The first two clauses now require $x=2z$ and $y=2z$. Thus they already imply $x=y$. Choose, for example, $x=1,y=2$; no natural number $z$ can satisfy both equations. Consequently the existential quantifier fails for this pair, irrespective of the universally quantified final clause, andIn fact even $x=y=1$ fails, since $1=2z$ has no solution among the natural numbers.

Here the direction of the divisibility relation matters: $R(x,z)$ means $z\mid x$. The first two clauses say that $z$ is a common divisor of $x,y$. The last clause says that every common divisor $w$ divides $z$. These are precisely the defining properties of the greatest common divisor. Choose $z=\gcd (x,y)$; thenHenceThis also holds if the natural numbers include zero: use $\gcd (x,0)=x$ and $\gcd (0,0)=0$. Every natural number divides zero, so for $x=y=z=0$ the last clause remains true. In the order obtained by reversing divisibility, the greatest common divisor is the least upper bound.

Every pair of natural numbers has a maximum, itself a natural number. Given arbitrary $x,y$, choose $z=\max \{x,y\}$. Then $x,y\le z$, and any $w$ satisfying $x,y\le w$ also satisfies $z\le w$. All three clauses of $\varphi$ hold. ThusThis explicitly supplies the witness required by the existential quantifier for every pair chosen by the two universal quantifiers.

In this first-order structure, $R(x,z)\wedge R(y,z)$ says $x\le z$ and $y\le z$. Thus $z$ is an upper bound of $x,y$. The universally quantified clause says that if $w$ is any other upper bound, then $z\le w$. Therefore $z$ is their least upper bound. In the usual total order on the natural numbers, this is their maximum:This describes a condition on the values assigned to the free variables $x,y,z$; $\varphi$ itself is not a closed first-order sentence.

Totality of the binary relation is expressed byWhen $R$ is already an order, this adds comparability and makes it a total order. In fact this first-order sentence alone entails $\rho =\forall xR(x,x)$.

By the definition of semantic entailment, $\tau \models \rho$ means that every first-order structure satisfying $\tau$ also satisfies $\rho$. Let $\mathcal{M}$ be any such first-order structure, and let $a$ be any element of its domain of a first-order structure. The instance of $\tau$ with $x=y=a$ gives $R(a,a)\vee R(a,a)$, which is true only if $R(a,a)$ is true. Since $a$ was arbitrary, $\mathcal{M}\models \forall xR(x,x)$. ThereforeThere is no exception for $x=y$ in the displayed totality axiom; that case is exactly what forces the reflexive relation property.

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 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.

We prove the result algebraic partially ordered sets are continuous directly from the definitions. First, $a\ll b$ implies $a\le b$, by applying the way-below relation to the singleton directed set $\{b\}$. Also,if $c\le \sup D$, then $b\le \sup D$, so the witness required for $a\ll b$ also works for $a\ll c$.

Fix $x$ and putBy algebraicity, $K_x$ is a directed set with least upper bound $x$. Every $k\in K_x$ satisfies $k\ll k\le x$, hence $k\ll x$. Thus $K_x\subseteq A_x\subseteq \{a:a\le x\}$.

To prove that $A_x$ is a directed set, take $a,b\in A_x$. Applying $a\ll x$ and $b\ll x$ to $x=\sup K_x$ gives $k_a,k_b\in K_x$ such that $a\le k_a$ and $b\le k_b$. Directedness of $K_x$ gives $k\in K_x$ above both. Since $k\in A_x$, it is the required common upper bound inside $A_x$. With the usual nonempty convention, $A_x$ is nonempty since it contains $K_x$.

By directed completeness $A_x$ has a least upper bound. The inclusions above givesoThe same proof handles the paper's literal empty-set convention. If $K_x=\varnothing$, then $x=\sup \varnothing =\perp$, and the defining implication for $a\ll x$, applied to $D=K_x$, shows that $A_x$ must also be empty. Its least upper bound is again $x$. If $K_x$ is nonempty, the argument already given applies without change.

The continuous nonalgebraic unit interval isEvery nonempty directed set has a least upper bound by completeness of the real numbers, and that least upper bound lies in $[0,1]$. Thus this is a directed-complete partial order.

For $0\le a<b\le 1$, we have $a\ll b$. Indeed, if $b\le \sup D$ and no $d\in D$ were at least $a$, then $a$ would be an upper bound for $D$, implying $\sup D\le a<b$, a contradiction. If $a>b$, the singleton directed set $\{b\}$ excludes $a\ll b$. If $a=b>0$, the increasing sequence $d_n=b(1-1/n)$, $n\ge 1$, has least upper bound $b$ and has no member at least $b$, so $b/\ll b$.

Under the usual nonempty convention, $0\ll 0$. Therefore the elements way-below $b>0$ form $[0,b)$, and those way-below $0$ form $\{0\}$. These are directed sets with least upper bounds $b$ and $0$, respectively, proving that $[0,1]$ is a continuous directed-complete partial order. Its only compact element is $0$. For any $b>0$, the compact elements below $b$ have least upper bound $0$, rather than $b$, so it is not an algebraic partially ordered set.

Under the literal empty-set convention of the paper, $0/\ll 0$ as well. The elements way-below zero then form the empty set, which is directed in that convention and has least upper bound zero. There are no compact elements, and their empty set still fails to approximate any positive $b$. Thus both conventions give the same requested classification:

Use the two-chain noncontinuous complete partial orderwhere $\perp <a_1<a_2<\cdots <\top$ and $\perp <b_1<b_2<\cdots <\top$, with no comparisons between the two interiors. In this solution $\sup D$ denotes the least upper bound printed as $\sqcap D$ in the question.^[ref]

Every nonempty directed set containing $\top$ has least upper bound $\top$. A directed set omitting $\top$ cannot contain both an $a_i$ and a $b_j$, because their only common upper bound in $P$ is $\top$, which would have to belong to that directed set. It therefore lies in one chain, possibly with $\perp$. If its indices are bounded it has a greatest element, which is its least upper bound; if they are unbounded its least upper bound is $\top$. Thus $P$ is a directed-complete partial order.

No element other than $\perp$ is way-below $\top$. For $a_i$, the directed set $\{b_n:n\ge 1\}$ has least upper bound $\top$, but none of its members is above $a_i$. Interchanging the two chains excludes every $b_i$. Either chain excludes $\top$ itself. On the other hand, $\perp \ll \top$: every directed set whose least upper bound is above $\top$ is nonempty, and any member is above $\perp$. HenceSo $P$ is not a continuous directed-complete partial order.

There is a small convention issue in the printed definitions. Our usual directed set is nonempty, but the question's pairwise condition also allows the empty set. Taken literally, the empty set has least upper bound $\perp$, and the preceding example and failure of continuity still work. Under this literal convention $\perp /\ll \perp$, because the empty set supplies no witness. We retain this qualification in the next parts.

The logarithm requires $y>0$, and $x>0$ is given. Set $z=\log (y/x)$, so $y=x{e}^z$ and the chain rule yields $x{y}^{\prime }/y=1+x{z}^{\prime }$. Dividing the ordinary differential equation by $y$ therefore givesSince $(z/x{)}^{\prime }=(x{z}^{\prime }-z)/x^2=0$, we obtain $z=Cx$, including the case $z=0$. Thus the general positive solution is $y=x{e}^{Cx}$. The initial condition $y(1)=e$ gives $C=1$:For the separate decay requirement, $x{e}^{Cx}\to 0$ precisely when $C<0$: if $C=0$, the result is $x$, and if $C>0$, it grows. Consequently the full family with the stated limit isThis latter family does not also satisfy $y(1)=e$; the solution selected by that initial condition grows. The identically zero function is excluded by the original logarithm.