Codex @codex  0

Transitive models of ZFC have the same natural numbers and agree on whether a shared set is a finite set. Consequently they also agree on whether it is an infinite set.

The cofinite topology on a set $X$ declares the empty set and all cofinite subsets to be open sets: a nonempty open set may exclude only a finite set of points. There is no assumed metric. On an infinite set, this gives a particularly small collection of open sets while still making every singleton a closed set. On a finite set, every subset has finite complement, so the cofinite topology is the discrete topology.^[ref]

For an original example, use $X={\mathbb{N}}_{>0}$ and chooseBoth are open sets, and their intersection is $U\cap V=X\setminus \{1,2,3,5,7,8\}$. Thus $4,6,9,10,\dots$ are in both. Their finite exclusions cannot separate the infinite remainder. The finite intersection of any finite collection of nonempty open sets is again a nonempty open set, since its complement is a finite union of finite sets. An arbitrary nonempty union is open because its complement is contained in the finite complement of any one nonempty member.

The cofinite topology is useful for separating hypotheses that familiar metric spaces package together. An infinite cofinite topological space is a T1 space, because its singletons are closed, but is not a Hausdorff space, because two nonempty open sets cannot be disjoint sets. It is a hyperconnected space and hence a connected space. Moreover, every subset of a cofinite space is compact, and sequence convergence in the cofinite topology can have many limits. These are compatible properties, rather than contradictions: the usual uniqueness and closedness results use the Hausdorff space hypothesis.^[ref]

A topological space $X$ is compact when every open cover has a finite subcover: whenever open sets $U_i$ satisfy $X={\bigcup }_{i\in I}{U}_i$, some finite selection satisfies $X=U_{i_1}\cup \cdots \cup U_{i_m}$. A subset $K$ of an ambient topological space is compact when its subspace topology is compact; equivalently, every open cover of $K$ by ambient open sets has a finite subcover. The word every matters: finding one finite subcover of one open cover is not enough to establish compactness. ^[ref]

The closed interval $[0,1]$ is compact by the Heine-Borel theorem. For a concrete open cover, take the three open intervals $(-0.1,0.45)$, $(0.35,0.8)$, and $(0.7,1.1)$. Their union is $(-0.1,1.1)$, so they cover every point of $[0,1]$, including both endpoints. The overlaps have lengths $0.1$ and $0.1$, leaving no gaps. Each member is necessary for this particular open cover: removing them respectively leaves $0$, $0.5$, or $1$ uncovered.

Above, hollow markers denote the excluded endpoints of the covering open intervals, while the solid endpoints belong to the covered closed interval. Below, each displayed open interval is one member of an infinite open cover of $(0,1)$; finitely many rows always leave points near zero uncovered. The rows are vertically separated for readability and all refer to the same real line.

The open interval $(0,1)$ is a bounded set but is not compact: the open setsgive an open cover of $(0,1)$, since every $0<x<1$ has $1/n<x$ for sufficiently large $n$. These are also open sets for the subspace topology on $(0,1)$. Any nonempty finite selection has a largest index $N$, so its union is just $(1/N,1)$ and misses the point $1/(N+1)$, which still belongs to $(0,1)$. Thus no finite subcover exists, however many members we select. The same family is not an open cover of $[0,1]$, since it misses both endpoints.

In a metric space, compactness is equivalent to sequential compactness: every sequence in the subset has a convergent subsequence whose limit of a sequence belongs to that subset. This equivalence does not extend to all topological spaces. For example, $x_n=1/n$ for $n\ge 2$ has limit of a sequence zero in $[0,1]$. The same sequence lies in $(0,1)$, but every subsequence still has limit of a sequence zero on the real line, so no subsequence converges to a point of $(0,1)$. Merely approaching a boundary is harmless when the required point remains available. ^[ref]

Both panels show the sequence $x_n=1/n$ for $n\ge 2$ approaching zero. In the closed interval, zero is an allowed limit of a sequence; in the open interval, the hollow marker identifies the missing point. This single sequence disproves sequential compactness of $(0,1)$; compactness of $[0,1]$ still requires control of every open cover or every sequence, supplied by the Heine-Borel theorem.

An infinite set with the discrete topology similarly fails compactness, using the open cover by all its singletons. Every finite set is compact in any topology: select one covering open set for each point. The empty set is compact because the empty selection is a finite subcover. This open cover definition does not require the topological space to be Hausdorff; in algebraic geometry it is often called quasi-compactness.

The beauty of compactness is that information valid around each point can often be controlled using finitely many regions. A finite subcover lets us combine finitely many local bounds, or choose the smallest of finitely many positive tolerances, to obtain control across the whole compact space. This is why compactness repeatedly turns approximation into existence and why it is so useful in optimization through the extreme value theorem.

Every closed set in a compact space is compact, and a continuous image of a compact space is compact. The latter proof pulls an open cover back by inverse images, selects a finite subcover, and maps forward. This makes compactness useful for proving existence: the extreme value theorem guarantees that a continuous function from a nonempty compact space to the real line attains both a minimum and a maximum. For example, $q(x)=x(1-x)=1/4-(x-1/2{)}^2$ on $[0,1]$ has maximum $1/4$ at $1/2$ and minimum zero at its endpoints. On $(0,1)$ its infimum is still zero but is never attained. ^[ref]

If $X$ is an infinite set with its cofinite topology and $Y$ is a Hausdorff space, every continuous map $f:X\to Y$ is a constant function. Otherwise two distinct image points would have disjoint open neighbourhoods. Their inverse images would be nonempty disjoint open sets of $X$, contradicting its hyperconnected space property.

For example, $f(n)=n$ from ${\mathbb{N}}_{>0}$ with its cofinite topology into the real line with its usual Euclidean metric is not continuous: the inverse image of the open interval $(1/2,3/2)$ is $\{1\}$, which is not open in the infinite domain. A continuous real-valued readout on that domain can only report one value. Finite cofinite domains are different: their discrete topology allows arbitrary functions into any topological space.

Counting measure makes integration into summation by giving every measurable singleton unit weight. On any measurable space $(X,\mathcal{F})$, it is the measure $\#$ defined byThus $\#(\varnothing )=0$, a finite set has its usual cardinality, and every infinite set has the same value $\infty$ in the extended real number line. In particular, countably infinite sets and uncountable sets are not distinguished by their counting measure. Taking $\mathcal{F}$ to be the power set makes every subset measurable; restricting the same rule to a smaller sigma-algebra is also allowed. ^[ref]

Why is this a countably additive measure? For pairwise disjoint measurable sets $A_n$,If the union is finite, only finitely many $A_n$ are nonempty, and finite cardinality adds. If the union is infinite, either one $A_n$ is infinite or infinitely many $A_n$ are nonempty; in both cases the nonnegative series on the right is infinite. For example, $\{0,2\}$ and $\{1\}$ have counting measures $2$ and $1$, and their union has counting measure $3$. The singletons $\{n\}$ for positive natural numbers give the infinite case: $\#(\mathbb{N})={\sum }_{n=1}^{\infty }1=\infty$.

With the power set as sigma-algebra, counting measure is a finite measure exactly when $X$ is finite, and a sigma-finite measure exactly when $X$ is at most countable. For the latter assertion, enumerate $X$ by measurable singletons; conversely, every finite-measure set is finite, and a countable union of finite sets is countable. Specifying the sigma-algebra matters: if $X=\mathbb{N}$ and $\mathcal{F}=\{\varnothing ,\mathbb{N}\}$, counting measure is not a sigma-finite measure, since the only measurable finite-measure set is the empty set.

For a finite or countably infinite set $X$ equipped with its power set, every nonnegative function is a measurable function, andThe supremum allows an infinite result and avoids choosing an order for the terms. On a finite set, expand $f$ as a simple function using singleton indicator functions. On a countably infinite set, enumerate the points and apply the monotone convergence theorem to finite truncations. Real or complex functions are integrable functions for counting measure exactly when ${\sum }_{x\in X}|f(x)|<\infty$; then their integral equals the absolutely convergent sum. This is the discrete counterpart of Lebesgue integration, developed further in series as a Lebesgue integral against counting measure. ^[ref]

For example, $f(n)=2^{-n}$ on the positive natural numbers has counting measure integral one, by the geometric series. The alternating harmonic series has a finite ordered sum but is not the integral of an integrable function for counting measure: its positive and negative parts both have infinite sums. Absolute convergence is what makes the finite integral independent of enumeration. In particular, the Lp spaces for counting measure on $\mathbb{N}$ are the l-p sequence spaces, with their sums replacing continuous integrals.

Compare three measures on the same Borel sigma-algebra of the real numbers: counting measure $\#$, Dirac measure ${\delta }_1$, and Lebesgue measure $\lambda$.The rational numbers in the last row form a countably infinite set; their Lebesgue measure is zero by countable additivity applied to singletons, even though they are dense in the interval.

These are different answers to different questions: counting measure asks how many points occur, Dirac measure asks whether the selected point occurs, and Lebesgue measure asks how much length occurs. A null set therefore depends on its measure. For counting measure, the only null set is the empty set, so equality almost everywhere is equality everywhere. For ${\delta }_1$, a measurable set is a null set exactly when it misses $1$; for Lebesgue measure, even the infinite set $\mathbb{Q}\cap [0,2]$ is a null set. Integrating an indicator function recovers each table entry through $\int 1_{A}d\mu =\mu (A)$. This lets the same Lebesgue integral framework handle point counts, point evaluation, and geometric length.

A set $X$ is Dedekind-finite when no injective function $\mathbb{N}\to X$ exists. In classical Zermelo–Fraenkel set theory this is equivalent to $X$ not being in bijection with a proper subset: an injective non-surjective self-map generates distinct iterates from an element outside its image, and a countably infinite subset permits a shift fixing its complement. A finite set is Dedekind-finite. Without the axiom of choice, some infinite sets can also be Dedekind-finite.

On a measurable space $(X,\mathcal{F})$, the Dirac measure at $x\in X$ assigns unit mass according to whether a measurable set contains that point:Here $1_A$ is the indicator function. This is a probability measure, since ${\delta }_x(X)=1$. For pairwise disjoint measurable sets $A_n$, at most one contains $x$, so ${\delta }_x({\bigcup }_n{A}_n)={\sum }_n{\delta }_x(A_n)$: countable additivity reduces to a membership test. The definition works even when the singleton $\{x\}$ is not a measurable set. ^[ref]

To assert that its support of a measure is exactly $\{x\}$, assume that $X$ is a metric space and $\mathcal{F}$ is its Borel sigma-algebra. The singleton $\{x\}$ is then a closed set and a Borel set, with ${\delta }_x(\{x\})=1$. Every open neighbourhood of $x$ has mass one; every other point $y$ has an open ball of radius less than its distance from $x$, with mass zero. Thus $\mathrm{supp}{\delta }_x=\{x\}$. An abstract measurable space has no specified topology, so this topological conclusion is not part of the bare definition. Indeed, on the sigma-algebra $\{\varnothing ,X\}$, all Dirac measures coincide: the available measurable sets cannot distinguish points.

The Lebesgue integral against a Dirac measure is evaluation:This holds for every nonnegative measurable function, allowing an infinite value, and for real or complex measurable functions satisfying $|f(x)|<\infty$, which are precisely the integrable functions for ${\delta }_x$. For a nonnegative simple function, only the cell containing $x$ contributes to its integral; increasing simple functions and the monotone convergence theorem extend the identity. Splitting into positive and negative parts, and then real and imaginary parts, gives the remaining cases. This is why a Dirac measure turns integration into a point query.

For a measurable function $T:(X,\mathcal{F})\to (Y,\mathcal{G})$, its pushforward measure satisfiesConsequently ${\int }_{Y}gd(T_{\ast }{\delta }_x)=g(T(x))$ for nonnegative or integrable measurable functions $g$. For example, on the real numbers with their Borel sigma-algebra, $T(t)=2t+1$ sends ${\delta }_1$ to ${\delta }_3$, so integrating $g(y)=y^2$ against that pushforward measure gives $9$. ^[ref]

Finite weighted sums of Dirac measures give useful measures with finitely many point masses. For finite nonnegative weights $w_j$,for nonnegative measurable functions $f$, or whenever ${\sum }_j{w}_j|f(x_j)|<\infty$, with zero-weight terms omitted. Repeated locations have their weights added. On the real numbers, this is the finite version of the measure underlying the left-continuous cumulative function of an atomic measure; a positive-weight singleton is an atom of a measure. Unit weights at the distinct points of a finite set recover counting measure on that set. ^[ref]

The same function can have sharply different integrals for different measures. On $[0,2]$ with its Borel sigma-algebra, defineHere $\lambda$ is Lebesgue measure and $\#$ is counting measure. The singleton $\{1\}$ is a Lebesgue null set, so its changed value leaves the Lebesgue integral equal to ${\int }_0^2(t+1)dt=4$. For counting measure on the whole interval, $f\ge 1$ on an infinite set, forcing an infinite integral. On the smaller measure space $D=\{0,1,2\}$ with its power set and counting measure, the restricted function instead has integral $1+5+3=9$.

This also explains the almost everywhere convention in Lp spaces: equality almost everywhere for ${\delta }_x$ requires equality at $x$, whereas equality almost everywhere for Lebesgue measure allows different values at that singleton. A point value therefore cannot be read unambiguously from an equivalence class in an Lp space formed using Lebesgue measure.

On the real numbers, a Dirac measure gives the Dirac delta distribution by $⟨{\delta }_x,\phi ⟩=\int \phi d{\delta }_x=\phi (x)$ for every test function. The name Dirac delta function does not mean an ordinary function that is zero away from $x$ and infinitely high at $x$. Such a nonnegative measurable function still has Lebesgue integral zero, because its nonzero values occur on a Lebesgue null set. More generally, no nonnegative density $h$ with respect to Lebesgue measure can represent ${\delta }_x$: it would give ${\delta }_x(\{x\})={\int }_{\{x\}}hd\lambda =0$, contradicting unit mass. The measure and its associated distribution act on different kinds of inputs, namely measurable sets and test functions, but express the same point evaluation. ^[ref]

The discrete metric sets $d(x,x)=0$ and $d(x,y)=1$ for distinct points. It induces the discrete topology, and its Cauchy sequences are exactly the eventually constant sequences.

Every nonempty set equipped with the discrete metric is a bounded set, even if it is an infinite set. For $0<r<1$, both the open ball and closed ball about $a$ are the singleton $\{a\}$. At $r=1$, the open ball is still $\{a\}$, but the closed ball is the whole space; for $r>1$, both are the whole space. Every subset is both an open set and a closed set. Consequently the boundary of the radius-one closed ball is empty, even though every point other than $a$ is at distance exactly one. Also, the closure of $B_d(a,1)$ is $\{a\}$, rather than the radius-one closed ball.

An infinite Dedekind-finite set is an infinite set with no subset that is a countably infinite set. The name Dedekind set is sometimes used for this combination of properties. Under the axiom of choice no such set exists; results about these sets must avoid obtaining arbitrary enumerations of finite subsets by unstated choices.

The infinite pigeonhole principle says that if an infinite set has a partition of a set into finitely many parts, at least one part is an infinite set. Equivalently, assigning infinitely many objects to a finite number of boxes forces some box to contain infinitely many objects. For example, separating the natural numbers into their two residue classes modulo $2$ forces at least one class to be an infinite set; here both are, as the lists $0,2,4,\dots$ and $1,3,5,\dots$ show.

Proof outline. A union of finitely many finite sets is a finite set. If all the parts were finite sets, their union could not be the original infinite set.

The assumption of finitely many parts is essential. With one box labelled by each natural number, assigning $n$ to box $n$ puts exactly one object in every box. Infinitely many objects and infinitely many boxes therefore need not produce even a shared box. The infinite pigeonhole principle is useful whenever finitely many labels organize an endless collection: one label must keep occurring infinitely often, providing an infinite set on which that label is constant.

Write $[\mathbb{N}{]}^{\omega }$ for the space of infinite subsets of the natural numbers, with each subset identified with its increasing enumeration. In the homogeneous-cone convention, a family $\mathcal{A}\subseteq [\mathbb{N}{]}^{\omega }$ is a Ramsey family in the homogeneous-cone sense when some infinite $M$ satisfiesThe stronger Ramsey set of infinite subsets convention asks for such an infinite $M\subseteq A$ inside every given infinite ground set $A$. We will prove the open-set assertion in this stronger form, and our Ramsey-but-not-completely-Ramsey example will satisfy it as well. Thus the distinction in conventions does not affect any of the conclusions below.

For a non-Ramsey example, declare $X\sim Y$ if their symmetric difference is finite, and choose a representative $R$ from each equivalence class using the axiom of choice. Setfor the representative of the class of $X$. Removing one point reverses this colour, because its symmetric difference with $R$ changes by exactly one point. Consequently every cone $[M{]}^{\omega }$ contains the oppositely coloured sets $M$ and $M\setminus \{\min M\}$. Either colour class of this finite-symmetric-difference parity colouring is not a Ramsey family in the homogeneous-cone sense. No definability is claimed for the representative selection.

The ordinary topology on infinite subsets, denoted $\tau$, has basic cylinders $[s]=\{X:s⊏X\}$, where $s$ is a finite stem of the increasing enumeration. Let $O$ be a $\tau$-open family and fix an arbitrary infinite ground set $A$. We give a complete fusion proof of a homogeneous cone inside $A$.

Use the convention $\max \varnothing =0$. For finite $s$ and an infinite reservoir $B$ above $\max s$, say that $B$ accepts $s$ if $[s,B]\subseteq O$, whereSay that $B$ rejects $s$ if no infinite subset of $B$ accepts $s$. There is always a refinement deciding $s$: choose an accepting subreservoir if one exists, and otherwise retain $B$, which rejects. Both decisions persist under further thinning.

A first fusion produces an infinite $H\subseteq A$ deciding every finite $s\subseteq H$ on its tail $H/s=\{h\in H:h>\max s\}$. First decide the empty stem. Choose the first point, and thin the remaining reservoir successively to decide every subset of the selected prefix. Repeat after choosing each new point. There are finitely many subsets to decide at each stage. The diagonal sequence of chosen points has the required property, since its tail after the largest point of any stem lies in the reservoir where that stem was decided. The empty-stem decision persists too. This is deciding all finite stems by fusion.

If $H$ accepts the empty stem, $[H{]}^{\omega }\subseteq O$ and we are done. Suppose it rejects. For a rejected finite $s\subseteq H$, only finitely many $h\in H/s$ can have $H/h$ accepting $s\cup \{h\}$. Otherwise let $D$ be their infinite set. Every infinite $X\subseteq D$ starts with some such $h$, and its remaining tail lies in $H/h$, so $s\cup X\in O$. Then $D$ accepts $s$, contradicting rejection. This proves finitely many accepting extensions of a rejected stem.

A second fusion constructs $K\subseteq H$ whose every finite subset is rejected. If a finite prefix of chosen points has been selected, all its subsets are rejected by induction. Avoid the union of their finitely many accepting successor sets when choosing the next point. The first fusion already decides every successor, so every new subset is rejected. An infinite reservoir remains after each finite exclusion.

If some $X\in [K{]}^{\omega }$ belonged to $O$, openness would give an initial segment $s⊏X$ with $[s]\subseteq O$. Its reservoir $H/s$ would then accept $s$, contradicting the second fusion. Therefore $[K{]}^{\omega }\cap O=\varnothing$. In either case we have a homogeneous cone inside the arbitrary ground set $A$. Every $\tau$-open family is a Ramsey set of infinite subsets, under both conventions.

A completely Ramsey set requires more: for every finite $s$ and infinite reservoir $A$ above $\max s$, there is an infinite $B\subseteq A$ such thatThe finite stem must be kept fixed. These $[s,A]$ are the basic neighbourhoods of the Ellentuck topology, also called the star topology.

Let $E$ be the even positive integers, and construct the finite-symmetric-difference parity colouring $\epsilon$ on $[E{]}^{\omega }$. DefineFor any infinite ground set $A$, the set $A\setminus \{1\}$ is still infinite and its whole cone misses $\mathcal{R}$. Thus $\mathcal{R}$ is a Ramsey set of infinite subsets even in the stronger convention. But inside $[\{1\},E]$, every infinite refinement $B\subseteq E$ gives the two sets $\{1\}\cup B$ and $\{1\}\cup (B\setminus \{\min B\})$ of opposite colours. Hence no such refinement is homogeneous. The family $\mathcal{R}$ is Ramsey but not completely Ramsey. The stem-supported Ramsey family need not be completely Ramsey construction explains this example: forgetting the stem $\{1\}$ is precisely what loses the obstruction.

Finally take $D=[E{]}^{\omega }$. It is nonempty and star-open, since it equals $[\varnothing ,E]$. It is $\tau$-closed: if an infinite set contains an odd integer, an initial segment witnessing that integer has a cylinder disjoint from $D$. Its $\tau$-interior is empty: after any finite stem of even integers, append an arbitrarily large odd integer and then infinitely many further points. ThusThis is the cone on an infinite coinfinite ground set. Choose its subset $Y=\{X\in [E{]}^{\omega }:\epsilon (X)=0\}$. Since ${\overline{Y}}^{\tau }\subseteq D$, $Y$ is also $\tau$-nowhere dense. Nevertheless every $[\varnothing ,B]$ with infinite $B\subseteq E$ contains both $B$ and $B\setminus \{\min B\}$, of opposite colours. Hence $Y$ is not completely Ramsey. We have proved explicitly that ordinarily nowhere-dense sets need not be completely Ramsey; the topology in the nowhere-dense hypothesis cannot be silently changed to the star topology.

Use the bounded successor characterization of finite ordinals. Let $\mathrm{Ord}(x)$ assert that $x$ is transitive, each of its members is transitive, and any two members are equal or comparable by membership. This is a bounded first-order formula; foundation makes its linear membership order well-founded, so it is precisely the Von Neumann ordinal predicate. Defineand putFor an ordinal, having a greatest member $y$ says exactly $x=y\cup \{y\}$. The displayed definition makes no reference to an infinite inductive set or to $\omega$; all quantifiers are bounded within the candidate and its members. Occurrences of $x=\varnothing$ can also be expressed as $(\forall u\in x)u\ne u$, so no unbounded quantifier is hidden in empty-set notation.

Every usual natural number satisfies this predicate, by induction: it is a finite ordinal, and every nonzero initial segment is a successor. Conversely, suppose an ordinal $x$ is not a usual natural number. Ordinal comparability gives $x\ge \omega$. If $x=\omega$, it has no greatest member, contradicting ${\mathrm{Succ}}_0(x)$. If $x>\omega$, then $\omega \in x$ has no greatest member, contradicting the corresponding bounded requirement on members of $x$. Hence $\mathrm{Nat}(x)$ holds exactly for the usual natural numbers. The axiom of infinity and separation therefore collect this class as the usual set $\omega$. This is a bounded definition of membership in the natural-number class; it does not claim that the infinite set $\omega$ itself can be produced without the infinity axiom.

Work in a ground first-order model of set theory satisfying ZF, and let $\pi$ be a definable bijection of its universe, whose inverse is also definable. Write $\psi ={\pi }^{-1}$ and replace membership byIn this Rieger-Bernays permutation model, an object having prescribed extension $b$ is represented by $\psi (b)$. The Axiom schema of replacement in the ground model guarantees that images of sets under $\pi$ and $\psi$ are sets. The axiom of extensionality follows from injectivity of $\pi$; the new empty set is $\psi (\varnothing )$, and the new pair of $x,y$ is $\psi (\{x,y\})$.

For the Axiom of union and Axiom of power set, representatives are respectivelyIndeed, $z$ is an $E$-subset of $x$ exactly when $\pi (z)\subseteq \pi (x)$. Translate every first-order formula by replacing membership with membership in $\pi (y)$. Ground separation then forms each desired subextension of $\pi (x)$, and ground replacement forms each functional image; applying $\psi$ supplies its representative. To verify the axiom of infinity, define by ground recursionThen $\psi (\{a_n:n<\omega \})$ is an $E$-inductive set. Thus every ZF axiom except possibly the Axiom of foundation survives.

Interchange two distinct objects $a$ and $\{a\}$, fixing everything else. Then $\pi (a)=\{a\}$, so $a$ is a Quine atom: internally $a=\{a\}$. The nonempty set $a$ has no member disjoint from itself, violating the Axiom of foundation. The original membership model satisfies that axiom. Foundation is therefore relatively independent of the other ZF axioms.

A full Rieger-Bernays permutation model of a choice model still satisfies the axiom of choice: choose an old member of each nonempty extension $\pi (y)$ and encode the resulting choice graph with the new ordered-pair operations. To obtain failure of choice, an additional restriction by symmetries is necessary.

Start instead with ZFC and use the definable permutation interchanging every $a_n=\omega +n$ with $\{a_n\}$. These transpositions are disjoint and produce an infinite set $A$ of Quine atoms. Inside the resulting membership model build the cumulative universe over $A$, closing at successive stages under sets of previously constructed objects, represented by $\psi (X)$. Atomic objects are kept as the designated base objects; $\psi (\{a\})=a$ for $a\in A$. Every permutation of $A$ extends recursively to an automorphism of this universe, with the atomic self-loops fixed by the recursive convention.

An object has finite support if every permutation fixing some finite $S\subseteq A$ pointwise fixes it. Retain the objects which have finite support and whose members recursively also have finite support, interpreting the atomic self-loops as the base case. This is the hereditarily finite-supported Quine-atom model. It is transitive for the new membership relation. Extensionality is inherited. Pairs and unions have supports contained in the union of the finitely many relevant supports. For a supported set $x$, the collection of all retained subsets of $x$ is a ground set, by the full model's power set axiom, and is fixed by every permutation fixing a support of $x$. Its members are retained, so this collection supplies the internal power set. A first-order formula with retained parameters is invariant under their common stabilizer. Consequently its definable subset of $x$, and the range of a functional relation on $x$, have that same finite support; ground separation and replacement first ensure that they are sets. The pure natural-number hierarchy supplies infinity. These arguments verify all of ZF except Axiom of foundation, which fails at the Quine atoms.

The family $[A{]}^2$ of all two-element subsets is retained and has empty support. If it had a retained choice function $f$, choose two distinct Quine atoms $a,b$ outside a finite support of $f$. The transposition of $a,b$ fixes $f$ and fixes the unordered pair $\{a,b\}$, but moves either possible value $f(\{a,b\})$. This is impossible. Thus ZF without Foundation has both choice models and models in which Choice fails, relative to the usual consistency assumption. The construction also explains why twisting membership alone was insufficient.

Use rooted-tree homeomorphic embedding: edges may be stretched into downward paths, different child branches stay disjoint, and the source root may map to any target vertex. Write $S⪯T$ for this relation. More generally allow labels in a well-quasi-ordering $Q$ and require every source label to be at most its image label. We prove this labelled version, which includes the unlabelled case by taking one label.

First, every infinite sequence in a well-quasi-ordering has an infinite nondecreasing subsequence. In every infinite set of indices there is an index with infinitely many later successors above its value. Otherwise repeatedly discard the finitely many successors of each chosen index, obtaining a bad sequence. Applying this observation successively within the infinite successor set produces the required subsequence.

We also need the Higman lemma, and prove it here. If finite words over $Q$ admitted a bad sequence, choose one $w_0,w_1,\dots$ with each word of least possible length among choices permitting a bad continuation of its fixed prefix. No word is empty. Write $w_i=v_i{a}_i$, and extract indices $n_0<n_1<\cdots$ with $a_{n_0}\le a_{n_1}\le \cdots$. The sequenceis bad: an earlier $w_i$ embedding into some $v_{n_j}$ would embed into $w_{n_j}$, while $v_{n_i}⪯v_{n_j}$ together with $a_{n_i}\le a_{n_j}$ would give $w_{n_i}⪯w_{n_j}$. This contradicts the minimal length of $w_{n_0}$, since $v_{n_0}$ is shorter. Hence finite words are well-quasi-ordered under subsequence embedding with increased labels.

Now suppose labelled rooted trees admitted a bad sequence $T_0,T_1,\dots$, chosen minimally by number of vertices at each successive position. Let $B$ consist of all immediate child subtrees of all the $T_i$, with the inherited labels. We claim $B$ is a well-quasi-ordering. Otherwise take a bad sequence of members of $B$. Only finitely many distinct rooted trees occur among the children of any finite initial list of $T_i$, so, by discarding terms and selecting a subsequence, we may arrange its members $S_0,S_1,\dots$ to come from strictly increasing parent indices $n_0<n_1<\cdots$. Thenis bad. A prefix rooted tree embedding into $S_j$ would embed into its parent $T_{n_j}$; embeddings between the $S_j$ are excluded by their choice. But $S_0$ has fewer vertices than $T_{n_0}$, contradicting minimality. This proves the claim.

List the child subtrees of each $T_i$ in any order. By the Higman lemma these child lists are well-quasi-ordered. Extract an infinite subsequence with nondecreasing root labels, and then compare two child lists on that subsequence. Their subsequence embedding gives distinct target child branches into which the source child subtrees embed. Map the source root to the target root and join each embedded child root to it by the corresponding downward path. The paths are disjoint except at the root, and labels increase. Thus $T_i⪯T_j$ for some $i<j$, contradicting badness. This proves Kruskal's tree theorem and its labelled version. If root preservation is required, mark the root with a special label incomparable with all ordinary labels; the same proof then forces root to map to root.

For Friedman's finite form of Kruskal's theorem, fix $k\in \omega$ and suppose there are arbitrarily long bad lists satisfying $|T_i|\le k+i$ for $i=1,2,\dots$. Make a finite bad-sequence tree whose nodes are these lists, using one representative of each finite rooted-rooted tree isomorphism type. At each position only finitely many choices satisfy the size bound, so the rooted tree is finitely branching. An infinite path exists: at every stage choose a child having extensions of arbitrarily large length, which must exist because there are finitely many children. The path is an infinite bad sequence, contradicting Kruskal's tree theorem. ThereforeThe identical argument works with finitely many labels and any fixed finite size bound at each position.

Take $fg=f\circ g$. The assertion can fail on an infinite set. On $X={\mathbb{N}}_0$, set $g(n)=n+1$, $f(0)=0$, and $f(n)=n-1$ for $n\ge 1$. Then $f(g(n))=n$ for every $n$, but $g(f(0))=1\ne 0$. Thus a right inverse need not be a left inverse.

On a finite set it is always true. From $fg={\mathrm{id}}_X$, $g$ is an injection: equality of two $g$-values gives equality after applying $f$. An injective self-map of a finite set is a bijection. Composing $fg={\mathrm{id}}_X$ with $g^{-1}$ shows $f=g^{-1}$, and hence $gf={\mathrm{id}}_X$. This includes the empty set.

Pull the given finite coloring $\chi :\mathbb{N}\to [k]$ back to a finite coloring of unordered pairs by puttingThe ordering $a<b$ makes this a well-defined function on the edges of a complete graph. By Ramsey's theorem, some infinite set $X=\{x_1<x_2<\cdots \}$ has all its pairs of one color $r$ under $\tilde{\chi }$. ConsequentlyPossible coincidences between these products cause no difficulty, since the original finite coloring assigns each positive integer a single color.

Write $\mathbb{N}=\{1,2,\dots \}$ and let $\chi$ be the given finite coloring of the edges of the complete graph on the positive integers. Construct an increasing sequence $(v_i)$ and nested infinite sets $A_0\supset A_1\supset \cdots$ as follows. Set $A_0=\mathbb{N}$. Having chosen $A_{i-1}$, take $v_i=\min A_{i-1}$, partition $A_{i-1}\setminus \{v_i\}$ by the color of $\{v_i,w\}$, and use the infinite pigeonhole principle to choose an infinite set $A_i$ on which this color is constant, say $r_i$. Every element of $A_i$ exceeds $v_i$.

The infinite pigeonhole principle applied again to $(r_i)$ gives an infinite set of indices $I$ and a color $r$ such that $r_i=r$ for all $i\in I$. For $i<j$ in $I$, nesting gives $v_j\in A_i$, so $\chi (\{v_i,v_j\})=r$. Thus the infinite set $X=\{v_i:i\in I\}$ is a monochromatic complete graph:This proves the required infinite case of Ramsey's theorem; finiteness of the color set is used at both applications of the infinite pigeonhole principle.

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.

True. If $x$ is a nonempty finite set, thenThis maximum is a successor ordinal, whereas the empty set has rank $0$. Hence a set of nonzero limit ordinal rank cannot be finite and must be an infinite set.