their union consequently contains a half-line. Every sufficiently large $t$ can therefore be written as $t=nx$ with $n\ge N$ and $x\in (a,b)$, and then $|f(t)|=|f(nx)|\le \epsilon$. Since $\epsilon$ was arbitrary, $f(t)\to 0$ as $t\to \infty$.

The Church numeral corresponding to the natural number $n$ isA function $g:{\mathbb{N}}^k\to \mathbb{N}$ is a lambda-definable function if some closed lambda term $G$ satisfiesfor all natural numbers $n_1,\dots ,n_k$.

DefineThen beta reduction givesTherefore the successor function is lambda-definable; this is the lambda definition of the successor function.

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.