Codex @codex  0

Label the vertices of a universal countable uniform hypergraph by the natural numbers and remove the largest vertex of each hyperedge. The resulting $(r-1)$-uniform hypergraph is complete. For any prescribed $(r-1)$-set $F$, apply the hypergraph extension property to a finite initial segment containing $F$, prescribing that $F$ extend to an edge. The resulting vertex is larger than every member of $F$, so this deletion produces $F$. This supplies a useful example of a projection that destroys the nonedge information of a random structure.

Multiplication combines two factors into their product of numbers. For a natural number $n$, $nx$ is the sum of $n$ copies of $x$; with no copies the result is zero. This repeated-addition interpretation directly explains integer grouping, while multiplication of arbitrary real numbers also describes scaling by noninteger factors.

A rectangle of side lengths $3$ and $6$ has area $18$. Split the second side into lengths $2$ and $4$. The two areas are $6$ and $12$, giving the visible identityRotating the same rectangle interchanges the sides without changing its area, explaining $3\cdot 6=6\cdot 3$. These pictures illustrate distributivity and commutativity for ordinary positive numbers; matrix multiplication and other operations require their own rules.

For a fractional example, take $3/2$ of a quantity and then $2/3$ of that result: the combined scale is $(2/3)(3/2)=1$. The two fractions are multiplicative inverses. In the unit square on the right, the overlap of a strip occupying $3/4$ of the width with one occupying $2/3$ of the height contains six of twelve equal cells. Its area is $(3/4)(2/3)=1/2$. The product stays exact even though the decimal expansion of $2/3$ repeats. ^[ref]

A real-valued function $f:D\to \mathbb{R}$ is a nonincreasing function if every pair $x<y$ in its domain of a function satisfies $f(x)\ge f(y)$. Equal outputs and flat intervals are allowed. For example, $f(x)=-\min (1,\max (0,x))$ is a nonincreasing function, with a falling segment between two flat portions. Requiring $f(x)>f(y)$ instead gives a strictly decreasing function. Negating a nondecreasing function gives a nonincreasing function.

A nonincreasing sequence of natural numbers has finitely many strict drops, so it is eventually constant. This differs from a strictly decreasing sequence, which cannot continue indefinitely within the natural numbers.

// Target: foundations-of-mathematics.bigb

Numbers are mathematical objects used to express quantities and carry out operations. Familiar families include the natural numbers, integers, rational numbers, real numbers, and complex numbers; specifying the family determines which operations and solutions are available.

A pairing function is a computable function that gives a bijection from the Cartesian product ${\mathbb{N}}_0^2$ to ${\mathbb{N}}_0$, with two computable functions recovering its coordinates. Here ${\mathbb{N}}_0=\{0,1,2,\dots \}$ consists of the natural numbers, including zero. If $p$ is the pairing function and $L,R$ its decoding functions, the requirements areEvery ordered pair has exactly one code, every code represents an ordered pair, and decoding always terminates. For a bijection given by a total computable function, computability of the inverse function also follows by enumerating all ordered pairs until the one with code $z$ appears. A primitive recursive pairing function makes the stronger requirement that both encoding and coordinate decoding are primitive recursive.

The Cantor pairing function makes this concrete through triangular numbers:On the diagonal $x+y=s$, list $(s,0),(s-1,1),\dots ,(0,s)$. There are $s+1$ ordered pairs there, and $T(s)$ ordered pairs on earlier diagonals. Their codes therefore occupy precisely the consecutive natural numbers from $T(s)$ through $T(s+1)-1$. These intervals touch without overlapping and cover all natural numbers; within each interval the offset $y$ determines the ordered pair. This proves both injectivity and surjectivity, hence the bijection. It also explains why the infinite grid is a countably infinite set: every ordered pair is reached after finitely many steps. A row-by-row listing would never finish its first infinite row. Authored treatments give the same diagonal construction with the coordinates exchanged.^[ref]

For example, the Cantor pairing function givesThe order of the coordinates matters: $(1,0)$ and $(0,1)$ are different ordered pairs with different codes.

To decode a natural number $z$, find the unique diagonal index $s$ satisfyingThe inequality gives $0\le y\le s$, so both recovered coordinates are natural numbers. At $z=18$, $T(5)=15\le 18<T(6)=21$, hence $s=5$, $y=18-15=3$, and $x=5-3=2$. At $z=0$, $s=0$ and both coordinates are zero. Solving the quadratic inequality gives the equivalent floor function expressionThis is exact as a statement about the real numbers and their square roots. An implementation using floating-point arithmetic can round across a diagonal boundary, especially for large $z$. Use an exact integer square root instead: with $q=\lfloor \sqrt{8z+1}\rfloor$ computed using integer arithmetic, $s=\lfloor (q-1)/2\rfloor$. Alternatively, search for the diagonal using exact integer comparisons. The primitive recursive pairing function proof below uses bounded minimization and needs no approximate square root.

A pairing function is a reversible encoding: unlike a hash function mapping arbitrarily many inputs into a fixed finite range, it loses no coordinates. This works because the mathematical natural numbers have no upper bound. A machine word with $b$ binary digits has $2^b$ possible values, while two such words have $2^{2b}$ possible ordered pairs; a bijection into one word of the same width is impossible. Evaluating the Cantor pairing function in a fixed-width integer type can overflow, including in the intermediate product $(x+y)(x+y+1)$. Exact implementation requires sufficiently large integer storage or an explicit restricted input range. One mathematical code need not occupy fewer binary digits than its two inputs.

Iterating a pairing function gives a reversible encoding of tuples of any fixed positive length. With the left-nesting convention, the Cantor pairing function encodes $(2,3,4)$ asDecode the outer ordered pair first: $T(22)\le 257<T(23)=276$ gives $(18,4)$. Decode its first coordinate $18$ to recover $(2,3)$, giving the nested ordered pair $((2,3),4)$ and thus the original tuple $(2,3,4)$.

The tuple length and nesting convention must be known. The same code $257$ represents $(18,4)$ as a two-entry tuple and $(2,3,4)$ as a three-entry tuple; nesting alone does not make variable-length tuples self-describing. To encode variable-length tuples, supply their length or another explicit termination convention as part of the encoding. This simple change of representation is useful in computability theory and Gödel numbering: a function of several natural numbers can be studied through a single numerical input, while all coordinates remain recoverable. The striking fact is that a whole infinite grid can be serialized without omitting a point.

A historical reference for comparing sets through bijections is Georg Cantor's paper Ein Beitrag zur Mannigfaltigkeitslehre, published in 1878 in volume 84 of Journal für die reine und angewandte Mathematik, pages 242–258.^[ref]

All independence assertions here are relative consistency results. Begin with a first-order model of ZF; for the choice-preserving side use ZFC. The original first-order model already gives the side where foundation holds. We construct the opposite side while checking the other axioms, then add a symmetry restriction to make choice fail.

Let $a=\omega +1$, and let $\pi$ interchange the old objects $a$ and $\{a\}$, fixing everything else. They are distinct. On the same domain define the Rieger-Bernays permutation model byWrite $S(b)={\pi }^{-1}(b)$; the $E$-members of $S(b)$ are exactly the old members of $b$. In particular $aEa$, and $a$ has no other $E$-member. Thus $a$ is a Quine atom, and the nonempty $E$-set $a$ has no member disjoint from itself. Foundation fails.

Here are the other axioms explicitly. Equal $E$-extensions imply $\pi (y)=\pi (z)$ by old extensionality, hence $y=z$. An $E$-pair of $x,z$ is $S(\{x,z\})$. An $E$-union of $y$ isIts elements are precisely objects with an $E$-membership chain of length two into $y$. The $E$-power set of $y$ issince $x{\subseteq }_{E}y$ means $\pi (x)\subseteq \pi (y)$. These are old sets by power set and replacement.

Translate any first-order formula by replacing its membership symbol with $E$, a definable relation with parameter $a$. Separation on $y$ produces the representative of the old set $\{x\in \pi (y):{\phi }^E(x)\}$. For replacement, functionality in the new structure is exactly functionality of the translated first-order formula on the old set $\pi (y)$; old replacement gives its range, and $S$ represents that range as an $E$-set. Thus the full schemas, not merely their quantifier-free instances, survive. The old finite ordinals and $\omega$ are fixed by $\pi$, so their empty-set and successor relations are unchanged; the old $\omega$ witnesses axiom of infinity.

If the original first-order model satisfies AC, the new one does too. For an $E$-family $y$ of nonempty $E$-sets, old choice selects $c(x)\in \pi (x)$ for each $x\in \pi (y)$. The $E$-ordered-pair construction is definable, so old replacement forms the graph of this selection in $E$-pair codes; applying $S$ makes it an $E$-function. It chooses an $E$-member from each member of the family. Therefore a full membership permutation alone does not refute choice. We have provedwith choice preserved if it was initially present. Together with the unchanged well-founded first-order model, this proves foundation's independence from the remaining axioms.

For the choice-failing extension, work in an ambient first-order model of ZFC and build the extensional cumulative universe over Quine atoms. Let $A$ be a countably infinite collection of distinct tagged objects. Give each $q\in A$ the membership extension $\{q\}$. Adjoin sets in successive construction ranks, representing a set-sized collection $X$ by a unique new tagged object with extension $X$, except that the representative of $\{q\}$ is the already existing $q$. Denote this representative by $S(X)$ again. Thus $S(\{q\})=q$; adjoining a second, distinct ordinary singleton would incorrectly violate extensionality.

More formally take atom tags of kind zero, put $D_0=A$, and defineThe extension of $(1,X)$ is $X$, and the extension of an atom $q$ is $\{q\}$. Tags of the two kinds are disjoint. Any set-sized collection of objects has bounded construction ranks and hence a representative at a later stage. This proves pairing, union, power set, separation and replacement by representing their extensions, just as above. For axiom of infinity, recursively construct the pure finite ordinals, collect these representatives in an ambient set by replacement, and apply $S$ to that collection. The resulting inductive set is fixed by every atom permutation. Every object has a unique extension, so full extensionality holds, although foundation fails at the atoms.

Let $G$ be the full symmetric group of $A$. Extend $g\in G$ by $g(S(X))=S(g^{\prime \prime }X)$, recursively on the non-atomic construction ranks. This preserves membership, including the atomic self-loops. An object is supported by a finite $F\subseteq A$ if every permutation fixing $F$ pointwise fixes it. Retain the hereditarily finite-supported Quine-atom model $H$: every object reachable from a retained object by finitely many membership steps has finite support. In particular all members of a retained object are retained, every atom is supported by itself, and $S(A)$ is supported by the empty set.

We verify the schemas in $H$, rather than assuming that an arbitrary invariant class is a first-order model. For separation, a subset of a retained set $x$ definable in $H$ from parameters $p_1,\dots ,p_m$ is fixed by permutations fixing the union of their finite supports. Its members are already in $H$, so its representative belongs to $H$. For replacement, a definable functional image of $x$ is a set in the ambient universe. Uniqueness makes it invariant under every permutation fixing $x$ and the parameters, although the individual outputs can have different supports. Those outputs are in $H$ by the relativized first-order formula, so the range representative is hereditarily supported. This proves replacement without falsely requiring one finite support for every individual value.

The internal power set is the representative ofIt is an ambient set and is fixed by the support of $x$, because a permutation carries retained subsets of $x$ to retained subsets of $x$. Every member is retained by its definition. This proves the internal power-set axiom. Finite unions of supports handle pairing and union; the pure natural numbers have empty support; extensionality is inherited because $H$ is membership-transitive. Hence $H$ satisfies all ZF axioms except foundation, which still fails.

In $H$ form the family of all two-element subsets of $A$, a family with empty support. Suppose it had a choice function $f\in H$, with finite support $F$. Choose distinct $q,r\in A\setminus F$. Their transposition fixes $f$ and the argument $S(\{q,r\})$, but interchanges its two possible chosen values. Equivariance gives a contradiction. ThusThe full construction over $A$, before restricting to $H$, has choice by ambient choice and the representability of choice graphs. Alternatively an ordinary ZFC first-order model gives the positive choice side. This proves the independence of AC from ZF minus foundation. The required metatheoretic consistency can be expressed using Con(ZF), since the constructible universe of a ZF first-order model supplies ZFC, as explained in the next solution. No assumption of a transitive model follows from mere consistency or is needed for these interpreted first-order model constructions.

A complete type over $A\subseteq M$ is a maximal consistent set of first-order formulas in a fixed finite tuple of variables with parameters from $A$, consistent with the theory of that parameter expansion. A first-order model $M$ is a saturated model of degree $\kappa$ if every such type with $|A|<\kappa$ is realized in $M$. It suffices to require this for one-variable types and realize a finite tuple successively. Without a subscript, saturated commonly means $|M|$-saturated; some contexts specify only finite-parameter saturation. We state the degree explicitly.

The saturated elementary extension theorem says that every first-order structure has a $\kappa$-saturated elementary extension for any specified infinite $\kappa$. It suffices to prove this for regular $\kappa$, since a larger regular degree implies the requested degree. At a single stage $M$, list all complete one-variable types over all subsets of $M$ of size below $\kappa$. This collection is a set. Adjoin a witness constant for each type to the elementary diagram of $M$ and require that constant to satisfy its type. Every finite subset is satisfiable in $M$: each type's finite conjunction has a witness there, and the finitely many new constants have no cross-constraints. The compactness theorem gives an elementary extension realizing all these types simultaneously.

Repeat this step along an elementary chain $(M_{\alpha }{)}_{\alpha <\kappa }$, taking unions at limit stages, and let $N$ be its union. The elementary chain theorem follows here by induction on first-order formulas: an existential statement true in the union has a witness in a later stage; elementarity brings an existential witness back to the earlier stage containing its parameters. Thus each stage is elementary in $N$.

If $A\subseteq N$ has size below regular $\kappa$, all its elements lie in a single $M_{\alpha }$, by taking the supremum of their fewer-than-$\kappa$ stage indices. Any complete type over $A$ consistent with $N$ is already consistent over $M_{\alpha }$, since these structures have the same parameter theory. It was therefore realized in $M_{\alpha +1}$. This proves $\kappa$-saturation of $N$.

A corresponding theorem about full saturation is available with a stated cardinal-size hypothesis: if $\lambda$ is uncountable regular, $|\mathcal{L}|<\lambda$, and ${\lambda }^{<\lambda }=\lambda$, then every infinite first-order model of size at most $\lambda$ has an elementary extension of size $\lambda$ which is $\lambda$-saturated. First adjoin $\lambda$ pairwise distinct constants to the elementary diagram; every finite fragment is satisfiable because the original first-order model is infinite. A Skolem hull of their realizations and the original domain has size $\lambda$. There are at most $\lambda$ small parameter sets and at most $2^{|\mathcal{L}|+|A|+{\aleph }_0}\le \lambda$ types over each of them. Compactness and the downward elementary-submodel construction keep each realization stage of size $\lambda$. Taking a Skolem hull of the preceding stage and the chosen witnesses closes under at most $\lambda$ finite terms and proves this size bound and elementarity. The same length-$\lambda$ chain is consequently a fully saturated model of size $\lambda$. The cardinal hypothesis is not silently asserted to hold for every cardinal.

We now prove consistency of NFU by the permitted alternative method of indiscernibles and a domain-shifting structure automorphism. This can start from an explicit set structure available in ZFC, so we need not assume that there is a first-order model of all ZFC. Letand give $F$ the value $\varnothing$ off the natural numbers. Regard $(H,\in ,F,\omega )$ as a first-order structure, with a named object $\omega$. The graph of $F$ is an external set in the metatheory; it need not be a member of $H$. This structure is extensional and satisfies every separation instance in the pure membership language: any subset of $x\in H$ has rank below $\omega +\omega$ and belongs to $H$. It also satisfies, for natural $m<n$,since a subset of $V_{\omega +m}$ has rank at most $\omega +m$ and hence belongs to $V_{\omega +n}$.

Apply the indiscernible construction proved in question 7(i) to a Skolem expansion of this set structure, requiring $(c_i{)}_{i\in \mathbb{Z}}$ to be increasing members of its named natural-number object. For every finite fragment use an increasing sequence of ordinary natural numbers in the initial structure in the Ramsey argument. The resulting generated elementary hull $M$ has an external structure automorphism $j$ with $j(c_i)=c_{i+1}$, preserving $F$ and the named $\omega$. Put $D_i=F^M(c_i)$ and $D=D_0$. Then $j(D_i)=D_{i+1}$, and $M$ satisfies the displayed subset-containment assertion between successive domains. No claim that $M$ satisfies replacement is required.

All membership and subset statements about these objects are interpreted inside $M$. The hull is necessarily externally nonstandard: the natural-number objects $\dots ,c_{-2},c_{-1},c_0$ give an external descending chain. The external structure automorphism is not an object that must be definable within that first-order model.

On the external domain $D$ defineThis is the rank-shifting NFU model with explicit set predicate. The guard $S(y)$ is essential: an object not contained in $D$ could still have some ordinary members in $D$, and must not thereby become a nonempty atom. The objects failing $S$ have no $E$-members. If two objects satisfying $S$ have the same $E$-members, their images under $j$ are subsets of $D$ with the same members, so extensionality in $M$ and injectivity of $j$ make the original objects equal. Thus set-extensionality and the atom condition hold.

It remains to prove every comprehension instance. Use the domains $D_i$ just constructed, equivalently $D_i=j^i(D)$, for integer sorts. The map $j$ is a bijection from the objects of $D_i$ to the objects of $D_{i+1}$, and every subset of $D_i$ that exists in $M$ is an element of $D_{i+1}$. In sort $i$ define the set predicate by $y\subseteq D_{i-1}$; from sort $i$ to $i+1$ define membership by ordinary membership guarded by $y\subseteq D_i$.

For a stratified formula, assign integer types to its variables with equal types for equality and adjacent types for membership. Shift all types uniformly if necessary so the finitely many used types are nonnegative. Replace a variable of type $i$ by $j^i$ of its untyped value, quantify it over $D_i$, and use the typed relations just described. Applying $j^i$ to the untyped definition shows that each atomic first-order formula has the same truth value under this translation; induction on connectives and quantifiers proves the same for the whole first-order formula. Importantly the translated first-order formula is an ordinary first-order formula of $M$ with finitely many domain and transformed-parameter constants, not a first-order formula containing the external structure automorphism as a predicate.

If the comprehension variable has type $i$, separation in $M$ produces its extension $X\subseteq D_i$. The subset-containment assertion gives $X\in D_{i+1}$. Put $y=j^{-(i+1)}(X)\in D$. Then $j(y)=j^{-i}(X)\subseteq D$, hence $S(y)$, and for every $x\in D$,This verifies stratified comprehension with arbitrary parameters. The empty extension gives a distinguished empty set; the always-true first-order formula gives a universal set. The unstratified Russell predicate is not an allowed comprehension instance. ThusStarting from the actual set structure $H$ and applying compactness constructs a set first-order model, so this is a consistency proof in ZFC, not merely an assumption of Con(ZFC). It proves the requested consistency with an explicit interpretation; it does not depend on treating the much stronger full extensionality of NF as already consistent. The external $j$ is a tool of the construction, not an extra symbol permitted in NFU's comprehension scheme.

In the game $G(A)$, where $A\subseteq {\omega }^{\omega }$, the players alternately choose natural numbers and produce a sequence $x$. Player I wins exactly when $x\in A$. A strategy in an infinite game assigns a move to every finite position where its player is to move; it is winning if every play following it is won by that player. A game is determined if one player has a winning strategy. Both cannot have winning strategies, since their joint play would have to be won by both. The axiom of determinacy asserts

Without assuming AD, one can prove determinacy for finite games by backward induction, and for games with open or closed payoffs by the following argument. Give ${\omega }^{\omega }$ its product topology of discrete move coordinates. If $A$ is open, let $W_0$ be the finite positions whose entire cylinder of extensions lies in $A$. Form a winning-position attractor in an infinite game by transfinite recursion: add an I-position if some child has already been added, and add an II-position if all its children have already been added. Take unions at limits. The process stabilizes because there are only set-many positions and each genuine successor change adds a new one.

An added nonterminal position receives its least entry ordinal. At an I-position in the stable attractor, choose a child of smaller rank; at an II-position every child has smaller rank. Thus every play following I's rank-decreasing strategy must reach $W_0$: an infinite strictly descending sequence of ordinals cannot exist. The resulting play is in $A$, so I wins from any position in the attractor.

Outside the attractor, every child of an I-position remains outside, and an II-position has some child outside. Otherwise the ranks of all its children would have a set-sized supremum and the parent would have been added at a later stage. II chooses such a child and keeps the play outside $W_0$ forever. If its outcome were in the open set $A$, some finite prefix would have its entire cylinder in $A$, contrary to this avoidance. Therefore II wins from outside. With natural-number moves the least suitable child may be chosen at each position, so no choice axiom is needed for these strategies. This proves open determinacy. For a closed payoff, apply the same argument to its open complement with the players' winning roles interchanged. Clopen payoffs are included in both results.

Now assume AC and construct an undetermined game. The set of strategies for either player has cardinality $\mathfrak{c}=2^{{\aleph }_0}$, since a strategy is a function on a countably infinite set of positions. Each fixed strategy is compatible with $\mathfrak{c}$ plays: the other player's sequence of freely chosen moves determines distinct full plays. Use choice to enumerate I's strategies as $({\sigma }_{\alpha }{)}_{\alpha <\mathfrak{c}}$ and II's as $({\tau }_{\alpha }{)}_{\alpha <\mathfrak{c}}$, and well-order the space of plays.

Recursively choose $x_{\alpha }$ following ${\sigma }_{\alpha }$ and $y_{\alpha }$ following ${\tau }_{\alpha }$, with all selected plays distinct. At stage $\alpha$ fewer than $\mathfrak{c}$ plays have been used, while each compatible-play set has size $\mathfrak{c}$, so fresh choices exist; take the least ones in the fixed well-order. This does not require regularity of $\mathfrak{c}$, since $|\alpha |<\mathfrak{c}$ for every $\alpha$ below its initial ordinal.

Put $A=\{y_{\alpha }:\alpha <\mathfrak{c}\}$. Every I-strategy ${\sigma }_{\alpha }$ has the losing compatible play $x_{\alpha }\notin A$, and every II-strategy ${\tau }_{\alpha }$ has the losing compatible play $y_{\alpha }\in A$. Therefore neither player has a winning strategy. This choice diagonalization of an undetermined game provesThe elementary determinacy proofs above concern specified definable payoff classes; the diagonal construction explains why extending them to every arbitrary subset of the play space is incompatible with choice.

Number the natural numbers from one. If the power set could be enumerated as $A_1,A_2,\dots$, form $D=\{j:j\notin A_j\}$. For each $j$, membership of $j$ in $D$ is opposite to its membership in $A_j$, so $D\ne A_j$. This contradicts completeness of the enumeration. Thus the Cantor diagonal argument proves that the power set is an uncountable set.

For a finite subset $A$, define $b(A)={\sum }_{j\in A}{2}^{j-1}$. This is a nonnegative integer and the map is injective: if two finite subsets differ, let $k$ be their largest differing element. Its contribution $2^{k-1}$ exceeds the sum $1+2+\cdots +2^{k-2}$ of all smaller possible contributions, so their codes differ. There are infinitely many singleton subsets. Hence the countability of finite subsets of a countable set here is countable infinitude.

To treat all bijections, associate to every subset $A\subseteq \mathbb{N}$ the permutation that swaps $2j-1$ and $2j$ when $j\in A$, and fixes both when $j\notin A$. These disjoint pair swaps define a bijection of $\mathbb{N}$. Different subsets give different permutations, so the map from the power set into $X$ is injective. Therefore $X$ is an uncountable set.

Every finitely supported permutation fixes all elements above some $k$. A bijection fixing that tail must permute $\{1,\dots ,k\}$: it cannot send a smaller element into the fixed tail without violating injectivity. There are exactly $k!$ such permutations. Thusis a countable union of finite sets. Enumerate first by $k$, then by one of the finitely many permutations at that $k$; duplicates can be omitted. Infinitely many distinct pair swaps belong to $Y$, so $Y$ is countably infinite. The full family is uncountable, while the finitely supported family is countably infinite.

A countable set is a set that is finite or can be put in bijection with the natural numbers; equivalently, it admits an injective function into the natural numbers. To enumerate the rational numbers, list pairs $(p,q)$ with $p\in \mathbb{Z}$, $q\ge 1$ in increasing order of $|p|+q$, and then list $p/q$, discarding repetitions. Each level is finite and every rational eventually appears. An enumeration $q_1,q_2,\dots$ of the rationals then enumerates $\mathbb{Q}\times \mathbb{Q}$ by listing $(q_i,q_j)$ in diagonals of constant $i+j$. ThusTo prove that the real numbers are uncountable, suppose $(0,1)$ had an enumeration $x_1,x_2,\dots$. Write each $x_n$ in its decimal expansion that does not end in an infinite string of nines. Define a new decimal $y=0.d_1{d}_2\cdots$ by taking $d_n=2$ if the $n$th digit of $x_n$ is $1$, and $d_n=1$ otherwise. Then $y\in (0,1)$, has no ambiguous terminating/nines representation, and differs from $x_n$ at digit $n$. The Cantor diagonal argument contradicts the supposed enumeration. Consequently the real numbers are uncountable.

For the discs, use the enumeration of ${\mathbb{Q}}^2$ just constructed. Every positive-radius disc contains a rational-coordinate point: its interior contains a small open rectangle, and each coordinate interval contains a rational number. For each disc $D$, let $j(D)$ be the least index of an enumerated rational-coordinate point in its interior. This minimum exists and requires no arbitrary simultaneous choice. Disjoint discs cannot receive the same point, so $D\mapsto j(D)$ is an injective function into the natural numbers. This proves the countability of pairwise disjoint open disks and also the same conclusion for closed discs, by using their disjoint interiors.

For circles, take all concentric circles centred at the origin with radii $r\in (1,2)$. Two different radii give disjoint circles, and every radius is positive. The radius map is a bijection from the uncountable interval $(1,2)$ to this family. An uncountable disjoint family of nontrivial circles is therefore possible: the rational-point argument for discs relies on their nonempty interiors, which circumferences do not have.

The axiom of determinacy, $\mathsf{AD}$, says that every infinite game of perfect information in which two players alternately choose natural numbers and Player I wins according to a specified payoff $A\subseteq {\omega }^{\omega }$ is a determined infinite game. A strategy specifies a move at every finite position where its player moves; it is winning if every compatible infinite play is won by that player.

Finite games are determined by backward induction. For infinite games, open determinacy has a direct proof. Let $T$ consist of the finite positions whose extension cylinder sets lie in the open payoff $A$. Build the winning-position attractor in an infinite game by transfinite recursion: start with $T$, add an I-position if some successor has already entered, add a II-position if every successor has already entered, and take unions at limit stages. The process stabilizes because there are only countably many finite positions.

At a position in the attractor but outside $T$, I can choose a successor with strictly smaller entry rank, and every move of II has smaller entry rank. There is no infinite descending sequence of ordinals, so I forces entry into $T$ and wins. Outside the attractor, every I-move stays outside it, while II can choose a successor outside it; that strategy avoids $T$ forever. Since $A$ is open, a play belongs to $A$ exactly when some prefix belongs to $T$. Thus II wins from outside the attractor. This proves determination for open payoffs; swapping the players proves it for closed payoffs too. The stronger Borel determinacy theorem gives determination for every Borel set payoff, whereas $\mathsf{AD}$ asserts this for arbitrary payoffs.

To contradict $\mathsf{AD}$ from choice, enumerate both players' strategies as $⟨{\sigma }_{\alpha }:\alpha <\mathfrak{c}⟩$ and $⟨{\tau }_{\alpha }:\alpha <\mathfrak{c}⟩$, where $\mathfrak{c}=2^{{\aleph }_0}$. There are $\mathfrak{c}$ strategies because the set of finite positions is countable; there are also $\mathfrak{c}$ plays following any fixed strategy because the other player's moves are arbitrary. At stage $\alpha$ of the choice diagonalization of an undetermined game, choose a fresh play $x_{\alpha }$ following ${\sigma }_{\alpha }$ and then a different fresh play $y_{\alpha }$ following ${\tau }_{\alpha }$, avoiding all earlier reservations. This is possible because fewer than $\mathfrak{c}$ plays have been reserved at every $\alpha <\mathfrak{c}$; regularity of $\mathfrak{c}$ is not required.

Set $A=\{y_{\alpha }:\alpha <\mathfrak{c}\}$. Every ${\sigma }_{\alpha }$ loses on $x_{\alpha }\notin A$, and every ${\tau }_{\alpha }$ loses on $y_{\alpha }\in A$. Thus neither player has a winning strategy, and AD and AC are inconsistent together.

An ultraproduct assembles a family of nonempty first-order structures $⟨M_i:i\in I⟩$ in one fixed language, using an ultrafilter $U$ to specify which coordinate statements hold almost everywhere. Its universe isFunctions are interpreted coordinatewise, and an $r$-ary relation holds on $[f_1],\dots ,[f_r]$ when its coordinate truth set belongs to $U$. Changing representatives changes such a truth set only off a $U$-large set, so these definitions are well defined. A principal ultrafilter simply recovers its distinguished factor; a nonprincipal ultrafilter can produce new elements and structures.

The fundamental result is the Łoś theorem:Prove it by structural induction on first-order formulas. Atomic formulas are the definition. Intersections handle conjunction, while exactly one of a set and its complement belongs to an ultrafilter, handling negation. For the existential step, if the coordinate existential truth set is in $U$, use choice to select witnesses there and arbitrary values elsewhere; their product function is an ultraproduct witness. Conversely a product witness makes the coordinate existential truth set contain a $U$-large set. Other connectives and universal quantification follow.

In an ultrapower, all factors are one structure $M$. The diagonal map $a\mapsto [\text{constant }a]$ is an elementary embedding by the Łoś theorem. For a nonprincipal $U$ on $\omega$, the class of $n\mapsto n$ in an ultrapower of the standard natural numbers is larger than every constant natural number, since each tail belongs to $U$. This gives a nonstandard model of Peano arithmetic. Similarly, an ultraproduct of finite fields ${\mathbb{F}}_{p_i}$ with distinct primes tending to infinity has characteristic zero: for every fixed positive integer $m$, the equation $m1=0$ fails in all sufficiently late factors. An infinite structure can therefore arise from finite factors.

Ultraproducts prove the compactness theorem. If each finite subset $\Delta$ of a theory $T$ has a model $M_{\Delta }$, index these models by $[T{]}^{<\omega }$. The cones $\{\Delta :\phi \in \Delta \}$ have the finite intersection property. Extend their filter to an ultrafilter using the ultrafilter lemma. The Łoś theorem makes the resulting ultraproduct satisfy every $\phi \in T$. This construction also gives elementary extensions realizing prescribed finitely satisfiable types.

A useful stronger property is countable saturation of an ultraproduct over omega. Let a countable type $\{{\phi }_n(x,{\bar{a}}_n):n<\omega \}$ be finitely satisfiable in an ultraproduct over a nonprincipal $U$ on $\omega$. Choose coordinate representatives of its parameters. Let $E_n$ be the coordinates where the first $n$ formulas have a simultaneous witness; then $E_n\in U$, the $E_n$ decrease, and we can replace $E_n$ by $E_n\cap [n,\infty )$. At coordinate $i$, choose a witness to the first $m(i)$ formulas, where $m(i)$ is the largest $n\le i$ with $i\in E_n$, taking an arbitrary value if there is none. For each fixed $n$, the set where $m(i)\ge n$ is $U$-large. Its class realizes the entire type. This proof uses countable incompleteness of $U$, not arbitrary completeness or automatic saturation at larger cardinals.

Finally, complete ultrafilters connect ultraproducts to large cardinals. A countably complete ultrafilter prevents an externally infinite descending membership chain in an ultrapower of a well-founded universe: intersect the countably many coordinate truth sets and obtain an impossible descending chain at one coordinate. A kappa-complete nonprincipal ultrafilter on $\kappa$ consequently supplies the ultrapower embedding associated with a measurable cardinal. Thus the same construction supports both elementary compactness arguments and large-cardinal embeddings, with the completeness hypothesis deciding which well-foundedness and closure properties survive.