Lecture 3 - Chapter 1: Preorders

John Baez

Okay, let's get started!

Fong and Spivak start out by explaining preorders, which is short for "preordered sets". Whenever you have a set of things and a reasonable way deciding when anything in that set is "bigger" than some other thing, or "more expensive", or "taller", or "heavier", or "better" in any well-defined sense, or... anything like that, you've got a poset. When \(y\) is bigger than \(x\) we write \(x \le y\). (You can also write \(y \ge x\), of course.)

What do I mean by "reasonable"? We demand that the \(\le\) relation obey these rules:

1. reflexivity: \(x \le x\)

2. transitivity \(x \le y\) and \(y \le z\) imply \(x \le z\).

A set with a relation obeying these rules is called a preorder.

This is a fundamental concept! After all, humans are always busy trying to compare things and see what's better. So, we'll start by studying preorders.

But I can't resist revealing a secret trick that Fong and Spivak are playing on you here. Why in the world should a book on applied category theory start by discussing preorders? Why not start by discussing categories?

The answer: a preorder is a specially simple kind of category. A category, as you may have heard, has a bunch of 'objects' \(x,y,z,\dots\) and 'morphisms' between them. A morphism from \(x\) to \(y\) is written \(f : x \to y\). You can 'compose' a morphism from \(f : x \to y\) with a morphism from \(g: y \to z\) and get a morphism \(gf : x \to z\). Every object \(x\) has an 'identity' morphism \(1_x : x \to x\). And a few simple rules must hold. We'll get into them later.

But a category with at most one morphism from any object \(x\) to any object \(y\) is really just a preorder! If there's a morphism from \(x\) to \(y\) we simply write \(x \le y\). We don't need to give the morphism a name because there's at most one from \(x\) to \(y\).

So, the study of preorders is a baby version of category theory, where everything gets much easier! And when Fong and Spivak are teaching you about preorders, they're sneakily getting you used to categories. Then, when they introduce categories explicitly, you can always fall back on preorders as examples.

I've posted four puzzles on preorders here. Look at them! I just answered Puzzle 3. Puzzle 4 has millions of answers - come up with another! Also look at Puzzle 5 here. And people who already know the definition of a category, and want to ponder how preorders are a special case of categories, should try Puzzles 6 and 7 here.

To read other lectures go here.

Joe Moeller

In your first sentence, did you mean to say "preorder" rather than "partial order"?

Joe Moeller

Or is that part of not trying to "send a hundred students out into the world talking funny."?

John Baez

Everyone, no matter how funny they talk, must admit that the "po" in "poset" comes from "partially ordered". So "poset" is short for "partially ordered set", even though Fong and Spivak have cruelly decided to redefine the term "poset" to mean "preorder".

Grant Roy

What's the reason the word morphism is used rather than binary relation?

John Baez

A morphism is not a binary relation. In a preorder, \(\le\) is a binary relation: for any pair of elements \(x\) and \(y\), \(x \le y\) is either true for false. When \(x \le y\) we can say there's a exactly one morphism from \(x\) to \(y\), if we so desire. But in a general category, there could be many (or no) morphisms \(f : x \to y\).

My remarks on morphisms were aimed only at people who know a bit of category theory and wonder why we're talking about preorders instead. So, if that stuff - or what I just said - makes no sense, don't worry about it too much.

Jacob MacDonald

Grant#4, John#5: Trying to make this make sense to me in my current state of very limited understanding. Binary relations apply to specific elements while morphisms are like functions in that they apply to categories and not specific elements. Is that correct?

Grant Roy

John#5 Thanks! I believe I see the point your making.

Jacob#6 My understanding is that morphisms apply to objects, and categories consist of objects. It seems that in the case of a preorder, we have distinct elements, which may be objects, however, there is no transformation being done on them, they are merely 'put into some relation or equivalence'...the result of which can be a simple fact or ordering relation. When they are 'transformed', 'mapped', 'sent to', they are 'morphed'. I think \( f(x) <= f(y) \) could be a monomorphism if certain other conditions hold, but I admittedly know 0 category theory, so I'm just trying to understand the basic terminology as well, and I could certainly be mistaking things.

Jacob MacDonald

Grant#7. Thanks. The part I still don't understand is "there's exactly one morphism from x to y" in John#5. But I suppose I'll have to wait for John to clarify his wording there.

John Baez

Jacob wrote:

Binary relations apply to specific elements while morphisms are like functions in that they apply to categories and not specific elements. Is that correct?

Not quite. Let's start by saying what a binary relation is. Given a set \(S\), a binary relation is something that can be either true or false for any pair of elements of \(S\). An example is equality: for any pair of elements of \(S\) the statement \(x = y\) is either true or false.

I suppose you could summarize this by saying "binary relations apply to specific elements of a set".

Next for something much harder: a taste of category theory. A category has a set of "objects", and for any pair of objects \(x\) and \(y\), there is a set of "morphisms from \(x\) to \(y\)". There's more to a category than this - see my lecture above for a small taste - but this is the first part of the story.

So, I hope you see why it's wrong to say "morphisms are like functions in that they apply to categories and not specific elements".

If this still doesn't make sense, please ignore all my remarks about categories until I get around to actually explaining this subject!

Xah Lee

Greatly enjoyed this one!

Here's my understanding of the gist. Place 4 stones on the floor, and place ONE arrow (e.g. a pen) between any 2 stones. (we'll have placed a total of 6 pens.)

The stones are the “objects” in category theory. The arrows represent the morphisms (or just think of morphism as function). But, since now there is only 1 arrow between any 2 stones, we can relook the whole picture, and now think of the arrow as relation between 2 stones!

So, if S1 → S2, we can think of it as S1 ≥ S2. So, the whole thing, becomes another notion, namely, what mathematicians calls “preorder” in order theory, where, the stones is the set, and each stone is a element.

right John?

Jonatan Bergquist

Re: #9 John 1. What is the difference between an element and an object?

Re: #5 John 2. Are all morphisms in a category of the same type or can a category have, for example, objects {\(x,y,a,b\)} where morphisms \(f: x \rightarrow y\) and \(g: a \rightarrow b\) are different?

Thanks!

Jesus Lopez

There are two levels I had difficulty putting apart when starting. Most of the time categories are so-called concrete, and the objects are sets with relations, operations, constants ("structure"), and morphisms are functions between them preserving that. For instance the category of vector spaces over a field. Thinking about it means you think about all vector spaces (over the field). This lives at a considerable level of abstraction (despite the "concrete" technical term). It occurs to me that one could phrase this as saying that the quinean ontological commitment is similar as that required in second order logic where one quantifies over relations, which is a huge commitment.

On the other hand, the axioms of the definition of category can be applied at much much more concrete level to whatever user defined set of objects and of arrows one can came out with, as far as the axioms hold. In our case of an order \((X,\leq)\), the set of objects of the category is \(X\), so the objects of the category are the elements of \(X\), (i.e, things as opposed to sets of things above) and the morphisms are merely pairs (as opposed to functions).

DavidTanzer

Morphisms in a category are typed: every morphism has a domain (source object) and codomain (target object).

In Jonatan’s question, f and g are necessarily distinguishable by the functions which map morphisms to their domains and codomains, respectively.

John Baez

Xah wrote:

So, if S1 → S2, we can think of it as S1 ≥ S2.

Yes, you can do that. But notice that I'm not doing that! I'm saying that if that given a category with a morphism \(f : S_1 \to S_2\) , I can create a preorder with \(S_1 \le S_2\).

Why \(\le\) instead of \(\ge\)? It's an arbitrary choice:if you have a preorder where you say \(x \le_{\textrm{Xah}} y\), I can make up a preorder where I say that \( x \ge_{\textrm{John}} y \). This is called the opposite preorder.

But there's a reason I'm making my choice! It's very nice to say that the inclusion of sets {1,2,3} \(\to\) {1,2,3,4,5,6} is another way of saying \( 3 \le 6 \). So, everyone in category theory does it the way I'm doing it.

Chris Nolan

So, this is a bit of a dumb question but I'm trying to understand why it's called a "partially ordered set" and not just an "ordered set," so according to old, handy Wikipedia/Partially ordered set:

the word "partial" in the names "partial order" or "partially ordered set" is used as an indication that not every pair of elements need be comparable. That is, there may be pairs of elements for which neither element precedes the other in the poset. Partial orders thus generalize total orders, in which every pair is comparable.

So, one example of this might be the set {\(a, b, c, 1, 2, 3\)} where the first subset {\(a, b, c\)} is ordered alphabetically ascending and the second subset {\(1, 2, 3\)} is ordered numerically ascending? If that's the case then does that mean since the first subset can't be compared to the second subset, then set {\(a, b, c, 1, 2, 3\)} is equivalent to {\(1, 2, 3, a, b, c\)}?

Joe Moeller

The partial orders and preorders that are being discussed are not the order that you write the elements inside braces. Writing a set as a list of elements inside braces is not intended to imply an order. But the order you are talking about (letters ordered alphabetically and numbers ordered in the normal way, and no comparison between the two types) is a valid partial order, and it has incomparable elements. In particular, \(b\) and \(3\) are incomparable.

John Baez

Jonatan wrote:

What is the difference between an element and an object?

I hope you know a bit about sets: if you feel shaky about them, click the link and read! A set \(S\) has elements, and we write \(x \in S\) when \(x\) is an element of \(S\).

A category \(C\) consists of two sets: a set \(\mathrm{Ob}(C)\) of objects, and a set \(\mathrm{Mor}(C)\) of morphisms. There's also more to a category (re-read my lecture for some of this), but that's where we start.

I emphasize that 'element', 'object' and 'morphism' are undefined terms: that is, they don't really mean anything except insofar as we demand that they obey certain rules.

Summarizing my answer to your first question: an object of a category \(C\) is an element of the set \(\mathrm{Ob}(C)\).

Note that you're forcing me to talk about categories, which is not at all what I want to be doing now!. I'm trying to explain preorders. My remark about categories was just a whispered comment to those who already understand categories - to explain why I'm starting a course on categories by explaining preorders!

Focus on preorders: objects and morphisms can wait.

John Baez

Chris Nolan wrote:

So, this is a bit of a dumb question but I'm trying to understand why it's called a "partially ordered set" and not just an "ordered set" [....]

I think Joseph gave a fine answer, namely: in a partially ordered set we can't compare every pair of elements, i.e. there are incomparable elements \(x\) and \(y\) where neither \(x \le y \) nor \(y \le x\). There's something 'partial' about this.

This comment of mine from over here may help explain some of the history:

---

Bob Haugen wrote:

Question: forgive my ignorance, but why is it called "preorder" and not just "order"? What is the significance of the "pre"?

The original concept of totally ordered set or order, still dominant today, obeys a bunch of rules:

1. reflexivity: \(x \le x\)

2. transitivity: \(x \le y\) and \(y \le z\) imply \(x \le z\)

3. antisymmetry: if \(x \le y\) and \(y \le x\) then \(x = y\)

4. trichotomy: for all \(x,y\) we either have \(x\le y\) or \(y \le x\).

The real numbers with the usual \(\le\) obeys all these. Then people discovered many situations where rule 4 does not apply. If only rules 1-3 hold they called it a partially ordered set or poset. Then people discovered many situations where rule 3 does not hold either! If only rules 1-2 hold they called it a preordered set or preorder.

Category theory teaches us that preorders are the fundamental thing: see Lecture 3. But we backed our way into this concept, so it has an awkward name. Fong and Spivak try to remedy this by calling them posets, but that's gonna confuse everyone even more! If they wanted to save the day they should have made up a beautiful brand new term.

Chris Nolan

So, can we compare posets to other posets? So, if all corresponding comparable subsets of one poset are equal to the other then can we say the two posets are equal?

Alex Kreitzberg

Chris Nolan#19

If all the comparable elements of two sets are equal, then the sets are equal. But posets also include an order, and the order of two sets could be different, even though the elements are the same.

Now, if by poset, you mean posets whose order is subset inclusion, if the corresponding sets are equal, then they would be equal as posets (because they are equal as sets, and have the same \( \leq \) ).

This is similar to saying that {1, 4, 6} and {1, 4, 6} are equal as sets with the usual order for integers.

Does that answer your question?

Scott Finnie

Chris Nolan #15: I was wondering that too, thanks for asking. Testing my understanding of the answers then:

For any pair of objects \(x, y\) in the set of objects, exactly one of the following must hold:

1. \(x\) is comparable to \(y\), or
2. \(x\) is not comparable to \(y\).

Case 2 fails trichotomy per John's note @ #18. Stated alternately: if all \((x, y)\) are comparable, then the set of objects has a total order provided reflexivity and transitivity hold.

That then raises a question about antisymmetry. But I think I read somewhere that the concept of equality is non-trivial in category theory. So I'll leave that one for now!

It does, however, raise another minor question for me: why use \(\le\) to mean "is less than" instead of \(<\)? I can see that reflexivity gets troublesome if I use a conventional interpretation of "less than": \(x\) can't be less than \(x\) in any meaningful way. \(x\) "is less than or equal to" \(x\) does have intuitive meaning. But that gets us back into equality again!

Robert Figura

Preorder, Poset, and totally ordered – Definitions, examples, and their degeneration.

We all know about totally ordered sets. They are sets which can be put on a line, like a chain or a necklace, a string of pearls. These can be captured by four axioms like this:

1. reflexivity: x≤x

2. transitivity: x≤y and y≤z imply x≤z

3. antisymmetry: if x≤y and y≤x then x=y

4. trichotomy: for all x,y we either have x≤y or y≤x.

And the canonical example are the integers with the ordering relation ≤ as we know it. Now, a poset is a structure where 4 doesn't hold, and a preorder is one where neither 3 nor 4 hold. Then what do these axioms mean, what do they do?

Axiom 4 makes sure that every pair of elements is related. It excludes parallel structures and this axiom is what causes the set to collapse to a string. When Axiom 4 doesn't hold we may get a directed acyclic graph (DAG) instead. Well, sort of, as a directed acyclic graph does not necessarily imply transitivity, a poset does. More on transitivity below.

Given a total order (like the integers with ≤) we can degenerate it by adding 'parallel structure' (e.g. an element ε which is just like 3 but not identical to it).

Axiom 3 suppresses cycles. It says that the only cycles are indentity arrows.

Given a total order (like the integers with ≤) we can create a cycle by adding a backward relation like (5≤2). Note that if axiom 3 / antisymmetry does not hold, axiom 4 cannot hold either, as the 'and' is incompatible with the 'either-or'.

What does the transitivity axiom 2 do? That's a funny one. It implies associativity, and says that if you find two arrows that can be combined because their ends match, you're allowed to talk about the combined arrow. It actually says that the combined arrow has to be among our collected arrows.

You can break this structure by taking out a combined arrow, e.g. insisting that 2 and 4 cannot be compared, even though 2≤3 and 3≤4. Strike out the corresponding arrow, mark it red, you don't have enough gas to drive two routes...

As I said earlier, a poset is usually drawn as a DAG without outlining each and every matching combination of arrows. That's because people (and algorithms) have no trouble to understand that once we find a connected path between two points that those points are also connected. So people will often draw a DAG when they actually want to talk about posets.

Alexander Telfar

So, I am not sure if I understood this correctly. Could a ring be a partially ordered set? $$1 \le 2, 2 \le 3, 3 \le 4, 4 \le 1$$ If so, it seems more intuitive to me to call it a locally ordered set??

Edit: Ok, the above comment clears things up. Was writing this comment and hadnt refreshed.

Narasimha Mujumdar

In the first chapter, it is mentioned that: "In other words A ≤ B in the poset if “A implies B”, often denoted A ⇒ B." An example could be: A: If it rains. B: The match wont happen. A => B. It can be seen that A = True then B = False. But if A = False, then B = True or False. Based on this we can say A <= B. Are there other explanations for this statement?

Marcello Seri

The web, where the objects are the web pages and the relation is given by ‘links to’, seems to me a preorder as we cannot avoid cycles, thus invalidating antisymmetry. If we try to generate a partial order from it, do you know if we get just a set with one element or if there is more structure? I believe the latter but I don’t really have the data to check it

Grant Roy

John Baez wrote:

Fong and Spivak try to remedy this by calling them posets, but that's gonna confuse everyone even more! If they wanted to save the day they should have made up a beautiful brand new term.

How about \(apriorder \)? :) Obviously leaning heavier on the Latin, and the idea of it being fundamental or 'from before'. Anyway, more to the point, for posts on the forum should we always consider poset to mean partially ordered set, and translate the terms from the book? From most contexts, it's probably pretty clear, but it is an unfortunate source of confusion.

ThomasRead

Marcello #25: Presumably there are many pages with no links, and many which are never linked to, so you'd definitely get some sort of non-trivial structure. I'd love to see what it actually looks like though.

Christina Vasilakopoulou

Narasimha #24: the way you formulated your propositions, it seems to me that A true gives B true! Unless you meant A=it rains B=the match is happening , in which case A cannot imply B (since true cannot imply false). You considered the negative statement instead!

To sum up: A=it rains, B=match is on, then A does not imply B and B does not imply A. However \(A\Rightarrow\neg B\).

Chris Nolan

Alex Kreitzberg #20: Yeah, I should have specified that I meant the subset with ordering.

Can we define a hierarchy of ordering within posets in order to make a totally ordered set? So given a poset made up of two types of objects, one letters and one numbers, both with the natural ordering for each type, then can we define a higher ordering to say letters are always less than numbers to make a totally ordered set, and would that still be a poset? Or does that not make sense within the concept of a poset?

Robert Figura

You can add arrows to flatten a poset to a total order, Chris Nolan. Posets are more general and less specific than total orders, every total order is a poset, but not every poset is a total order.

Alex Kreitzberg

Chris Nolan #29: The answer to all your questions is yes. For your example:

Let poset A = ( {a, b}, { (b, a), (a, a), (b, b) } ) and B = ( {0, 1}, { (0, 1), (0, 0), (1, 1) } ) In abuse of notation, I'll also write A = {a, b}, B = {0, 1}, and \( \leq_A \) = { (b, a), (a, a), (b, b) }, and \( \leq_B \) = { (0, 1), (0, 0), (1, 1) }.

You can form a new poset by taking a pairwise union of sets:

C = (A \(\cup\) B, \(\leq_A \cup \leq_B\)) = ( {a, b, 0, 1}, { (0, 1), (0, 0), (1, 1), (b, a), (a, a), (b, b) } )

Like you said, this isn't a total order. The order can be flattened by saying "all letters are less than numbers". In notation, this is done by setting \(\leq_C \) to \(\leq_C \cup \) { (a, 0), (a, 1), (b, 0), (b, 1), (c, 0), (c, 1) }

For what it's worth, I don't really like thinking of relations as certain special types of sets. But it's consistent with this book. Your question seemed like a nice way to practice manipulating posets as sets.

Edit And it seems that although in this case we do get a poset by performing a union, this isn't always true. David Tanzer gave an example of where a normal union fails, and some more typical examples of products between posets.

John Baez Gave insights for how subtle these operations can be. Taking the union as I did is apparently complicated in general. So the procedure I described doesn't generalize well. A "disjoint union" is apparently a much safer operation. So I think your basic relations combined nicely, because the underlying sets are different.

Jesus Lopez

Marcelo #25: The data shows that there are some islands disconnected of each other, and also if you collapse pages joined by reversible paths there still remain other irreversible paths.

Scott Finnie

Posting this as someone who learns best from concrete examples and geometric representations, on the basis that trying to explain something is a good way to test one's understanding. If it's inappropriate for the forum please just say. I ask forbearance from those with more natural mathematical ability. Corrections or comments gratefully received.

The examples below all consider the subject matter of completing tasks. Assume order means "must be completed before": so \(x \le y\) means "task \(x\) must be completed before task \(y\)". Note I've used 'arrow' in the geometric sense, not the categorical one.

The Landscape

First, a picture:

This says:

• All Orders are also Posets
• All Posets are also Preorders
• All Preorders are also Sets

Why the subset relations among Preorders, Posets and Orders? Because of the rules that they must obey. We'll look at that below; but first, let's cover off sets without any order.

No Order

Consider cleaning the house. Let's assume there are 3 rooms: bathroom, bedroom and kitchen. To clean the house, we need to clean all three rooms. It doesn't matter what order they're cleaned in: there's no dependence.

Preordered Sets (Preorders)

Preorders must obey 2 rules:

1. reflexivity: \(x \le x\)

2. transitivity \(x \le y\) and \(y \le z\) imply \(x \le z\).

As an example, consider a (very simplified) iterative approach to writing software:

The arrows are straightforward: we must Define Requirements before Write Code, and Write Code before Test Solution. We can also combine those according to rule (2) - so Define Requirements before Test Solution.

What about the loop? That's OK according to rule 1. It says any task must be "less than or equal to" itself. Define Requirements equals itself, so it satisfies rule 1 (as do all the others in the loop, for the same reason).

Note also there's no arrow connecting Write Code and Write Tests. Again, that's fine: there's nothing in the rules to say every task must be related to every other.

Partially Ordered Sets (Posets)

Posets obey both rules for Preorders, and add a third:

1. antisymmetry: if \(x \le y\) and \(y \le x\) then \(x = y\)

Our iterative process above fails this rule. Why? Substitute Define Requirements for \(x\) and Test Solution for \(y\) in the antisymmetry rule. Define Requirements does come before Test Solution so \(x \le y\). Test Solution also comes before Define Requirements. So \(y \le x\). But Define Requirements is a different task to Test Solution. So \(x = y\) is not true. So the iterative process isn't antisymmetric: it's not a poset. We can make it so by removing the loop, and creating a waterfall process:

Ordered Sets (Orders)

Orders obey all the rules for Posets, and add yet another:

1. trichotomy: for all \(x,y\) we either have \(x\le y\) or \(y \le x\).

Antisymmetry removed loops in our task ordering. Trichotomy removes parallel paths. Why? Let \(x\) be Write Code and \(y\) be Write Tests. There's no arrow, or sequence of arrows, that connect the two. So we don't have \(x \le y\), nor do we have \(y \le x\). Trichotomy fails.

We can resolve that by linearising the process. Let's follow Test-Driven Development and write tests before the code:

Marcello Seri

Jesus #25: nice! Do yiu know if the relationships data is available somewhere for use?

John Baez

Scott Finnie #21 wrote:

It does, however, raise another minor question for me: why use ≤ to mean "is less than" instead of <?

We don't. We use \(\le\) to mean "less than or equal to". In a preorder, \(x = y\) implies \(x \le y\). So, the case of equality is always included in \(\le\).

John Baez

Chris Nolan #19 wrote::

So, can we compare posets to other posets?

Yes. The best way is with monotone functions. There's a lot to say about this, but we can use monotone mappings to define when two posets are "isomorphic".

So, if all corresponding comparable subsets of one poset are equal to the other then can we say the two posets are equal?

Alex has already answered your question quite nicely but I'll just say a bit more.

A mathematician wouldn't ask that question unless they had first told us that the two posets under discussion are equal as sets. They'd say something like this:

Suppose I have a set \(S\) with two different rrelations on it, \(\le_1\) and \(\le_2\), which give me two posets \((S,\le_1)\) and \((S,\le_2)\). Suppose \(x \le_1 y\) if and only if \(x \le_2 y\) for all \(x,y \in S\) . Does that mean these two posets are equal?

And the answer is yes: the posets \((S,\le_1)\) and \((S,\le_2)\) are equal if and only if

$$ x \le_1 y \textrm{ if and only if } x \le_2 y \textrm{ for all } x,y \in S. $$ However, equality of posets is ultimately less interesting than other ways of comparing posets, using monotone mappings.

Note: everything I just said would still be true if I replaced the word "poset" by "preorder" throughout this comment.

John Baez

Scott Finnie # wrote::

Posting this as someone who learns best from concrete examples and geometric representations, on the basis that trying to explain something is a good way to test one's understanding. If it's inappropriate for the forum please just say.

No, this is GREAT!!! This is just what I'm hoping people will do: add their own material to this course. If just 5% of the people here do this, we'll have 10 people explaining things in different ways, and students will get a multi-faceted view of the course material that's far better than anything I could produce on my own.

Like you, I learn best from pictures. If I had more time, every lecture of mine would be packed with pictures, because that's how I think. I just don't have time to draw all those pictures - there are too many in my mind. Fong and Spivak's book has some of the missing pictures, but I'm very happy for people to draw more!

Scott Finnie

@John #35 & #37: thanks. Have updated #33 to use "less than or equal to", which makes the description for preorders less clumsy. Great that you're open to alternative descriptions. I'll try to add as and when time allows.

Jesus Lopez

Marcelo #34: don't know about the raw data, I read about the islands in some paper about Google PageRank algorithm. It is based in random walks on the web graph.

John Baez

Scott: here's one interesting nuance. Your original chart:

showed unordered sets (i.e., just sets) as disjoint from ordered ones. A mathematician would prefer to see a bunch of nested blobs, as in your new chart:

The point is that a preordered set is a special kind of set, a partially ordered set is a special kind of preordered set, and an ordered set is a special kind of partially ordered set.

Category theory goes much further with this line of thought. I don't want to get into it now, but each of these blobs is actually a category. There's a category of sets, a category of preordered sets, and so on.

Scott Finnie

@John #40: thanks. I've updated the diagram accordingly, though a side effect is your post now shows the new version (apologies). You can use https://raw.githubusercontent.com/sfinnie/CategoryTheoryCourseNotes/6bbcc770474a8945ee4e47e273a65909f979b818/posets/img/orderVennDiagram.gif to access the original in your post.

It was something I considered in drawing the diagram. I decided on "Unordered" as a disjoint set, on the basis that there are some which are explicitly not ordered. So those that obey the rule:

• unordered: for all \(x\) and \(y\), there exists no \(x, y\) such that \(x \le y \).

I think the "Set" rectangle in the revised diagram now means "Possibly ordered". It still caters for unordered sets, but doesn't assert any properties about them. Is that correct?

Thanks.

Filip Vukovinski

I've decided to follow this course in Coq, and it already hit me in the head, meaning I have still much more to learn after not being able to formalize preorders (on the inductive type level as I presumed possible).

Still, I made a promise, so I dug deep in Coq's standard library to find it's default implementation. Those are available here: https://github.com/vukovinski/azimuth-act/blob/master/chapter-1/poset.v

My next update should solve some puzzles by providing instances to the classes defined in the file above. I'll also write a short readme.md on how to work with Coq.

Marcello Seri

@Filip #42 that is a really interesting experiment! I am no expert of coq, so what I say may be completely meaningless. But, could you use some of the typeclasses from the standard library for preorders, posets and ordered sets? I think the definition of preorder there is very close to the one you give in your file. Maybe staying close to the standard library can help to smooth some of the obstacles

John Baez

Thanks for updating your chart, Scott, and thanks for giving me a heads-up. I've updated my own comment accordingly.

You originally were imagining a thing called an "unordered set":

So those that obey the rule:

• unordered: for all x and y, there exists no x,y such that x≤y.

One could make up this notion, but it would be a useless notion: one is saying "take a set and equip it with a relation \(\le\), but also decree that this relation never holds: that is, \(x \le y\) is never true".

This is sort of like defining an adjective "cruncy" by saying

Cruncy (adj.): a property of plants that never holds: no plant is ever cruncy.

One could do it, but it's not much use.

More generally, mathematicians have learned over time, and especially since the invention of category theory, that's it's vastly better to make definitions contain only "positive" properties: that is, properties that don't involve "not" or "no". There are deep reasons for this, but it would take more category theory to explain them!

Scott Finnie

@John #44. Thanks for the explanation. Learning and enjoying!

Shaaz Ahmed

Could someone explain the skeletality criterion that Fong and Spivak talk about? They use the term poset to refer to a preorder, and state that a partially ordered set (the usual expansion of the term 'poset') has a requirement of 'skeletality'. From their description, and the one on Wikipedia, I understand that skeletality implies that 'isomorphic objects are necessarily identical'.

Can some one provide examples a non-skeletal preorder? I can't think of one, or I don't understand the concept well.

Keith E. Peterson

@Shaaz #46.

Think of skeletal as being the condition wherein a preorder (or even category), the only thing that matters is whether at least one morphism exists or not. It's skeletal because we're 'gutting out' all other structure except the barebones of the preorder (or category).

Anindya Bhattacharyya

@Shaaz #46 – if you imagine a preorder drawn with arrows between elements (so we draw \(x\rightarrow y\) whenever \(x\leq y\)) then a non-skeletal preorder would be one with a non-trivial cycle, eg \(x\rightarrow y\rightarrow z\rightarrow x\)

Shaaz Ahmed

Thanks @Keith and @Anindya, that makes sense.

Scott Finnie

@Keith @Shaaz @Anindya: thanks for the question and explanations.

Assuming I understand right then, this is just different terminology for the fundamental difference between a preorder and a poset. At least, using the common meaning for those things outside the book. Specifically:

• A poset must obey antisymmetry (= "skeletality").
• That's what differentiates posets from preorders.

Stated alternately, it's antisymmetry that prevents non-trivial cycles.

Is that correct? Thanks.