Partial ordering and joining

Question

This question is based on material from "An invitation to applied category theory" (ISBN 9781108711821) - I asked a closely related question before (Partial order of the Booleans true, false) but essentially didn't really understand the answer given, left the book and have now come back to it and this a much fuller question which, I hope, gets to the heart of my problem of understanding.

To establish its definitions the book uses an example of a set with three elements, and rather than try to replicate the symbols used in the book (page 5), I will use $x, y, z$.

The book states "$A \leq B$ if, whenever $x$ is connected to $y$ in $A$, then it is connected to $y$ in $B$."

Now the operator $\leq$ is never named in the book - but the obvious temptation is to say it means "less than or equal to" but the book consistently implies that it mean "less than" (as in "smaller") but then seems to use cases where it means "equal to" - it's extremely confusing, at least to me.

Let me give a concrete example and perhaps someone could clarify the meaning?

It pictures a Hasse diagram of this set $(x, y, z)$ of three objects under this $\leq$ ordering, where the joining of all three elements is at the top and having no elements joined is at the bottom and there is a middle layer - apologies for the quality of the text art here:

       (x, y, z)
       ^  ^  ^
       |  |  |
      /   |  \
     /    |    \
    /     |      \
(x, y) z  x(y, z)  (x, z) y
 ^       ^        ^
  \      |       /
   \     |      /
       x, y, z

And then states "the joined system $A \vee B$ is the smallest system that is bigger than both $A$ and $B$. That is $A \leq (A \vee B)$ and $ B \leq (A \vee B )$, and for any $C$, if $A \leq C$ and $B \leq C$ then $(A \vee B) \leq C$."

So this is the definition of the (partial) order - there is then an exercise which looks at the booleans $\mathbb{B}=\{true, false\}$... has an order $false \leq true$ adding "Thus $false \leq false$, $false \leq true$, and $true \leq true$ but $true \nleq false$. In other words $A \leq B$ if A implies B."

It then asks "what is $true \vee true$?" and gives the answer $true$. But how can (adapting the definition for join given above) $true$ be "the smallest system that is bigger than both $true$ and $false$"?

Adding: To take the $x,y,z$ example, the smallest system that is bigger than both, say $A:(x, y), z$ and $B:x, y, z$ is $(x, y, z)$ but what is $A \vee B$? Is it $(x, y), z$ or $(x, y, z)$? Or to put it another way: is $A \leq A$?

Answer 1

I think you're overthinking this. You write:

the book consistently implies that it ($\leq$) means "less than" (as in "smaller").

I guess this is because of phrases like

the joined system $A\vee B$ is the smallest system that is bigger than both $A$ and $B$"

But I think the authors write "bigger than" here, rather than "bigger than or equal to" just because it's inconvenient to wrote "or equal to" over and over again. It's quite common in mathematical writing to see "$x$ is smaller/bigger than $y$" include the case $x = y$, and to see "$x$ is strictly smaller/bigger than $y$" when the case $x = y$ is excluded. As is often the case with reading mathematics, deduction from the context about the precise meaning is sometimes required here.

In the passage you're referring to, the symbol $\leq$ is used to refer to a partial order (and remember that one of the axioms of partial order is reflexivity: for all $x$, $x\leq x$). So yes, you can think of $A \leq B$ as meaning "$A$ is less than or equal to $B$".

It's clear from the definition of $\leq$ in your specific example that for all $C$, $C\leq C$. Indeed, whenever $x$ is connected to $y$ in $C$, then it is connected to $y$ in $C$.