Remark 10.2.1.1.2 (00TN) — The Clowder Project

Table of Contents
Part 3: Relations
Chapter 10: Conditions on Relations
Section 10.2: Reflexive Relations
Subsection 10.2.1: Foundations
Remark 10.2.1.1.2: Unwinding Definition 10.2.1.1.1 (cite)

10.2.1 Foundations

Let $A$ be a set.

Definition 10.2.1.1.1 ▶ Reflexive Relations

A reflexive relation is equivalently:¹

•
An $\mathbb {E}_{0}$-monoid in $(\mathrm{N}_{\bullet }(\mathbf{Rel}(A,A)),\chi _{A})$.
•
A pointed object in $(\mathbf{Rel}(A,A),\chi _{A})$.

¹Note that since $\mathbf{Rel}(A,A)$ is posetal, reflexivity is a property of a relation, rather than extra structure.

Remark 10.2.1.1.2 ▶ Unwinding Definition 10.2.1.1.1

In detail, a relation $R$ on $A$ is reflexive if we have an inclusion

\[ \eta _{R}\colon \chi _{A}\subset R \]

of relations in $\mathbf{Rel}(A,A)$, i.e. if, for each $a\in A$, we have $a\sim _{R}a$.

Definition 10.2.1.1.3 ▶ The Po/Set of Reflexive Relations on a Set

Let $A$ be a set.

1.
The set of reflexive relations on $A$ is the subset $\smash {\mathrm{Rel}^{\mathrm{refl}}(A,A)}$ of $\mathrm{Rel}(A,A)$ spanned by the reflexive relations.
2.
The poset of relations on $A$ is is the subposet $\smash {\mathbf{Rel}^{\mathsf{refl}}(A,A)}$ of $\mathbf{Rel}(A,A)$ spanned by the reflexive relations.

Proposition 10.2.1.1.4 ▶ Properties of Reflexive Relations

Let $R$ and $S$ be relations on $A$.

1.
Interaction With Inverses. If $R$ is reflexive, then so is $R^{\dagger }$.
2.
Interaction With Composition. If $R$ and $S$ are reflexive, then so is $S\mathbin {\diamond }R$.

Proof of Proposition 10.2.1.1.4.

Item 1: Interaction With Inverses

Since $R$ is reflexive, we have $a\sim _{R}a$ for each $a\in A$. Since $a\sim _{R^{\dagger }}b$ iff $b\sim _{R}a$, it follows that $a\sim _{R^{\dagger }}a$ as well for each $a\in A$. Thus $R^{\dagger }$ is reflexive.

Item 2: Interaction With Composition

Since $R$ and $S$ are reflexive, we have $a\sim _{R}a$ and $a\sim _{S}a$ for each $a\in A$.

To prove that $a\sim _{S\mathbin {\diamond }R}a$ holds as well, we have to exhibit some $b\in A$ such that $a\sim _{R}b$ and $b\sim _{S}a$, following Chapter 8: Relations, Item 1 of Definition 8.1.3.1.1. Choosing $b=a$, we see that this condition is indeed satisfied. Thus $S\mathbin {\diamond }R$ is reflexive.

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