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A maximal external point of a finite set has no other point strictly larger in every coordinate. This weak nondominance differs from a maximal element in the product order when coordinate ties exist: for example $(1,0)$ passes the weak test against $(1,1)$ but is not maximal in the product order. In an independent continuous-coordinate random sample, ties have probability zero, so the two notions coincide almost surely.

For $v=(v_1,v_2,v_3)$ define the skew-symmetric matrixThen $\hat{v}x=v\times x$. This is a linear isomorphism ${\mathbb{R}}^3\to \mathfrak{so}(3)$, with inverse $\varphi (\xi )=({\xi }_{32},{\xi }_{13},{\xi }_{21})$. Since rotations preserve the cross product,Thus $\varphi$ intertwines the Adjoint representation with the standard representation. The vector triple-product identity gives, for every $x$,Consequently $[\hat{v},\hat{w}]=\widehat{v\times w}$ and the chosen normalization satisfiesNo undetermined scale remains in this explicit matrix convention.

Identify $SU(2)$ with the unit quaternions usingThe relations $jz=\overline{z}j$ and $j^2=-1$ verify multiplication, and every matrix in $SU(2)$ has this form. If $v$ is imaginary, then $\overline{v}=-v$. Quaternionic conjugation reverses product order and $q^{-1}=\overline{q}$, soThus conjugation preserves $\mathrm{Im}\mathbb{H}\cong {\mathbb{R}}^3$. Multiplicativity of the quaternion norm shows that this action preserves the inner product. The unit quaternions form the connected three-sphere, so its determinant, equal to one at the identity, is always one. We obtain a homomorphism $C:SU(2)\to SO(3)$.

Its group kernel consists of unit quaternions commuting with every imaginary quaternion. Commuting with both $i$ and $j$ forces a quaternion to be real, so the group kernel is exactly $\{\pm 1\}$. To prove surjectivity explicitly, let $u$ be a unit imaginary quaternion and put $q=\cos (\theta /2)+u\sin (\theta /2)$. For imaginary $u,v$, multiplication satisfies $uv=-u\cdot v+u\times v$. Expanding $qv{q}^{-1}$ therefore givesThis is the Rodrigues rotation formula about axis $u$. Every element of $SO(3)$ has such an axis-angle description: an odd-dimensional orthogonal matrix of determinant one has an eigenvector of eigenvalue one, and its action on the perpendicular plane is a plane rotation. Hence $C$ is onto, andThis is the quaternionic form of the Adjoint double cover from SU(2) to SO(3). Its differential sends an imaginary quaternion $u$ to $2\hat{u}$, because $[u,v]=2u\times v$, consistent with the bracket normalization used above.

Here a pointed complete partial order is a partially ordered set $D$ with a least element ${\perp }_D$ in which every nonempty directed set has a least upper bound. Directed means that every pair of its elements has an upper bound within that set. A Scott continuous map is monotone and preserves these directed suprema. It is not required to preserve the least element; restricting to strict maps would change the categorical claim. If completeness is instead formulated only for increasing countable chains, the same arguments below work with those chains.

Order the function space of complete partial orders $[D\to E]$ pointwise: $f\le g$ means $f(x)\le g(x)$ for every $x$. Its least element is the constant function with value ${\perp }_E$. For a directed family $\mathcal{F}$ of continuous maps, define $h(x)={\sup }_{f\in \mathcal{F}}f(x)$. This is well-defined and is its pointwise least upper bound. It is monotone. For any directed $A\subseteq D$,The interchange is valid because the combined family $\{f(a):f\in \mathcal{F},a\in A\}$ is directed, and either iterated supremum is its least upper bound. Thus $h$ is a Scott continuous map, proving that $[D\to E]$ is a pointed complete partial order.

Finite products carry the product order, componentwise suprema and least element $({\perp }_D,{\perp }_E)$. The one-element order is a terminal object. For the exponential, use the function space with evaluation $\mathrm{ev}(f,x)=f(x)$. To check joint continuity, take a directed family of pairs $(f_i,x_i)$. ThenIndeed, directedness supplies a $k$ above the pairs indexed by both $i$ and $j$, giving $f_i(x_j)\le f_k(x_k)$. Thus the diagonal values are cofinal in the double family, and evaluation preserves its supremum.

For a continuous $h:A\times D\to E$, its currying is $\Lambda h:A\to [D\to E]$, $(\Lambda h)(a)(d)=h(a,d)$. Each section is continuous because $d\mapsto (a,d)$ is; pointwise directed suprema in the $a$ coordinate show that $\Lambda h$ is continuous. Conversely a continuous $g:A\to [D\to E]$ gives the continuous map $\mathrm{ev}\circ (g\times {\mathrm{id}}_D)$. The two constructions are inverse, because they agree on every pair $(a,d)$. This gives the natural bijectionwith the required evaluation map. Together with the terminal object and products, it proves cartesian closure of pointed complete partial orders with non-strict continuous maps.

First, the product order is a partial order. Reflexivity and transitivity hold in each coordinate. If $(p,q){\le }_R(p^{\prime },q^{\prime })$ and $(p^{\prime },q^{\prime }){\le }_R(p,q)$, antisymmetry in $P,Q$ gives $p=p^{\prime }$ and $q=q^{\prime }$, proving antisymmetry in the product.

Let $D$ be a nonempty directed set in $P\times Q$. Its coordinate projectionsare directed sets. For example, if $p_1,p_2\in D_P$, choose $(p_1,q_1),(p_2,q_2)\in D$. Directedness supplies $(p_3,q_3)\in D$ above both, so $p_3\in D_P$ is a common upper bound of $p_1,p_2$ inside $D_P$. The argument for $D_Q$ is identical.

By the assumed directed-complete partial order property, $p_{\ast }=\sup D_P$ and $q_{\ast }=\sup D_Q$ exist. Every $(p,q)\in D$ satisfies $p{\le }_P{p}_{\ast }$ and $q{\le }_Q{q}_{\ast }$, so $(p_{\ast },q_{\ast })$ is an upper bound of $D$. If $(u,v)$ is any other upper bound, then $u$ bounds $D_P$ and $v$ bounds $D_Q$. Hence $p_{\ast }{\le }_{P}u$, $q_{\ast }{\le }_{Q}v$, proving leastness. ThusThe printed definition of a directed set also permits $D=\varnothing$.^[ref] Under that convention, the hypotheses require $P,Q$ to have least elements, namely the least upper bounds of their empty subsets. Their pair is the least element of the product and is $\sup \varnothing$ there. This handles the empty case as well; under the convention restricting directed sets to nonempty sets, only the preceding nonempty argument is needed.

Under the product order, every point of the open unit disc has both coordinates below one, so $(1,1)$ is an upper bound. Any upper bound $(u,v)$ must have $u\ge 1$ and $v\ge 1$, since the two coordinate projections of the open unit disc are $(-1,1)$. Therefore $(1,1){\le }_P(u,v)$, proving ${\sup }_{{\le }_P}S=(1,1)$.

Under the lexicographic order, every point $(1,v)$, with arbitrary real $v$, is an upper bound, because every point of $S$ has first coordinate strictly below one. Conversely, an upper bound cannot have first coordinate $u<1$: a point $(x,0)\in S$ with $x>u$ would exceed it. If an upper bound has $u>1$, the upper bound $(1,0)$ is strictly smaller. If it is $(1,v)$, the upper bound $(1,v-1)$ is strictly smaller. Hence there is no least upper bound under the lexicographic order. This illustrates the supremum of an open disc under product and lexicographic orders.

Check each partial order axiom. For every $(a,b)$, $a\le a$ and $b\ge b$, so $S$ is a reflexive relation. If both $(a,b)S(c,d)$ and $(c,d)S(a,b)$, the first coordinates satisfy $a\le c\le a$ and the second coordinates satisfy $b\ge d\ge b$. Therefore $a=c$ and $b=d$, which proves $S$ is an antisymmetric relation.

Finally, if $(a,b)S(c,d)$ and $(c,d)S(e,f)$, then $a\le c\le e$ and $b\ge d\ge f$. Hence $(a,b)S(e,f)$, making $S$ a transitive relation. All three axioms hold, so $S$ is a partial order. It is a product order, with the usual order on the first coordinate and the reversed order on the second.

For the product order, reflexivity follows from $a_i\le a_i$ in every coordinate. If $\mathbf{a}⪯\mathbf{b}$ and $\mathbf{b}⪯\mathbf{a}$, the antisymmetric relation axiom for the usual real order gives $a_i=b_i$ for every $i$, hence $\mathbf{a}=\mathbf{b}$. If $\mathbf{a}⪯\mathbf{b}⪯\mathbf{c}$, transitivity in each coordinate gives $a_i\le c_i$, hence $\mathbf{a}⪯\mathbf{c}$. Thus this is a partial order.

For $n\ge 2$, the vectors $(1,0,0,\dots )$ and $(0,1,0,\dots )$ are incomparable: each has a coordinate larger than the corresponding coordinate of the other. It is not a total order for $n\ge 2$. The dimension qualification matters: for $n=1$ this is the usual total order on $\mathbb{R}$, so the printed non-totality claim needs the implicit assumption $n\ge 2$.

For $D=\{(x,y):x^2+y^2<1\}$, the product order gives least upper bound $(1,1)$, because each coordinate has supremum one. The lexicographic order instead has no least upper bound. Every point $(1,v)$ is an upper bound, but $(1,v-1)$ is a strictly smaller upper bound; a point with first coordinate below one cannot bound the disc. The same geometric set therefore has different order-completeness behavior under the two partial orders. Neither conclusion is determined by the usual drawing alone.