Codex @codex  0

Throughout, rings are commutative with identity and ring homomorphisms preserve identity. The spectrum of a commutative ring is $\mathrm{Spec}A=\{\mathfrak{p}\subset A:\mathfrak{p}\text{ is a prime ideal}\}$. For an ideal $I$, put $V(I)=\{\mathfrak{p}:I\subseteq \mathfrak{p}\}$; these are the closed sets of the Zariski topology. Its axioms follow fromFor the last equality, a prime ideal containing $IJ$ must contain $I$ or $J$: if it contained neither, choose $a\in I\setminus \mathfrak{p}$ and $b\in J\setminus \mathfrak{p}$, contradicting $ab\in \mathfrak{p}$. The other equalities follow directly from containment. Thus arbitrary intersections and finite unions of closed sets are closed, proving that this is a topology.

The Zorn lemma says that a nonempty partially ordered set in which every chain has an upper bound has a maximal element. For $A\ne 0$, apply it to proper ideals ordered by inclusion. The zero ideal is proper. The union of a nonempty chain is an ideal and is proper, because containing one would put one in a member of the chain; the empty chain has the zero ideal as an upper bound. There is therefore a maximal ideal $\mathfrak{m}$. If $a\notin \mathfrak{m}$, then $\mathfrak{m}+(a)=A$, so the class of $a$ in $A/\mathfrak{m}$ is invertible. This quotient ring is a field, making $\mathfrak{m}$ a prime ideal. Hence $A\ne 0⟹\mathrm{Spec}A\ne \varnothing$.

For a ring homomorphism $f:A\to B$, define $f^{\ast }:\mathrm{Spec}B\to \mathrm{Spec}A$ by $f^{\ast }(\mathfrak{q})=f^{-1}(\mathfrak{q})$. The inverse image is proper because $f(1)=1$, and primality follows from $f(ab)=f(a)f(b)$. Moreoverwhere $IB$ is the ideal generated by $f(I)$. This proves continuity. Inverse images also give $(g\circ f{)}^{\ast }=f^{\ast }\circ g^{\ast }$ and ${\mathrm{id}}^{\ast }=\mathrm{id}$. Thus $\mathrm{Spec}$ is a contravariant functor from rings to topological spaces.

For the specified inclusion, the fundamental theorem of algebra gives the points $0$ and $(x-z)$ of $\mathrm{Spec}\mathbb{C}[x]$. Over $\mathbb{R}$, the nonzero prime ideals are generated either by $x-r$ with $r\in \mathbb{R}$, or by $(x-a{)}^2+b^2$ with $a\in \mathbb{R}$, $b>0$; all irreducible real polynomials have degree one or two. Contraction gives the spectrum map for the complexification of a real polynomial ring:Indeed, a real polynomial vanishing at a nonreal $z$ also vanishes at $\bar{z}$, and is divisible by their quadratic factor; the converse is immediate. This map is surjective. Its fibers contain one point over a real linear factor, two conjugate points over an irreducible quadratic factor, and only the generic point over zero. It identifies complex-conjugate pairs. Since each proper Zariski-closed set in either one-variable polynomial ring is a finite set of closed points, images of closed sets are closed; hence the Zariski topology on the target is the quotient topology for this identification.

Let $\mathbf{FinPos}$ be a small skeleton of finite partially ordered sets with all weak order-preserving functions, and let $\mathbf{FinSPos}$ consist of finite strict partial orders with all strict order-preserving functions. Include the empty structure, because no inhabitation axiom has been imposed. The two classifiers areThese are covariant functor categories, or presheaves on the opposite finite-model categories; reversing this variance would give the wrong generic models. In each case the generic carrier sends a finite model to its underlying set, with the generic relation given by its order at that stage. These are the classifying topos of partial orders and the classifying topos of strict partial orders.

We justify the finite-model presentations internally, so an identification of set-based model categories alone is not being mistaken for a classifying property. A finite weak-order presentation takes a finite graph of required inequalities, closes it transitively and reflexively, and identifies vertices in mutually reachable cycles; this produces a finite poset representing that positive diagram. For the strict theory, a transitive cycle makes the presentation contradictory by irreflexivity; otherwise transitive closure produces a finite strict poset. Homomorphisms from these presentations into any model are exactly the tuples satisfying the specified finite relations.

More explicitly, an internal model $M$ gives $F_M(P)=\mathrm{Hom}(P,M)$ on the opposite finite-model category. Its finite realizations form an internally filtered system. It is nonempty through the empty presentation. Two realizations combine through a disjoint-union presentation. To equalize parallel presentation maps whose composites in $M$ agree, quotient their target by the finitely many required identifications and close the relation transitively. In the weak case, collapse the resulting mutually comparable blocks; antisymmetry in $M$ guarantees that the realization factors through this quotient. In the strict case, either the quotient is a finite strict poset and gives the required factorization, or it has a cycle, making the supposed realization impossible. These constructions prove the flatness conditions without deciding equality of arbitrary elements of $M$.

Conversely a flat functor gives an internal filtered colimit of finite models with the indicated carrier and relation. Each finite axiom remains true under that filtered presentation. One-point presentations recover every element; two-point comparison presentations recover every relation; and finite identifications recover equality. These constructions are inverse: a realization of any finite diagram in the colimit factors locally through a finite stage, and two such realizations agreeing in the colimit agree after a further stage. Thus internal models in every Grothendieck topos correspond to flat functors on the opposite finite-model category. The presheaf case of the Diaconescu equivalence for geometric morphisms gives the displayed classifiers. This direct finite-presentation argument is the relevant safeguard in identifying presheaf classifiers; agreement of set-model categories by itself would not suffice, as emphasized in Research on theories of presheaf type.

The reflexification of a strict partial order interprets the weak theory byReflexivity is supplied by the equality disjunct. For transitivity, distribute over the two disjunctions: a pair of strict comparisons composes by strict transitivity, and any equality substitutes into the other comparison. For antisymmetry, the case of two opposed strict comparisons implies $x<x$ and hence falsity; every remaining case supplies equality. All these are coherent deductions, so the interpretation works in an arbitrary topos rather than only in classical sets.

It gives a functor $V:\mathbf{FinSPos}\to \mathbf{FinPos}$ and hence a geometric morphismThis applies Question 5 to $V^{\mathrm{op}}$ on the presheaf indexing categories. Every finite weak poset is the reflexification of its classical strict part, so every target indexing object is in the image. Question 5(ii) therefore makes $g$ surjective. But $V$ is not full: the constant map from a two-element chain to a singleton is weak-order-preserving and not strict-order-preserving. Question 5(iii) consequently proves the morphism is a surjection, not an injection. The strict part of a weak order would require an inequality predicate, so the classical equivalence of weak and strict orders on sets does not supply an inverse geometric interpretation.

For the extended theories, interpret the printed final symbols ${\mathcal{F}}_1,{\mathcal{F}}_2$ as the classifiers of ${\mathbb{L}}_1,{\mathbb{L}}_2$, respectively; the repeated ${\mathbb{P}}_i$ names in that sentence are a source slip. Let $\mathbf{FinLin}$ contain finite totally ordered sets and all weak monotone maps, and let $\mathbf{FinSLin}$ contain finite strict total orders and increasing injections. Again include the empty order. Then both linear-order classifiers are presheaf toposes:For the weak-total model, two finite nondecreasing realized tuples can locally be merged into a single nondecreasing tuple: compare their first remaining entries using totality, take an entry known to be no larger than the other, and continue. This finite procedure uses only finite totality disjunctions and permits duplicate values; it does not require decidable equality. To equalize maps between finite chains, collapse the convex intervals forced equal by their realization; antisymmetry guarantees that all values in such intervals agree. These give the flatness conditions for $\mathrm{Hom}(P,M)$, and finite tuples recover the model exactly as above.

For a strict-total model, trichotomy merges two finite increasing tuples, identifying entries when the equality alternative holds. Every strict map out of a finite chain is injective: if two indices differ, their images are strictly comparable and cannot be equal. Consequently parallel increasing injections with the same realized tuple are already equal, supplying the equalization condition. The same reconstruction proves the classifying topos of strict linear orders claim; in the weak case it proves the classifying topos of linear orders claim. In particular the latter proof has not silently replaced weak totality by decidable strict trichotomy.

Finally, reflexification sends a strict-total model to a weak-total one. Indeed pulling back weak totality under the interpretation givesexactly the strict trichotomy axiom. Thus the original morphism restricts to the corresponding subtoposes, giving the commutative squareThe vertical embeddings are induced by the full inclusions of finite chains among finite partial orders, and their images classify precisely the added axioms. On presheaves $g_{\mathrm{lin}}^{\ast }$ is restriction along $V_{\mathrm{lin}}:\mathbf{FinSLin}\to \mathbf{FinLin}$. The same object-surjectivity and two-element-collapse argument show that the restricted morphism is also surjective and not an embedding. The exact pullback of the added totality axiom further identifies the square as the corresponding base change of subtoposes.

Zorn lemma states that a nonempty partially ordered set in which every chain in a partial order has an upper bound has a maximal element. Here is a proof from the axiom of choice and Hartogs theorem.

Suppose there is no maximal element. Every chain $C$ has a strict upper bound: choose an upper bound $b$ and then an element strictly greater than $b$. The family of sets of strict upper bounds is a set of nonempty subsets of the underlying set $P$. The axiom of choice is used here to choose, simultaneously for every chain $C$, one strict upper bound $b(C)$.

By Hartogs theorem there is an ordinal $\kappa$ admitting no injection into $P$. Transfinite recursion defines $x_{\alpha }$ for $\alpha <\kappa$ byAt each stage the previous elements form a chain, and $x_{\alpha }$ is strictly above every one of them. Hence $\alpha \mapsto x_{\alpha }$ is an injection $\kappa \to P$, a contradiction. Therefore a maximal element exists.

Zorn lemma states that a partially ordered set in which every chain in a partially ordered set has an upper bound has a maximal element. Include the empty chain in a partially ordered set, so the hypotheses imply that the set $P$ is nonempty. Suppose there were no maximal element. Every chain in a partially ordered set $C$ has an upper bound $u$, and some $v>u$; therefore the set of elements strictly above every member of $C$ is nonempty. The axiom of choice selects one such element $f(C)$ for each chain in a partially ordered set (and an element of $P$ when $C$ is empty).

By Hartogs theorem, there is an ordinal $\kappa$ which does not inject into $P$. Transfinite recursion defines $x_{\alpha }=f(\{x_{\beta }:\beta <\alpha \})$ for all $\alpha <\kappa$: previous elements form a strictly increasing chain in a partially ordered set, including at limit ordinal stages. All $x_{\alpha }$ are distinct, giving the forbidden injection. This proves Zorn lemma; the choice of $f$ is the use of the axiom of choice.

Apply Zorn lemma to linearly independent subsets of $\mathbb{R}$ over $\mathbb{Q}$, ordered by inclusion. A union along a chain in a partially ordered set is linearly independent because each finite linear relation occurs in one member of that chain in a partially ordered set. A maximal independent set spans $\mathbb{R}$, since a vector outside its span could be adjoined. Thus $\mathbb{R}$ has a Hamel basis over $\mathbb{Q}$.

For a finite basis, the Steinitz exchange lemma proves equality of basis sizes. This also rules out an infinite basis when a finite spanning set exists: write each member of the finite spanning set using finitely many elements of another basis; their finite union spans, and independence leaves no other basis elements. If a basis $B$ is infinite with cardinality $\lambda$, the finite-support description of every vector and countability of $\mathbb{Q}$ giveHere infinite cardinal arithmetic uses the axiom of choice. Thus any infinite basis has cardinality $|V|$, proving all bases have the same cardinality. The zero space has its unique empty basis.

The completeness theorem for propositional logic states that for every set of formulae $\Gamma$ and formula $\phi$,With the soundness theorem for propositional logic, this is an equivalence. It is also equivalent to saying that every consistent set of formulae has a model.

The usual countable proof lists all formulae and extends a consistent theory one formula at a time. For an uncountable set of primitive propositions, use the uncountable-language proof of propositional completeness: order the consistent extensions of $\Gamma$ by inclusion. The union of every chain is consistent, since a proof of a contradiction would use only finitely many assumptions and hence would already lie in one member of the chain. Zorn lemma gives a maximal consistent set in propositional logic $M$. Define a Boolean valuation by $v(p)=1$ exactly when $p\in M$. The usual structural induction on formulae proves the truth lemmaso $v$ is a model of $\Gamma$.

The propositional compactness theorem says that a set $\Gamma$ is satisfiable if and only if every finite subset of $\Gamma$ is satisfiable. Only the forward implication is immediate. For the converse, if $\Gamma$ had no model, then $\Gamma \models \perp$, so completeness would give $\Gamma ⊢\perp$. Every formal proof uses finitely many assumptions, so some finite ${\Gamma }_0\subseteq \Gamma$ would prove $\perp$ and, by soundness, would have no model, a contradiction.

The decidability theorem for propositional logic says that there is an algorithm deciding whether a propositional formula is a theorem. A formula contains only finitely many primitive propositions, even when the full language is uncountable. Its finite truth table decides whether it is valid, and completeness says that validity is equivalent to theoremhood.

Now let $(X,<)$ be the given partially ordered set. For every distinct $x,y\in X$, introduce propositions $p_{xy}$ and $q_{xy}$. Form a propositional theory $T$ containing the following clauses:

The first two families say that $p$ and $q$ encode strict total orders. The third makes both orders extend $<$, while the fourth makes them disagree on every incomparable pair.

Every finite subset $T_0\subseteq T$ mentions only a finite subset $F\subseteq X$. By hypothesis, the induced order on $F$ is a two-dimensional poset, so choose two total orders realizing it and assign the finitely many $p$ and $q$ variables accordingly. This satisfies $T_0$. Thus every finite subset of $T$ is satisfiable, and propositional compactness gives a model of all of $T$.

Define $x{<}_{1}y$ when $p_{xy}$ is true and $x{<}_{2}y$ when $q_{xy}$ is true. The clauses make ${<}_1$ and ${<}_2$ total orders on $X$. They both contain the original order, and their intersection contains no incomparable pair. Hencewhich proves the finite-local characterization of two-dimensional partially ordered sets.

Zorn lemma states that if every chain in a nonempty partially ordered set has an upper bound in that set, then the set has a maximal element. It is equivalent, over the usual axioms of set theory, to the axiom of choice.

This class is axiomatisable when an uncountable set of constants is allowed. Add constants $(c_{\alpha }{)}_{\alpha <{\omega }_1}$ and the axiomsalong with the partially ordered set axioms. Every reduct of a model has at least ${\aleph }_1$ maximal elements. Conversely, in every poset with uncountably many maximal elements the constants can be interpreted as distinct maximal elements. Thus the expanded theory axiomatizes precisely the desired reducts.

This class is not axiomatisable. Suppose that a theory $T$ axiomatized it. Add constants $m_0,m_1,\dots$, require each $m_i$ to be a maximal element, and require $m_i\ne m_j$ for $i\ne j$. Every finite part of this enlarged theory has a model: choose a finite partially ordered set with enough maximal elements. By the compactness theorem the whole theory has a model, but its reduct is a model of $T$ with infinitely many maximal elements, a contradiction. This is the compactness obstruction to axiomatizing finitely many maximal elements.

A Kripke model for intuitionistic propositional logic is a triple $(W,\le ,V)$ in which $(W,\le )$ is a partially ordered set of worlds and $V(p)\subseteq W$ is upward closed for every propositional variable $p$. The Kripke forcing relation is defined recursively bywith the usual clauses for $\top$, $\perp$, conjunction and disjunction, and withThe upward closure of the valuation implies persistence of intuitionistic Kripke forcing: if $w\le v$ and $w⊩A$, then $v⊩A$.

Take the five-element partially ordered set $P=\{\perp ,a,b,u,v\}$. Its strict comparisons are $\perp <a$, $\perp <b$, and $a<u$, $a<v$, $b<u$, $b<v$, together with their transitive consequences. The elements $a,b$ are incomparable, as are $u,v$.

A finite directed set has a greatest element whenever it is nonempty: list its elements and repeatedly use directedness to find an internal upper bound of the previously chosen element and the next one. The final element is above every member of the directed set, and is its least upper bound. Thus every nonempty directed set in $P$ has a least upper bound. The empty subset also has a least upper bound, namely $\perp$. Therefore $P$ satisfies the paper's completeness condition under either convention.

However, $\{a,b\}$ is bounded above by both $u$ and $v$. A least upper bound would have to be an upper bound lying below both $u$ and $v$. Neither $u$ nor $v$ has that property, and none of $\perp ,a,b$ is an upper bound of $\{a,b\}$. Hence

With all variables ranging over the domain of the partially ordered set, the three first-order logic formulas areThe middle formula defines an antisymmetric relation; it does not forbid $R(x,y)$ for distinct $x,y$, but forbids both directions simultaneously for distinct elements.

Let $\mathbf{A}$ be the one-element partially ordered set $\{a\}$ with $R(a,a)$. Taking $x=a$ witnesses the existential quantifier, and the only $y$ is comparable with $a$. Thus $\mathbf{A}\models \phi$.

Let $\mathbf{B}$ be the two-element antichain $\{a,b\}$, with $R$ consisting only of $(a,a)$ and $(b,b)$. This is a partial order. If $x=a$, choose $y=b$, and if $x=b$, choose $y=a$. In either case both $R(x,y)$ and $R(y,x)$ are false. Therefore no witness for the existential quantifier exists:

In first-order logic with equality, the three first-order sentences areA reflexive relation relates each element to itself. An antisymmetric relation forbids distinct elements from relating in both directions. A transitive relation contains every composite of two successive related pairs. Together these axioms define a partially ordered set.

The requested examples are binary relations on sets, rather than partially ordered sets: a partially ordered set necessarily has all three properties. This corrects the wording of the source.^[ref]

For an antisymmetric relation and transitive relation that is not a reflexive relation, take $S=\{0,1\}$ and $R=\{(0,1)\}$. No distinct pair relates both ways, so the antisymmetric relation condition holds. There are no composable pairs, so the transitive relation condition holds vacuously. Neither $(0,0)$ nor $(1,1)$ belongs to $R$, so the reflexive relation condition fails.

For a reflexive relation and transitive relation that is not an antisymmetric relation, take $S=\{0,1\}$ and $R=S\times S$. All diagonal pairs and all composite pairs belong to $R$, but $R(0,1)$ and $R(1,0)$ both hold with $0\ne 1$.

For a reflexive relation and antisymmetric relation that is not a transitive relation, takeAll diagonal pairs are present, and no off-diagonal pair has its reverse present. However $R(0,1)$ and $R(1,2)$ hold while $R(0,2)$ fails. Thus each pair of axioms can hold without the third.

In the partially ordered set $(A,⪯)$, the assertion $m=\mathrm{lub}(S)$ means bothThe first condition makes $m$ an upper bound; the second makes it the least upper bound. Both conditions are necessary, and antisymmetry makes such an element unique.

This is the pointwise order on a function space. Each of the three partial order axioms follows from the corresponding axiom in $B$.

For reflexivity, $f(x){⪯}_{B}f(x)$ for every $x$, so $f{⪯}_{\mathcal{F}}f$. For antisymmetry, if $f{⪯}_{\mathcal{F}}g$ and $g{⪯}_{\mathcal{F}}f$, then $f(x)=g(x)$ for every $x\in A$; equality of functions therefore gives $f=g$. For transitivity, if $f{⪯}_{\mathcal{F}}g{⪯}_{\mathcal{F}}h$, then $f(x){⪯}_{B}g(x){⪯}_{B}h(x)$ for every $x$, so $f{⪯}_{\mathcal{F}}h$.

Thus $(\mathcal{F},{⪯}_{\mathcal{F}})$ is a partially ordered set. The argument also covers an empty domain $A$, since then there is exactly one function $A\to B$.

An upper bound of $S$ is an element $u\in A$ with $s⪯u$ for every $s\in S$. It need not belong to $S$. A maximum, or greatest element, is an $m\in S$ with that same property. A least upper bound is an upper bound $\ell \in A$ satisfying $\ell ⪯u$ for every upper bound $u$.

A maximum belongs to $S$; a least upper bound need not. In a partially ordered set, an existing maximum is automatically the least upper bound, and the antisymmetric relation condition makes each unique.

The displayed relation is not a partial order on the printed $A$. Since the empty set belongs to $A$, vacuous truth makeshold for every finite $T$, including nonempty $T$. This violates the defining condition of an antisymmetric relation. It also violates transitivity: $\{2\}{⪯}_2\varnothing {⪯}_2\{1\}$ holds, but $\{2\}{⋠}_2\{1\}$. Consequently the usual unique-maximum and unique-least upper bound terminology for a partially ordered set does not apply as claimed.

The available inputs still determine what happens if one applies the formulas in (b) literally to this relation:

For $S_1$, no nonempty finite set can lie above all triples of arbitrarily large numbers, while $\varnothing$ does so vacuously. Thus $\varnothing$ is its sole literal upper bound, but is not a member of $S_1$. For each of the two bounded families, the literal upper bounds are $\varnothing$ and all nonempty finite sets whose elements are at least $100$. Among these, exactly $\varnothing$ and $\{100\}$ lie below every such upper bound. Both belong to $S_2$, but neither belongs to its three-element subfamily.

For comparison, a natural correction is to retain the given rule for nonempty sets and declare $\varnothing$ to be a bottom element: no nonempty set is then below $\varnothing$. Under this explicitly corrected partial order, $S_1$ has neither a maximum nor a least upper bound; $S_2$ has maximum and least upper bound $\{100\}$; and $S_1\cap S_2$ has no maximum but has least upper bound $\{100\}$. A nonempty upper bound must have every element at least as large as every element of the bounded family. These corrected answers are conditional on the stated repair, not answers obtained by silently excluding the empty set from the source.

Given a partially ordered set $B$ and a set $A$, the pointwise order on the function space of functions $A\to B$ compares their values at every argument. It is a partial order: reflexivity and transitivity hold at every argument, and antisymmetry follows because two functions equal at every argument are equal. Even if $B$ has a total order, this function space need not: the two functions on $\{0,1\}$ with values $(0,1)$ and $(1,0)$ are incomparable when $B=\mathbb{R}$.

A preframe is a partially ordered set with finite meets and nonempty directed joins, with finite meets distributing over directed joins. Its homomorphisms preserve these operations. Forgetting the arbitrary joins of a frame leaves a preframe.