Defining 'free monoid' without Nat? - MathOverflow

Defining 'free monoid' without Nat?

2010-03-08T15:54:30Z

Is there a definition of what is a 'free monoid' which does not pre-suppose that the natural numbers has already been defined? The definitions that I have been able to track down all use the natural numbers (since sequences/words need them).

In other words, I am trying to define the theory of free monoids (as a signature with sorts, operations and axioms), and I would like to know if I can do this without having the theory of the naturals already defined. For exhibiting/constructing a free monoid, I do expect to need Nat.

Note that my ambient logic is higher-order, so I am fine with a second-order axiomatization. If dependent-types are needed, that would also be acceptable.

Answer by Wlog for Defining 'free monoid' without Nat?

2010-03-08T16:28:52Z

Can you give a good example of this in another instance?

The Wikipedia definition seems to imply that if you do not have natural numbers defined yet, they follow as the lengths of the finite strings.

In particular, given any Free Monoid, the unique word length function maps elements of that monoid into the natural numbers.

Answer by Pete L. Clark for Defining 'free monoid' without Nat?

2010-03-08T16:33:37Z

Well, yes. You define it in terms of its universal mapping property: the free monoid on a set $S$ is a monoid $FM(S)$ and a set map $\iota: S \rightarrow FM(S)$ such that for any monoid N and set map $f: S \rightarrow N$, there exists a monoid map $F: FM(S) \rightarrow N$ such that $f = F \circ \iota$.

This definition doesn't mention natural numbers at all. A construction of $FM(S)$ might involve sequences, but this could probably be avoided, e.g. by applying some general nonsense to show that the forgetful functor from monoids to sets has a left adjoint.

Or is this not what you had in mind?

Answer by Dan Piponi for Defining 'free monoid' without Nat?

2010-03-08T18:53:29Z

Could we mimic the definition of a list in a programming language?

So if you want $M$ to be the free monoid on $A$ you can say $M$ is a structure equipped with a function $m:A\times M\rightarrow M$ and an identity element $e\in M$.

You'd have some axioms like $m(x,y)=m(x',y')$ iff $x=x'$ and $y=y'$.

You'd also need some kind of inductive axiom. So for any property $P$, if we have ($P(e)$ and $P(x)\Rightarrow\forall a P(m(a,x))$ ) then $\forall x P(x)$.

This inductive axiom now implies that the elements of $M$ are finite strings of elements of $A$, and we can define the usual monoid multiplication the way we'd append two lists in a programming language. The inductive axiom should guarantee that appending is well-defined.

Answer by François G. Dorais for Defining 'free monoid' without Nat?

2010-03-09T13:32:16Z

As I pointed out in the comments, the theory of free monoids is somewhat ill defined. It is still unclear what your logic is and what you really want, but you have two basic options which were proposed by Pete Clark and sigfpe. Here are a few additional remarks that may help you sort things out.

Universal Property à la Pete Clark. The free monoid functor $F:Set\to Mon$ is left-adjoint to the forgetful functor $G:Mon \to Set$. This adjunction completely determines the isomorphism classes of free monoids. Moreover, many of the essential properties of free monoids follow directly from properties of sets and adjoint functors.

Alternately, the bijection between $Set(A,GB)$ and $Mon(FA,B)$ can be described as Pete Clark suggested, without any reference to adjoint functors. This latter option preferable if your logic can handle the categories $Set$ and $Mon$, but not functors between them.

The obvious advantage of this approach is that this is the true freeness property of free monoids (the fact that word monoids are free is nothing but an interesting accident from this point of view). Other than the fact that $F1$ is a natural number object, there is no mention of natural numbers here. However, I always thought that there was a lot of unnecessary tedium to show that free monoids can be viewed as word monoids in this context.

Induction Axiom à la sigfpe. The idea here is to attempt to capture freeness via induction. Here is a variant that fixes the minor problem I pointed out in a comment to sigfpe's answer -- free monoids are structures $(F,A,e,{\cdot})$ where $(F,e,{\cdot})$ is a monoid and $A \subseteq F$ (understood as a unary predicate) is a distinguished set of generators. The axioms to describe this setup are the usual monoid axioms together with the obvious modifications of sigfpe's two axioms

$A(a) \land A(a') \land x\cdot a = x' \cdot a' \to x = x' \land a = a'$ ,
$\forall P\,(P(e) \land \forall x,a\,(P(x) \land A(a) \to P(x\cdot a)) \to \forall x\,P(x))$ ,

where $P$ varies over all unary predicates $P \subseteq F$. In first-order logic, quantification over such $P$ is not possible and the induction axiom must be replaced by the corresponding induction scheme where $P$ varies over all formulas of the language instead. (In higher-order logics without predicate types, you can usually replace $P(x)$ by $f(x) = g(x)$ for appropriate $f$ and $g$.)

In the standard semantics for second-order logic, these axioms completely describe free monoids. However you must be careful since not all formulations of higher-logic are equal. Much like first-order logic, some of these formulations will implicitly allow for models of the above theory which are not free in the categorical sense. In truth, this is unlikely to be a problem for you but I am compelled to warn you of this possibility.

Answer by Benjamin Steinberg for Defining 'free monoid' without Nat?

2011-06-23T21:45:26Z

I am not this is what you want, but there is an old theorem of Levi that a monoid is free iff

It is cancellative.
It has only trivial group of units.
It is equidivisible (you can google the definition; the key thing is that it is first order).
There are only finitely many principal right ideals containing any given principal right ideal.

1-3 are first order. Probably 4 is a problem for what you want.