Generalized Kneser superfunction trick (the iterated limit definition)

[JmsNxn]

1. By closure in the monoid, I mean specifically closure under Cauchy sequences using uniform convergence on all compact subsets. By this I mean,

Ok, I was suspecting it but I was misguided by you emphasis on " closure of B AS a monoid". With closure as a monoid I think algebraically. I take a subset of the ambient monoid and perform the closure. Instead you are considering the closure as a subset of a topological space or a kind of completion in respect to the ambient topology, i.e. that of C^1( R).

This should still be a monoid; and will look close enough to \( \mathcal{B} \) for our purposes.

If your \( \mathcal B \), as you defined it, is a monoid, and I can se an informal proof at page 2, I don't see why the Cauchy-completion of the set of elements of this monoid should be a monoid too. Even if it is, the union of two monoids need not to be a monoid: we must do set-union on their sets and then generate the monoid from that union, i.e. perform the, this time monoid-theoretic, closure of the set under composition. From now on I'll denote with \( \overline{\mathcal B} \) your Cauchy-completion and with \( \langle \mathcal B\rangle \) the monoid-completion of the set (the monoid freely generated by that elements).

My proof, I'm enough sure of it is the following corollary 2:
Definition 1: Let \( M \) be a monoid, let \( X\subseteq M \) be a generic subset. Define the submonoid \( \langle X\rangle \) generated by \( X \)

\( \langle X\rangle:=\{\Omega_{i=0}^nf_i\in M \,| f_i\in X;n\in{\mathbb N} \} \)

Definition 2: Let \( M \) be a monoid, let \( X\subseteq M \) be a subset of invertible elements of \( M \). Define the set

\( X^{-1}:=\{f^{-1}\in M \,| f\in X\} \)

Observation: Let \( M \) be a monoid, let \( X,Y\subseteq M \)  be subsets or submonoids. The union \( X\cup Y \)   is not generally a monoid.
Corollary 1: Let \( M \) be a monoid, let \( X\subseteq M \)  be a subset. If \( X \) is already a (sub)monoid then \( \langle X\rangle=X \).
Corollary 2: Let \( M \) be a monoid, let \( X\subseteq M \) be a subset of invertible elements of \( M \). The submonoid \( \langle X\cup X^{-1}\rangle \) is a group. (proof in the previous post).

Questions: let \( X\subseteq {\mathcal C}^1({\mathbb R}) \) a submonoid.

1. When \( \overline{X} \) is monoid?
   
2. If \( X \) is made of invertible functions, does  \( \overline{(X^{-1})}=(\overline{X})^{-1} \) ?
   

Answers to those questions can inform us on the properties of objects like \( \langle \overline{X}\cup \overline{X}^{-1}\rangle \), \( \langle \overline{X\cup X^{-1}}\rangle \) or \( \overline{\langle X\cup X^{-1}\rangle} \).

A2. Yes, I did this construction in hopes this would be a diffeomorphic group (whatever the hell you call it), a subgroup of \( \text{Diff}(\mathbb{R}) \). I avoided this language largely because I'm not familiar enough with the language.

Rememeber that I would not call a group diffeomorphic like it is a property of the group. I call it a group of diffeomorphism, like a group of apples or a group of permutations. "The diffeomorphisms groups of R" just means "the group made of diffeomorphism of R to itself". I'm not aware if a grop can be diffeomorphic in any sense (and idc right now).

Lol, yes diffeomorphic group is nonsense, I meant group of diffeomorphisms . I'm horrible with jargon.

The reason I call the closure of \( \mathcal{B} \) a monoid is because it should look something like,

\(
f : \mathbb{R} \to \mathbb{R}\,\,\text{bijectively}\\
f'(x) \ge 0\\
\lim_{x\to\infty} f'(x) \ge 1\\
\lim_{n\to\infty} f^{-n}(x) = -\infty\\
\)

Which is certainly still a monoid; but pulling out inverses will become difficult to complete it as a group, because \( f'(x) = 0 \) is a possibility. And we may lose the last property (I'm not certain, I hope not), but at worst only a weakening of it. And we should also ensure convergence of the derivatives in the Cauchy sequence. For that reason we'll want \( f'_n(x) \) to be Cauchy on compact subsets too.

You're right about the union too; in the general case. But again, the union should look like,

\(
f : \mathbb{R} \to \mathbb{R}\,\,\text{bijectively}\\
f'(x) \ge 0\\
f(x) \to \infty\\
\lim_{x\to\infty} f'(x) \ge 0\\
\)

So at least near infinity these things should have inverses and form a group. It'll be hard to ensure it works on all of \( \mathbb{R} \) though, we'll need \( f'(x) > 0 \) which will require more complicated tricks than just completing with Cauchy; so you're right there. I think the closure of this monoid will have to be done more carefully--you're right. I'm going to focus my attention on simply strengthening everything I put above; and covering all my bases.