Hey, so I thought it might help you out to give you what I mean by the real case looks something like the riemann-stieljtes construction.
First of all, if we take a sequence of functions \( h_{jn}(x) \) in \( j,n \in \mathbb{N} \) (where the idea is to limit to a continuous space, through some kind of enumeration),
\(
Y_{n} = \Omega_{j=1}^n h_{jn}(x)\bullet x\\
\)
Then, for \( n\to\infty \), to make \( Y_n \) converge, we get one of two cases.
\(
1.\,\,h_{jn} \to h_j \neq x\,\,\text{as}\,\,n\to\infty\\
2.\,\, h_{jn} \to x\,\,\text{as}\,\,n\to\infty\\
\)
Now if we are in the first case, then we should expect some kind of summability criterion,
\(
\sum_{j=1}^\infty |h_j(x) - x| < \infty\\
\sum_{j=1}^\infty |h_j(x) - L| < \infty\\
\sum_{j=1}^\infty |h_j(x_j) - L| < \infty\\
\)
Where here \( L \) is a point, and \( x_j \to L \) in a summable manner. These are three possible summability criterion, which provide weaker results the further down you go. The first one converges to a function; the second to a point; the third to a point, but it depends on how we choose \( x_j \). This is essentially the discrete case. Where we are not really enumerating anything.
The second way is much more complicated--but this is something John Gill wrote about a lot that I simplified using the Riemann Stiejltes construction. The main normality condition is that,
\( h_{jn}(x) = x + q_{jn}(x)\\ \)
Then, we have to get that \( q_{jn}(x) = \mathcal{O}(1/n)\\ \). If for instance \( q_{jn}(x) = \mathcal{O}(1/n^{1+\epsilon}) \) then \( Y_n \to x \) which is the trivial sequence. If \( q_{jn}(x) = \mathcal{O}(1/n^{1-\epsilon}) \) then \( Y_n \to \infty \), we get the divergent case. Now when \( q_{jn} = \mathcal{O}(1/n) \), its the sweet spot, and not necessarily, but probably will converge to a function,
\(
\lim_{n\to\infty} Y_n = \int_a^b f(s,x)\,ds\bullet x\\
\)
For some \( f(s,x) \). Now this is not my result, I used this result to reframe everything in the Riemann Stieljtes language--where we start with \( f \) rather than derive it afterward. John Gill is the creditor of this result. Now I said, not necessarily that it will look like this, but as soon as we ask for normality of \( h_{jn}(x) \) then necessarily it'll look like this. So if we want to talk about holomorphic functions, then yes it will look like this. The trouble is, it may not be entirely obvious what the function \( f \) is.
Now of course there's a whole bunch of anomalous cases (if we added some cantor arguments or something)--but I don't do that because I want every sequence normal, and every function holomorphic. In which case, the Riemann Stieljtes construction suffices for most general purposes. The difference being, we care about \( f \) rather than \( h \)--and we start the construction with \( f \).
As to your point about the analogy between \( X,Y \) and derivation and integration. I'd like to throw in my own analogy.
\(
\sum \mapsto \Omega\\
\int...ds \mapsto \int...ds\bullet z\\
\)
Which is the analogy that discrete sums become discrete compositions. And continuous sums become continuous compositions. Where now, the beauty being, we still have the same difference relationship and differential relationship. Which I do see you seeing. Which can be expressed as,
\(
\Delta y = F(s) \mapsto \Delta y = F(s,y(s))\\
\frac{dy}{ds} = F(s) \mapsto \frac{dy}{ds} = F(s,y(s))\\
\)
So very much inherently we care about first order differential equations and first order difference equations.
The second analogy comes from something you're saying about superfunctions. Which I think might help. Correct me if I'm wrong, but I think it might be what you are driving at,
Suppose I write,
\(
F(x,z) = \int_0^x h(z)\,ds\bullet z\\
\)
Then,
\(
F'(x,z) = h(F(x,z))\\
F(x,F(y,z)) = F(x+y,z)\\
\)
So technically, for tetration, if I write,
\(
h(x) = (\frac{d}{dx} e\uparrow \uparrow x) \bullet \text{slog}(x)\bullet x\\
\)
Then trivially,
\(
\frac{d}{dx}e \uparrow \uparrow x= h(e \uparrow \uparrow x)\\
\)
And less trivially,
\(
e \uparrow \uparrow x = \int_0^x h(z)\,ds\bullet z |_{z=1}\\
\)
Which, although expressing the superfunction relation, is rather useless if we wanna try and solve using this. But we could definitely approximate using this method--I had a pretty rudimentary way of constructing \( \sqrt{\exp} \) using this, but I wasn't sure if it converges.
So yes, I definitely agree that the continuous case is deeply connected to super functions. Especially when we start taking contour integrations, and modding out by conjugations. Which was the central subject of my last paper--which allows you to construct things like TAylor series and the like.
EDIT:
And I was just joking about Tate's thesis, but what you are suggesting is very similar. If you look at haar measures and the sort it is very very similar to what you are suggesting. Except we're making the transition from \( \sum \mapsto \Omega \)--and haar measures are on groups rather than monoids, but the construction would be similar.
EDIT2 (BUT really EDIt 3 because I reedited the edit; sorry, I'm a fan of the edit functor):
Thought I'd explain this haar measure a bit better. Suppose we have a group \( \{\mathcal{G}, \cdot\} \) and we define functions \( f : \mathcal{G} \times \mathbb{C} \to \mathcal{G} \). Then we write,
\(
+ \mapsto \cdot\\
\)
And we derive a notion of "measure" within this group. Call \( \mu \) the haar measure.
Then instead of writing, as Tate would,
\(
\int f(x) d\mu\\
\)
We'll twist it by binding it to \( z \). We would write,
\(
\int f(x,z)d\mu\bullet z : \mathbb{C} \to \mathbb{C}\\
\)
Which, would look similar to what you are describing. It would simply mean something along the lines (NOT THIS BUT CLOSE),
\(
\Omega_{j=1} z + f_\mu(x,z)\bullet z\\
\)
Where \( f_\mu \) is horrible notation for the use of a measure across a function across groups, which is then measured in the complex plane.
The haar measure is central to preserving the algebraic structure of the group (with monoids you could definitely do at least half of what you can do with groups; I'm curious if there are "monoid haar measures"). We would get things like, for \( \mathcal{I},\mathcal{J} \subset \mathcal{G} \) and measureable.
Product identity (which is an additive condition with Tate; for us it'll be non-abelian so definitely more to unpack),
\(
\int_{\mathcal{I} \bullet \mathcal{J}} f(x,z)d\mu\bullet z = \int_{\mathcal{I}} f(x,z)d\mu\bullet \int_{\mathcal{J}} f(x,z)d\mu\bullet z\\
\)
And left/right invariance (because it's a haar measure)
\(
\int_{a \cdot \mathcal{J}} f(x,z)d\mu\bullet z = \int_{\mathcal{J}} f(ax,z)d\mu\bullet z\\
\int_{\mathcal{J}\cdot a} f(x,z)d\mu\bullet z = \int_{\mathcal{J}} f(xa,z)d\mu\bullet z\\
\)
But, it wouldn't look exactly like this. But what you are driving at with monoid indexes is very much a beast not too different.
First of all, if we take a sequence of functions \( h_{jn}(x) \) in \( j,n \in \mathbb{N} \) (where the idea is to limit to a continuous space, through some kind of enumeration),
\(
Y_{n} = \Omega_{j=1}^n h_{jn}(x)\bullet x\\
\)
Then, for \( n\to\infty \), to make \( Y_n \) converge, we get one of two cases.
\(
1.\,\,h_{jn} \to h_j \neq x\,\,\text{as}\,\,n\to\infty\\
2.\,\, h_{jn} \to x\,\,\text{as}\,\,n\to\infty\\
\)
Now if we are in the first case, then we should expect some kind of summability criterion,
\(
\sum_{j=1}^\infty |h_j(x) - x| < \infty\\
\sum_{j=1}^\infty |h_j(x) - L| < \infty\\
\sum_{j=1}^\infty |h_j(x_j) - L| < \infty\\
\)
Where here \( L \) is a point, and \( x_j \to L \) in a summable manner. These are three possible summability criterion, which provide weaker results the further down you go. The first one converges to a function; the second to a point; the third to a point, but it depends on how we choose \( x_j \). This is essentially the discrete case. Where we are not really enumerating anything.
The second way is much more complicated--but this is something John Gill wrote about a lot that I simplified using the Riemann Stiejltes construction. The main normality condition is that,
\( h_{jn}(x) = x + q_{jn}(x)\\ \)
Then, we have to get that \( q_{jn}(x) = \mathcal{O}(1/n)\\ \). If for instance \( q_{jn}(x) = \mathcal{O}(1/n^{1+\epsilon}) \) then \( Y_n \to x \) which is the trivial sequence. If \( q_{jn}(x) = \mathcal{O}(1/n^{1-\epsilon}) \) then \( Y_n \to \infty \), we get the divergent case. Now when \( q_{jn} = \mathcal{O}(1/n) \), its the sweet spot, and not necessarily, but probably will converge to a function,
\(
\lim_{n\to\infty} Y_n = \int_a^b f(s,x)\,ds\bullet x\\
\)
For some \( f(s,x) \). Now this is not my result, I used this result to reframe everything in the Riemann Stieljtes language--where we start with \( f \) rather than derive it afterward. John Gill is the creditor of this result. Now I said, not necessarily that it will look like this, but as soon as we ask for normality of \( h_{jn}(x) \) then necessarily it'll look like this. So if we want to talk about holomorphic functions, then yes it will look like this. The trouble is, it may not be entirely obvious what the function \( f \) is.
Now of course there's a whole bunch of anomalous cases (if we added some cantor arguments or something)--but I don't do that because I want every sequence normal, and every function holomorphic. In which case, the Riemann Stieljtes construction suffices for most general purposes. The difference being, we care about \( f \) rather than \( h \)--and we start the construction with \( f \).
As to your point about the analogy between \( X,Y \) and derivation and integration. I'd like to throw in my own analogy.
\(
\sum \mapsto \Omega\\
\int...ds \mapsto \int...ds\bullet z\\
\)
Which is the analogy that discrete sums become discrete compositions. And continuous sums become continuous compositions. Where now, the beauty being, we still have the same difference relationship and differential relationship. Which I do see you seeing. Which can be expressed as,
\(
\Delta y = F(s) \mapsto \Delta y = F(s,y(s))\\
\frac{dy}{ds} = F(s) \mapsto \frac{dy}{ds} = F(s,y(s))\\
\)
So very much inherently we care about first order differential equations and first order difference equations.
The second analogy comes from something you're saying about superfunctions. Which I think might help. Correct me if I'm wrong, but I think it might be what you are driving at,
Suppose I write,
\(
F(x,z) = \int_0^x h(z)\,ds\bullet z\\
\)
Then,
\(
F'(x,z) = h(F(x,z))\\
F(x,F(y,z)) = F(x+y,z)\\
\)
So technically, for tetration, if I write,
\(
h(x) = (\frac{d}{dx} e\uparrow \uparrow x) \bullet \text{slog}(x)\bullet x\\
\)
Then trivially,
\(
\frac{d}{dx}e \uparrow \uparrow x= h(e \uparrow \uparrow x)\\
\)
And less trivially,
\(
e \uparrow \uparrow x = \int_0^x h(z)\,ds\bullet z |_{z=1}\\
\)
Which, although expressing the superfunction relation, is rather useless if we wanna try and solve using this. But we could definitely approximate using this method--I had a pretty rudimentary way of constructing \( \sqrt{\exp} \) using this, but I wasn't sure if it converges.
So yes, I definitely agree that the continuous case is deeply connected to super functions. Especially when we start taking contour integrations, and modding out by conjugations. Which was the central subject of my last paper--which allows you to construct things like TAylor series and the like.
EDIT:
And I was just joking about Tate's thesis, but what you are suggesting is very similar. If you look at haar measures and the sort it is very very similar to what you are suggesting. Except we're making the transition from \( \sum \mapsto \Omega \)--and haar measures are on groups rather than monoids, but the construction would be similar.
EDIT2 (BUT really EDIt 3 because I reedited the edit; sorry, I'm a fan of the edit functor):
Thought I'd explain this haar measure a bit better. Suppose we have a group \( \{\mathcal{G}, \cdot\} \) and we define functions \( f : \mathcal{G} \times \mathbb{C} \to \mathcal{G} \). Then we write,
\(
+ \mapsto \cdot\\
\)
And we derive a notion of "measure" within this group. Call \( \mu \) the haar measure.
Then instead of writing, as Tate would,
\(
\int f(x) d\mu\\
\)
We'll twist it by binding it to \( z \). We would write,
\(
\int f(x,z)d\mu\bullet z : \mathbb{C} \to \mathbb{C}\\
\)
Which, would look similar to what you are describing. It would simply mean something along the lines (NOT THIS BUT CLOSE),
\(
\Omega_{j=1} z + f_\mu(x,z)\bullet z\\
\)
Where \( f_\mu \) is horrible notation for the use of a measure across a function across groups, which is then measured in the complex plane.
The haar measure is central to preserving the algebraic structure of the group (with monoids you could definitely do at least half of what you can do with groups; I'm curious if there are "monoid haar measures"). We would get things like, for \( \mathcal{I},\mathcal{J} \subset \mathcal{G} \) and measureable.
Product identity (which is an additive condition with Tate; for us it'll be non-abelian so definitely more to unpack),
\(
\int_{\mathcal{I} \bullet \mathcal{J}} f(x,z)d\mu\bullet z = \int_{\mathcal{I}} f(x,z)d\mu\bullet \int_{\mathcal{J}} f(x,z)d\mu\bullet z\\
\)
And left/right invariance (because it's a haar measure)
\(
\int_{a \cdot \mathcal{J}} f(x,z)d\mu\bullet z = \int_{\mathcal{J}} f(ax,z)d\mu\bullet z\\
\int_{\mathcal{J}\cdot a} f(x,z)d\mu\bullet z = \int_{\mathcal{J}} f(xa,z)d\mu\bullet z\\
\)
But, it wouldn't look exactly like this. But what you are driving at with monoid indexes is very much a beast not too different.