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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.

All assertions of independence below are relative consistency assertions. We construct structures satisfying the other ZF axioms but falsifying the indicated axiom; an ordinary model of ZF supplies the positive side. The convenient ground theory is ZFC, which has the same consistency strength as ZF by the constructible universe theorem.

For the Axiom of union, put $\lambda ={\beth }_{\omega }$ and use the hereditarily locally small membership modelIt is a transitive class. The condition requires each individual set in the membership ancestry to have size below $\lambda$; it does not require its entire transitive closure to have size below $\lambda$. Extensionality and foundation are inherited. Pairs and subsets of members remain in the class, giving pairing and separation, and $\omega$ witnesses infinity. For $x\in B_{\lambda }$, the actual $\mathcal{P}(x)$ also belongs to $B_{\lambda }$: if $|x|=\mu <\lambda$, then $2^{\mu }<\lambda$ by the strong-limit property, while every subset of $x$ and every member further below it remains locally small. This verifies the full power set axiom. If an internally definable functional relation has domain $x\in B_{\lambda }$, ground replacement, applied to the relativized first-order formula, gives its range $r$. Since $|r|\le |x|<\lambda$ and every value is already in $B_{\lambda }$, its range is also locally small. Thus every replacement instance holds.

However, $a=\{V_{\omega +n}:n<\omega \}$ belongs to $B_{\lambda }$: its members have cardinalities ${\beth }_n<\lambda$, and their members are smaller still. Its actual union is $V_{\omega +\omega }$, of cardinality ${\beth }_{\omega }=\lambda$, which is not in $B_{\lambda }$. Transitivity makes any internal union witness equal to this actual union. Hence Union fails while all the other ZF axioms hold.

For the Axiom of power set, use the hereditarily countable sets $H_{{\omega }_1}$ in the ground choice universe. This is transitive and contains $\omega$. Pairs, subsets and unions have countable transitive closure. For replacement, a functional image of a countable domain is countable, and the union of countably many countable transitive closures is countable; hence that image is again hereditarily countable. These observations, with inherited extensionality and foundation, verify all the remaining axioms. Every real, viewed as a subset of $\omega$, belongs to $H_{{\omega }_1}$. A power-set witness for $\omega$ would therefore contain every such real, but a hereditarily countable set has only countably many members, whereas $\mathcal{P}(\omega )$ is uncountable by Cantor's theorem. Power set fails.

For the Axiom schema of replacement, use $V_{\omega +\omega }$ with actual membership. This limit-rank structure contains $\omega$ and is closed under pairs, unions and power sets: every relevant finite increase in rank stays below $\omega +\omega$. A separated subset has rank no higher than that of the original set. Extensionality and foundation are inherited. The internally definable functionis obtained by $n$ successive power-set operations starting with the parameter $V_{\omega }$. Each finite iteration, including its finite history, belongs to the structure. Its range has rank $\omega +\omega$ and is not an element of $V_{\omega +\omega }$. Thus the replacement instance for this first-order formula fails, although every other ZF axiom holds.

For the axiom of extensionality, take the well-founded cumulative set universe over two distinct urelements, and forget the predicate distinguishing urelements from sets. Each urelement has no members, just as the ordinary empty set has no members, so extensionality fails. Every other axiom survives in this pure membership language. Pairing, union, separation and replacement are witnessed by the corresponding ordinary sets of urelements and sets; foundation follows from their well-founded ranks, and the pure $\omega$ supplies infinity. Here foundation is written with “has a member” as its nonemptiness antecedent, and disjointness means having no common member. Formulations using inequality to one designated empty object are not equivalent after removing extensionality. There is also one necessary adjustment to power set: an urelement is vacuously a subset of every object because it has no members. Thus the power-set witness for $x$ is the ordinary set of all usual subsets of its extension, together with both urelements. This is still a set, and contains exactly the objects satisfying the unrestricted subset first-order formula. Consequently the countermodel falsifies Extensionality alone, without silently weakening Power set to sets only.

Finally, the hereditarily finite sets $V_{\omega }$ satisfy extensionality, foundation, pairing, union, power set and separation. Every functional image of a finite domain is a finite set of hereditarily finite objects, hence also belongs to $V_{\omega }$, giving all replacement instances. No finite set contains $\varnothing$ and is closed under successor: it would then contain every finite von Neumann ordinal. Infinity fails while the remaining axioms hold.

An uncountable cardinal $\kappa$ is a measurable cardinal if there is a nonprincipal $\kappa$-complete ultrafilter $U$ on $\kappa$: intersections of fewer than $\kappa$ members of $U$ remain in $U$. Such an ultrafilter is uniform. No singleton belongs to it, and intersecting the complements of fewer than $\kappa$ singletons shows that no set of cardinality below $\kappa$ belongs to it.

Form the ultrapower of the universe by $U$, with classes $[f]$ of functions $f:\kappa \to V$, and $[f]{\in }_U[g]$ exactly when $\{\xi :f(\xi )\in g(\xi )\}\in U$. The Łoś theorem proves that the map taking $x$ to the constant function with value $x$ is elementary. The ultrapower relation is well-founded: from an infinite descending sequence $[f_0]{\ni }_U[f_1]{\ni }_U\cdots$, countable completeness gives a coordinate satisfying all the membership relations, producing an actual infinite descending membership sequence and contradicting foundation. It is also set-like. Predecessors of $[g]$ can be represented by functions whose value at $\xi$ lies in $g(\xi )\cup \{\varnothing \}$; these functions form a set. The Mostowski collapse theorem therefore gives a transitive class $N$ and an elementary embeddingThe usual class-ultrapower notation is understood via set-sized representatives and this set-like collapse.

For $\alpha <\kappa$, any function into $\alpha$ is constant on a $U$-large set: if none of its fewer-than-$\kappa$ fibers belonged to $U$, intersect their complements to get the empty set in $U$. Consequently every member of the ultrapower ordinal represented by the constant $\alpha$ is represented by a constant ordinal below $\alpha$. Induction on $\alpha$ now gives $j(\alpha )=\alpha$.

On the other hand the identity function $d(\xi )=\xi$ represents an ordinal below $j(\kappa )$. For every $\alpha <\kappa$, the tail $\{\xi :\xi >\alpha \}$ belongs to $U$, so the collapsed ordinal $[d]$ is above $j(\alpha )=\alpha$. HenceThis proves nonidentity explicitly. The collapsed identity class is at least $\kappa$; it need not equal $\kappa$ unless an additional normality convention is imposed on $U$.

Conversely, an elementary embedding $j:V\to N$ into a transitive class with critical point $\kappa$ yieldsFor the converse use the usual amenability assumption on the class embedding, so this collection is a set. Since $j(\kappa )>\kappa$, exactly one of $j(X)$ and $j(\kappa \setminus X)$ contains $\kappa$, proving the ultrafilter condition. Singletons are excluded because $j(\alpha )=\alpha <\kappa$. If $\lambda <\kappa$ and $X_i\in U$ for $i<\lambda$, then $j$ fixes $\lambda$ and every index below it, giving $j({\bigcap }_{i<\lambda }{X}_i)={\bigcap }_{i<\lambda }j(X_i)$; the intersection still contains $\kappa$. Thus $U$ is $\kappa$-complete and $\kappa$ is measurable.

In particular the first moved ordinal is a strongly inaccessible cardinal. If $\mathrm{cf}(\kappa )<\kappa$, partition $\kappa$ into fewer than $\kappa$ bounded intervals. Each is too small to lie in $U$, contradicting completeness. Thus $\kappa$ is regular. If $2^{\lambda }\ge \kappa$ for some $\lambda <\kappa$, choose distinct subsets $A_{\xi }\subseteq \lambda$ for $\xi <\kappa$. For each $\eta <\lambda$, choose the $U$-large side of the question $\eta \in A_{\xi }$. Intersect these fewer-than-$\kappa$ large sides. The intersection belongs to $U$, but all its indices have identical $A_{\xi }$, so it contains at most one index, contradicting nonprincipality. Therefore $2^{\lambda }<\kappa$ for every $\lambda <\kappa$, as required.

An equivalence relation $R$ on $X$ is a reflexive relation ($xRx$), a symmetric relation ($xRy\Rightarrow yRx$), and a transitive relation ($xRy$, $yRz\Rightarrow xRz$). Its equivalence class at $x$ is $[x{]}_R=\{y\in X:xRy\}$. Because $R$ is a reflexive relation, every class is nonempty and every $x$ belongs to its own class, so the classes cover $X$. If $[x{]}_R$ and $[z{]}_R$ share an element $y$, symmetry and transitivity give $xRz$. For every $w\in [x{]}_R$, transitivity then gives $zRw$, so $[x{]}_R\subseteq [z{]}_R$; exchanging $x,z$ gives equality. Thus distinct classes are disjoint. The equivalence classes form a set partition of $X$. On the empty set, the empty family is the corresponding partition.

For (1)(a), a primitive recursive function is obtained by a finite construction from the zero function, successor function and projection functions, using function composition in recursion theory and primitive recursion. The recursion rule, with parameters $\mathbf{x}$, isHere $g,h$ have already been constructed. All these functions are total on nonnegative integers: the initial functions are total, composition preserves totality, and ordinary mathematical induction proves totality of each recursion.

For (1)(b), use a shifted hyperoperation sequence, beginning with addition. Defineand, for each fixed $r\ge 0$, defineThe first three members are therefore addition, multiplication and exponentiation:The exponentiation convention includes $a^0=1$, hence $0^0=1$. The next member iterates exponentiation, with $H_3(a,0)=1$; there is no need for an exceptional domain restriction at $a=0$.

For (1)(c), $H_0$ is a primitive recursive function by its displayed recursion. If $H_r$ is a primitive recursive function, thenare primitive recursive functions, using constants, projection functions and function composition in recursion theory. Primitive recursion applied to $g_r,h_r$ gives $H_{r+1}$. Mathematical induction on the fixed rank $r$ proves every member of the sequence is primitive recursive. This does not assert that the joint three-variable evaluator $(r,a,b)\mapsto H_r(a,b)$ is a primitive recursive function.

For (2), as written, the answer is no: the empty set $X=\varnothing$ is semidecidable, whereas a total computable function from $\mathbb{N}$ to $\mathbb{N}$ always has a nonempty range. This is the only obstruction. To prove the stronger useful statement, suppose $X$ is a nonempty computably enumerable set and choose $x_0\in X$. Fix a Turing machine which halts exactly on the inputs in $X$. Its bounded halting predicateis primitive recursive: encode configurations arithmetically, iterate its total single-step operation $t$ times by primitive recursion, and check the halt state. A halted configuration is kept fixed. This is a bounded simulation, not unbounded waiting.

Fix a primitive recursive pairing function $⟨x,t⟩$ with primitive recursive inverse coordinate functions. For example, the Cantor pairing function has inverses obtained by bounded minimization. DefineThe two coordinate functions, the bounded predicate, and the finite case distinction are primitive recursive functions, so $f$ is a primitive recursive function. Its values always lie in $X$. Conversely, if $x\in X$, its computation halts within some finite $t$, and $f(⟨x,t⟩)=x$. ThusThe range need not be a computable set; deciding whether a value occurs anywhere in this total enumeration is an unbounded search.

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.

For a measurable map, each preimage ${\theta }^{-1}(A)$ lies in $\mathcal{E}$, so ${\mu }_{\theta }$ is well-defined and nonnegative. The preimage of the empty set is empty, and preimages commute with disjoint countable unions, givingfor pairwise disjoint measurable $A_j$. Thus ${\mu }_{\theta }$ is a measure, the pushforward measure ${\theta }_{\ast }\mu$. Its total mass is $\mu (E)$, because ${\theta }^{-1}(E)=E$. Finiteness is not needed for this construction, but is used in the uniqueness argument next.

The map swaps each adjacent pair of binary digits: $({\omega }_1,{\omega }_2,{\omega }_3,{\omega }_4,\dots )\mapsto ({\omega }_2,{\omega }_1,{\omega }_4,{\omega }_3,\dots )$. Canonical expansions with no eventual all-ones tail remain canonical under this swap, and ${\theta }^2=\mathrm{id}$.

Under Lebesgue measure, any specified values of $m$ distinct binary digits have probability $2^{-m}$. Indeed they describe a union of dyadic intervals at the largest of the specified positions, with that total length. The preimage of a cylinder prescribing its first $m$ output digits prescribes $m$ distinct input digits and therefore has the same measure. The half-open dyadic intervals, with the empty set, form a pi-system generating the Borel sigma-algebra. Apply the preceding uniqueness result to obtainThe choice of binary expansion matters at dyadic endpoints for the map's pointwise definition; the chosen convention removes that ambiguity and those endpoints are null in the measure calculation.

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.

Write $N=(n_1,n_2,n_3)$. Restriction to the plane $z=0$ givesThis is an equation of degree at most two in two Cartesian coordinates, the general algebraic form of a conic, with possible degenerate conics and an empty set of real points. No determinant hypothesis is needed for this substitution.

For an arbitrary plane, choose a point ${\mathbf{x}}_0$ and an orthonormal basis $\mathbf{e},\mathbf{f}$ of its direction linear subspace. Put $B=(\mathbf{e}\mathbf{f})$ and $\mathbf{x}={\mathbf{x}}_0+B\mathbf{\xi }$, where $\mathbf{\xi }=(s,t{)}^T$. With $S=(M+M^T)/2$, the symmetric part of a matrix, the restricted equation isThis proves the plane-section result whenever “conic” allows the appropriate degeneracies. A rigid motion taking the chosen plane to $z=0$ gives the same argument.

The source's assumption $\det M\ne 0$ does not guarantee a nondegenerate conic. Even with $M=I$ and $f=x^2+y^2+z^2-1$, the section $z=1$ is a single point, and $z=2$ has no real points. With the nonsingular symmetric matrix giving $f=2xz+y^2$, the section $z=0$ is the repeated straight line $y=0$.

There is a further literal omission: the source does not assume $M$ is symmetric. Forone has $\det M=2$ but $f=2xz$. Thus the section $z=0$ is the entire plane, rather than a conic. Only $S$ controls the quadratic form; invertibility of an arbitrary $M$ cannot remove this exception. If the intended hypothesis is that $M$ is symmetric and nonsingular, the quadratic restriction to a plane cannot vanish identically: otherwise $B^{T}MB=0$, so $M$ would map a two-dimensional linear subspace injectively into its one-dimensional orthogonal complement, which is impossible. Degenerate and empty sections still remain possible.

The original PDF places complement bars on the second $A$ in each parenthesis and on the final $A$. Thus the first parenthesis is the empty set, whereas the second is the whole sample space. Their union with any further subset is still the whole space:The bars are missing from the local OCR, so using its unbarred expression would give the wrong result.

The empty set is bounded. Otherwise choose $p$ in the compact metric space $X$. The open balls $B(p,n)$, $n=1,2,\dots$, form an open cover. A finite subcover exists by compactness; because the balls are nested, the largest one already covers $X$. Hence $X\subseteq B(p,N)$ for some $N$.

For $x,y\in X$, the triangle inequality gives $d(x,y)\le d(x,p)+d(p,y)<2N$. ThusThe proof uses only compactness and the metric, not a coordinate system.

A metric space is connected when it has no separation of a topological space: it cannot be written as the union of two disjoint nonempty open subsets. Equivalently, the only subsets that are both open and closed are the empty set and the whole space.The open sets here are those of the metric topology.

Use $z\overline{z}=|z{|}^2$ and $\overline{B}z+B\overline{z}=2\mathrm{Re}(\overline{B}z)$. When $A\ne 0$, completing the square givesThus, with $\Delta =|B{|}^2-AC$,If $\Delta >0$, this is a circle with centre $-B/A$ and radius $\sqrt{\Delta }/|A|$. If $\Delta =0$, it is the single point $-B/A$, sometimes called a degenerate circle of radius zero. If $\Delta <0$, it is the empty set, because a squared complex modulus cannot be negative.

When $A=0$ and $B\ne 0$, write $B=u+iv$ and $z=x+iy$. The remaining equation iswhich is a straight line with normal vector $(u,v)$. It passes through zero exactly when $C=0$. Finally, when $A=B=0$, the equation has no solutions if $C\ne 0$ and every complex number is a solution if $C=0$.

The complete classification of a real complex-circle equation therefore includes two degeneracies in addition to ordinary circles, straight lines and the empty set:For the three advertised alternatives to be exhaustive, one must exclude the identically zero equation and either exclude or explicitly allow radius-zero circles. This is a qualification of the unrestricted coefficients, not a reason to discard the genuine circle and straight line cases used next.

Newton's first law identifies a class of inertial frames in which isolated bodies move uniformly, while the second relates changes of momentum to resultant force. Both have definition-like roles: the first helps characterize an inertial frame, and the second can be used to calibrate force given independently measured mass and acceleration. That does not make the whole scheme an empty set of definitions.

If force meant nothing except whatever product $m\mathbf{a}$ happens to be observed, the second law alone would have little independent predictive content. Actual mechanics also specifies gravitational, elastic and other interactions with independently constrained parameters, superposition rules and reaction relations. The combined framework predicts the accelerations of new arrangements. One cannot arbitrarily change one body's mass or redefine every unexplained acceleration as a different force without affecting those cross-system comparisons.

Likewise, an inertial frame cannot be separately tailored to make each trajectory straight while retaining a common interaction account. A rotating frame introduces coordinated velocity-dependent accelerations for different freely moving bodies. The empirical claim is that a common family of frames exists with the prescribed relations, at least over the applicable domain. Tests involve judgments about isolation, clocks, geometry and interaction laws, so they are not independent of all auxiliary assumptions.

Thus definitions organize the theory but do not exhaust its content. Whether one calls $\mathbf{F}=d\mathbf{p}/dt$ a definition or a dynamical postulate depends partly on how the remaining force concepts are introduced. What is testable is the consistent package that connects independently characterized interactions, mass ratios and motion. Relativity and gravitation show how that package can be revised when its domain is exceeded.^[ref]

A topological space is connected if it has no separation: two disjoint nonempty open sets whose union is the whole space. Equivalently, its only subsets that are both open and closed are the empty set and the whole space.

A subset is connected when its subspace topology has this property. Thus the sets in a proposed separation of a subset need only be relatively open, rather than open in the ambient space.

If both factors are Hausdorff, two distinct product points differ in at least one coordinate. Disjoint open neighbourhoods in that coordinate, multiplied by the other whole factor, give disjoint open neighbourhoods of the product points. Thus the product is Hausdorff.

Conversely, assume the product is Hausdorff. Fix $y_0\in Y$, using nonemptiness. For $x\ne x^{\prime }$, choose disjoint open neighbourhoods of $(x,y_0)$ and $(x^{\prime },y_0)$ and refine them to basis rectangles $U\times V$ and $U^{\prime }\times V^{\prime }$. Since $y_0\in V\cap V^{\prime }$, their disjointness forces $U\cap U^{\prime }=\varnothing$. These separate $x,x^{\prime }$, proving that $X$ is Hausdorff. Fixing a point of $X$ proves the same for $Y$.The nonempty hypothesis is essential: a product with the empty set is empty regardless of the other factor. This is the Hausdorff property of a nonempty product.

Use the basis of a topology consisting of open rectangles:It covers $X\times Y$, since $X\times Y\in \mathcal{B}$. Whenever a point lies in two basis rectangles, their intersection is another basis rectangle containing it:These are precisely the two basis of a topology axioms. Arbitrary unions of the rectangles therefore form a topology, the product topology. To check the finite-intersection axiom directly, distribute the intersection of two unions over all pairs of rectangles and use the displayed formula. The empty union gives the empty set.

Every topological subspace $S\subseteq X$ again has the cofinite topology: its nonempty open subsets have finite complements in $S$, and every finite exclusion is obtained by intersecting with a cofinite open subset of $X$. As usual, the empty set is also open; this exception is implicit in the paper's shorthand definition.

If $S$ is infinite, any two nonempty relatively open sets intersect, since their complements are finite. Hence $S$ is hyperconnected space and therefore connected. If $S$ is finite, its cofinite topology is discrete topology. It is then connected exactly when it has at most one point. ThusThere are also infinite disconnected subsets of the product. Choose distinct $a,b\in X$ and put $D=\{a,b\}\times X$. This is infinite. The traces on $D$ of the ambient open setsare respectively $\{a\}\times X$ and $\{b\}\times X$. They are disjoint, nonempty and cover $D$, giving a separation. This answers the infinite-subset question, rather than only producing two isolated product points.

For the canonical projection of a quotient set $p$, defineThe inverse images of the empty set and whole target are $\varnothing$ and $X$. Inverse images commute with arbitrary unions and finite intersections, so the topology axioms follow from those of $\mathcal{T}$. This is the quotient topology, and it makes $p$ continuous by definition.

The crucial criterion is an equivalence: $U$ is open precisely when $p^{-1}(U)$ is open. This is stronger than continuity alone and is what makes the next part work.