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title: 'Lawvere-Cauchy Completeness, Examples'

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# Lawvere-Cauchy Completeness, Examples

This note goes along with the note on [Lawvere-Cauchy Completeness](https://hackmd.io/Z_pG2rcESGWgOr4kvQrHbg), which is a commentary on page 163 of Lawvere's [Metric spaces,generalized logic, and closed categories](https://raw.githubusercontent.com/mattearnshaw/lawvere/master/pdfs/1973-metric-spaces-generalized-logic-closed-categories.pdf)

For background see the note on [adjoint relations](https://hackmd.io/@alexhkurz/SJRU2jXJD).

## Introduction

**Definition:** A category is Lawvere-Cauchy complete if all left-adjoint relations (aka bimodules, profunctors, distributors) are induced by a function.

In analogy with lattice theory, we may also say that all left-adjoint relations (or ideals) are principal.

## In posets

We consider posets as $2$-enriched categories where $2 = \{\bot <\top\}$ is the two-element chain.

Let $Y$ be a poset.

Consider a pair of [adjoint monotone relations](https://hackmd.io/@alexhkurz/SJRU2jXJD)

$$L:1\to Y \quad\quad R:Y\to 1 \quad\quad L\dashv R$$

Spelling out the details, this means that we can think of $L$ and $R$ as monotone functions $L:Y\to 2$ and $R:Y^{op}\to 2$ satisfying

\begin{gather}
\exists y \in Y\,.\, L(y)\wedge R(y)\\[1ex]
R(y)\wedge L(y') \ \Rightarrow \ y \le y' 
\end{gather}

We see from this that $L$ is the principal upset and $R$ is the principal downset of some $y\in Y$:

The first condition says that there is a $y$ in the intersection of $L$ and $R$ ($L$ is 'total'), while the second condition says that this $y$ is unique ($L$ is 'single-valued').

Using that $2$ has meet and implication (internal hom), we can internalise in $2$ the two conditions above as follows.

\begin{gather}
\top \le \bigvee_{y \in Y} L(y)\wedge R(y)\\[1ex]
R(y)\wedge L(y') \ \le \ y \to y' 
\end{gather}

Note that the second form internalises the logical connectives $\exists$ and $\Rightarrow$ in the lattice $2$ of truth-values. This opens the way to generalise logic by replacing $2$ with other unital quantales.

Note that the two conditions above make it obvious that $L$ and $R$ determine each other uniquely.

## In metric spaces

We consider metric spaces as $[0,\infty]$-enriched categories, where $[0,\infty]$ is the set of non-negative real numbers with $0$ at the top, addition as the binary operation and truncated minus as internal hom ($[x,y]$ is $y-x$ if $x\le y$ and $0$ otherwise). 

I write $\sqsubseteq$, $\bigsqcup$, etc for the operations in $[0,\infty]$ and $\ge$, $\inf$, etc for the corresponding operations in the real numbers.

Consider again a pair of adjoint relations

$$L:1\to Y \quad\quad R:Y\to 1 \quad\quad L\dashv R$$

In the following we denote the element of $1$ by $\ast$.

**Example 1:** Let $Y$ be the non-negative rationals with distance $Y(y,y')$ being $y-y'$ if $y\ge y'$ and $0$ otherwise. 

Let $\ast Ly$ be $\pi-y$ if $\pi\ge y$ and $0$ otherwise. (Thus $\ast L$ is an enriched upset.)  We know from the [calculus of relations]() that the right-adjoint can be calculated as

$$\bigsqcup \{ R \mid R\cdot L \sqsubseteq Y\}$$

After working through a series of case distinction, one shows that $yR\ast = y-\pi$ if $y\ge\pi$ and $0$ otherwise. (Thus, $R\ast$ is an enriched downset.) [^R\ast]

**Example 2:** Let $Y$ be the non-negative rationals with distance $Y(y,y')$ the Euclidean distance. 

If we define $L$ as before, is there a right-adjoint at all? (I believe not, but I didnt check carefully).

So let $Ly$ be 0 for $10\le y\le 11$ and outside of this interval let $Ly$ be as big as it can be without violating $Y(y,y')\ge [0,\infty](Ly,Ly')$.

For $R$ to be a right-adjoint, $R$ must the smallest value (inf) in $[0,\infty]$ such that $Ry + Ly' \ge Y(y,y')$, in other words

$$ Ry = \sup_{y'} \{[Ly',Y(y,y')]\}$$

This should mean that for $Ly=0$ then $Ry$ is the longest distance one can travel from $y$ and stay inside $L$

[^R\ast]: $R\cdot L\sqsubseteq Y$ means that $yR\ast + \ast Ly'\ge Y(y,y')$ for all $y,y'$. Thus $yR\ast$ is the smallest value $v$ such that $v+\ast Ly'\ge Y(y,y')$ for all $y'$. This always holds if $y'\ge y$. Now suppose $y\ge y'$. We distinguish to cases. If $y'\ge \pi$, then $\ast Ly'=0$ and the smallest $v$ we can choose is $v= y-y'\le y-\pi$. If $\pi\ge y'$, then $\ast Ly'=\pi-y'$ and the smallest $v$ such that $v+\pi-y'\ge y-y'$ is $y-\pi$ if $y\ge\pi$ and 0 otherwise.