In this section, we will focus on the relations between **IS** and **CS**. In particular, we will propose a full subcategory of **CS**, consisted of arity 2 convex spaces and study its relations with interval spaces.

#### Definition 5.1

A convex space (*X*, 𝓒) is called arity 2 if it satisfies

(AR2) ∀*A* ∈ 2^{X}, ∀*x*, *y* ∈ *A*, *co*^{𝓒}({*x*, *y*}) ⊆ *A* implies *A* ∈ 𝓒.

Let **CS(2)** denote the full subcategory of **CS**, consisted of arity 2 convex spaces.

Next we will study the relations between **CS** (**CS(2)**) and **IS**.

#### Proposition 5.2

*Let* (*X*, 𝓒) *be a convex space and define* 𝓘^{𝓒} : *X* × *X* ⟶ 2^{X} *by*

$$\begin{array}{}{\displaystyle \mathrm{\forall}x,y\in X,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\mathcal{J}}^{\mathcal{C}}(x,y)=c{o}^{\mathcal{C}}(\{x,y\})=\bigcap _{x,y\in A\in \mathcal{C}}A.}\end{array}$$

*Then* 𝓘^{𝓒} *is an interval operator on* *X*.

#### Proof

The verifications of (I1) and (I2) are straightforward and omitted.□

#### Proposition 5.3

*If* *f* : (*X*, 𝓒_{X}) ⟶ (*Y*, 𝓒_{Y}) *is a CP mapping*, *then* *f* : (*X*, 𝓘^{𝓒X}) ⟶ (*Y*, 𝓘^{𝓒Y}) *is a IP mapping*.

#### Proof

Since *f* : (*X*, 𝓒_{X}) ⟶ (*Y*, 𝓒_{Y}) is a CP mapping, it follows that *f*^{←}(*B*) ∈ 𝓒_{X} for each *B* ∈ 𝓒_{Y}. Then for each *x*, *y* ∈ *X*, we have

$$\begin{array}{}{\displaystyle {f}^{\leftarrow}({\mathcal{J}}^{{\mathcal{C}}_{Y}}(f(x),f(y)))}& =& {\displaystyle \bigcap _{f(x),f(y)\in B\in {\mathcal{C}}_{Y}}{f}^{\leftarrow}(B)}\\ & \supseteq & {\displaystyle \bigcap _{x,y\in {f}^{\leftarrow}(B)\in {\mathcal{C}}_{X}}{f}^{\leftarrow}(B)}\\ & \supseteq & {\displaystyle \bigcap _{x,y\in A\in {\mathcal{C}}_{X}}A={\mathcal{J}}^{{\mathcal{C}}_{X}}(x,y).}\end{array}$$

This implies that *f*^{→}(𝓘^{𝓒X}(*x*, *y*)) ⊆ 𝓘^{𝓒Y}(*f*(*x*), *f*(*y*)), as desired.□

By Propositions 5.2 and 5.3, we obtain a functor ℍ as follows:

$$\begin{array}{}{\displaystyle \mathbb{H}:\left\{\begin{array}{ccc}\mathbf{C}\mathbf{S}& \u27f6& \mathbf{I}\mathbf{S}\\ (X,\mathcal{C})& \u27fc& (X,{\mathcal{J}}^{\mathcal{C}})\\ f& \u27fc& f.\end{array}\right.}\end{array}$$

#### Proposition 5.4

*Let* (*X*, 𝓘) *be an interval space and define* 𝓒^{𝓘} *as follows*:

$$\begin{array}{}{\displaystyle {\mathcal{C}}^{\mathcal{J}}=\{A\in {2}^{X}\mid \mathrm{\forall}x,y\in A,\text{\hspace{0.17em}}\mathcal{J}(x,y)\subseteq A\}.}\end{array}$$

*Then* (*X*, 𝓒^{𝓘}) *is an arity 2 convex space*.

#### Proof

(C1) is obvious. We need only verify (C2), (C3) and (AR2).

(C2) Take any {*A*_{i}}_{i∈I} ⊆ 𝓒^{𝓘}. Then for each *i* ∈ *I* and for each *x*, *y* ∈ *A*_{i}, 𝓘(*x*, *y*) ⊆ *A*_{i}. This implies that

$$\begin{array}{}{\displaystyle x,y\in \bigcap _{i\in I}{A}_{i}}& \u27fa& {\displaystyle \mathrm{\forall}i\in I,x,y\in {A}_{i}}\\ & \u27fa& {\displaystyle \mathrm{\forall}i\in I,\mathcal{J}(x,y)\subseteq {A}_{i}}\\ & \u27fa& {\displaystyle \mathcal{J}(x,y)\subseteq \bigcap _{i\in I}{A}_{i}.}\end{array}$$

Hence, ⋂_{i∈I} *A*_{i} ∈ 𝓒^{𝓘}.

(C3) Take any {*A*_{j}}_{j∈J} $\begin{array}{}{\displaystyle \stackrel{dir}{\subseteq}}\end{array}$ 𝓒^{𝓘}. Then for each *x*, *y* ∈ $\begin{array}{}{\displaystyle {\bigcup ^{dir}}_{j\in J}}\end{array}$ *A*_{j}, there exist *j*_{1}, *j*_{2} ∈ *J* such that *x* ∈ *A*_{j1} and *y* ∈ *A*_{j2}. Since {*A*_{j}}_{j∈J} is directed, there exists *j*_{3} ∈ *J* such that *A*_{j1} ⊆ *A*_{j3} and *A*_{j2} ⊆ *A*_{j3}. This implies that *x*, *y* ∈ *A*_{j3} ∈ 𝓒^{𝓘}. Then it follows that 𝓘(*x*, *y*) ⊆ *A*_{j3} ⊆ $\begin{array}{}{\displaystyle {\bigcup ^{dir}}_{j\in J}}\end{array}$ *A*_{j}. This means $\begin{array}{}{\displaystyle {\bigcup ^{dir}}_{j\in J}}\end{array}$ *A*_{j} ∈ 𝓒^{𝓘}.

(AR2) Take any *A* ∈ 2^{X} such that

$$\begin{array}{}{\displaystyle \mathrm{\forall}x,y\in A,\text{\hspace{0.17em}}c{o}^{{\mathcal{C}}^{\mathcal{J}}}(\{x,y\})\subseteq A.}\end{array}$$

In order to show *A* ∈ 𝓒^{𝓘}, take any *x*, *y* ∈ *A*. It follows that

$$\begin{array}{}{\displaystyle c{o}^{{\mathcal{C}}^{\mathcal{J}}}(\{x,y\})=\bigcap _{x,y\in B\in {\mathcal{C}}^{\mathcal{J}}}B\supseteq \mathcal{J}(x,y).}\end{array}$$

Then we have 𝓘(*x*, *y*) ⊆ *co*^{𝓒𝓘}({*x*, *y*}) ⊆ *A* for each *x*, *y* ∈ *A*. This implies that *A* ∈ 𝓒^{𝓘}, as desired.□

#### Proposition 5.5

*If* *f* : (*X*, 𝓘_{X}) ⟶ (*Y*, 𝓘_{Y}) *is a IP mapping*, *then* *f* : (*X*, 𝓒^{𝓘X}) ⟶ (*Y*, 𝓒^{𝓘Y}) *is a CP mapping*.

#### Proof

Since *f* : (*X*, 𝓘_{X}) ⟶ (*Y*, 𝓘_{Y}) is a IP mapping, it follows that

$$\begin{array}{}{\displaystyle \mathrm{\forall}x,y\in X,\text{\hspace{0.17em}}{f}^{\to}({\mathcal{J}}_{X}(x,y))\subseteq {\mathcal{J}}_{Y}(f(x),f(y)).}\end{array}$$

Then for each *B* ∈ 𝓒^{𝓘Y}, take any *x*, *y* ∈ *f*^{←}(*B*). It follows that *f*(*x*), *f*(*y*) ∈ *B* ∈ 𝓒^{𝓘Y}. This means that 𝓘_{Y}(*f*(*x*), *f*(*y*)) ⊆ *B*. Further we have

$$\begin{array}{}{\displaystyle {\mathcal{J}}_{X}(x,y)\subseteq {f}^{\leftarrow}({\mathcal{J}}_{Y}(f(x),f(y)))\subseteq {f}^{\leftarrow}(B).}\end{array}$$

This implies that *f*^{←}(*B*) ∈ 𝓒^{𝓘X}, as desired.□

By Propositions 5.4 and 5.5, we obtain a functor 𝕂 : **IS** ⟶ **CS** as follows:

$$\begin{array}{}{\displaystyle \mathbb{K}:\left\{\begin{array}{ccc}\mathbf{I}\mathbf{S}& \u27f6& \mathbf{C}\mathbf{S}\\ (X,\mathcal{J})& \u27fc& (X,{\mathcal{C}}^{\mathcal{J}})\\ f& \u27fc& f.\end{array}\right.}\end{array}$$

#### Theorem 5.6

(𝕂, ℍ) *is an adjunction between* **IS** *and* **CS**.

#### Proof

Since 𝕂 and ℍ are both concrete functors, we need only verify that 𝕂 ∘ ℍ ⩽_{𝓒} 𝕀_{CS} and ℍ ∘ 𝕂 ⩾_{𝓘} 𝕀_{IS}. That is to say, for each (*X*, 𝓒) ∈ |**CS**| and (*X*, 𝓘) ∈ |**IS**|, 𝓒^{𝓘𝓒} ⩽_{𝓒} 𝓒 and 𝓘 ⩽_{𝓘} 𝓘^{𝓒𝓘}.

On one hand, take any *x*, *y* ∈ *X*. Then

$$\begin{array}{}{\displaystyle \mathcal{J}(x,y)\subseteq \bigcap _{x,y\in A\in {\mathcal{C}}^{\mathcal{J}}}A=c{o}^{{\mathcal{C}}^{\mathcal{J}}}(\{x,y\})={\mathcal{J}}^{{\mathcal{C}}^{\mathcal{J}}}(x,y).}\end{array}$$

On the other hand, take any *A* ∈ 2^{X}. Then

$$\begin{array}{}{\displaystyle A\in \mathcal{C}}& \u27f9& {\displaystyle \mathrm{\forall}x,y\in A,c{o}^{\mathcal{C}}(\{x,y\})\subseteq A}\\ & \u27fa& {\displaystyle \mathrm{\forall}x,y\in A,{\mathcal{J}}^{\mathcal{C}}(x,y)\subseteq A}\\ & \u27fa& {\displaystyle A\in {\mathcal{C}}^{{\mathcal{J}}^{\mathcal{C}}}.}\end{array}$$

This means that 𝓒 ⊆ 𝓒^{𝓘𝓒}, that is, 𝓒^{𝓘𝓒} ⩽_{𝓒} 𝓒, as desired.□

By Propositions 5.2 and 5.4, we know 𝕂^{*} ≜ 𝕂 : **IS** ⟶ **CS(2)** and ℍ^{*} ≜ ℍ|_{CS(2)} : **CS(2)** ⟶ **IS** are still functors. Moreover, we have the following result.

#### Theorem 5.7

(𝕂^{*}, ℍ^{*}) *is an adjunction between* **IS** *and* **CS(2)**. *Moreover*, 𝕂^{*} *is a left inverse of* ℍ^{*}.

#### Proof

By Theorem 5.6, it suffices to show that 𝓒^{𝓘𝓒} = 𝓒 for each arity 2 convex space (*X*, 𝓒). Take any *A* ∈ 2^{X}. Then

$$\begin{array}{}{\displaystyle A\in \mathcal{C}}& \u27fa& {\displaystyle \mathrm{\forall}x,y\in A,c{o}^{\mathcal{C}}(\{x,y\})\subseteq A((X,\mathcal{C})\text{is arity 2})}\\ & \u27fa& {\displaystyle \mathrm{\forall}x,y\in A,{\mathcal{J}}^{\mathcal{C}}(x,y)\subseteq A}\\ & \u27fa& {\displaystyle A\in {\mathcal{C}}^{{\mathcal{J}}^{\mathcal{C}}}.}\end{array}$$

This means 𝓒^{𝓘𝓒} = 𝓒.□

#### Corollary 5.8

*The category* **CS(2)** *can be embedded in the category* **IS** *as a reflective subcategory*.

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