0
$\begingroup$

There is a very wide series of questions I have been thinking about and I am wondering if there is any literature on this type of structures.

Let's start with the set of all functions $\mathcal{F}: \mathbb{R} \to \mathbb{R}$. The subsets of this set form a lattice (based on set inclusion). Let $\mathbb{P}_{n}$, $\mathcal{A}$, and $\mathcal{C}$ be the rings of polynomials with degree at most $n$, analytic functions, and continuous functions respectively. So that in the lattice structure we have $\mathbb{P}_0 \subset \mathbb{P}_3 \subset \mathcal{A} \subset \mathcal{C} \subset \mathcal{F}$. For simplicity limit the elements of this lattice to those sets which form a ring of functions. I am not sure why, but this seems like a good restriction.

Suppose we are given a subset of $\mathcal{F}$ which is a ring, let's call this $\mathcal{R}$. Suppose we also a set of points $\mathbb{S} \subset \mathbb{R}$. Then we say $\mathbb{S}$ is a separating set of $\mathcal{R}$, if for any $f,g \subset \mathcal{R}$ there is an $x \in \mathbb{S}$, such that $f(x) \neq g(x)$. Essentially the values of a function on $\mathbb{S}$ uniquely determine the function in $\mathcal{R}$.

So, let's look at what some of these sets look like.

Ring of Functions Separating set
$\mathbb{P}_0$ $\{x_0 | x_0 \in \mathbb{R}\}$, any singleton
$\mathbb{P}_n$ $\{x_0, x_1, ..., x_n | x_i \in \mathbb{R}, x_i \neq x_j \text{ if } i \neq j \}$, any set of $n$ points
$\mathcal{A}$ $\mathbb{S} \subset \mathbb{R}$, such that $\mathbb{S}$ has a limit point.
$\mathcal{C}$ $\mathbb{S} \subset \mathbb{R}$, such that $\mathbb{S}$ is dense in $\mathbb{R}$
$\mathcal{F}$ $\mathbb{R}$

These sets seem to be capturing some kind of topological property. I am not sure about what kind of topological equivalence would capture these sets. On the other hand, these sets definitely form a lattice as well through inclusion. There is an anti-correspondence between these sets and their corresponding function rings.

Do these functions sets have to be rings? No, but it seems like a good idea.

Does the base field have to be $\mathbb{R}$? No, but it is a good starting place.

What I am interested in is if the topological property can be well defined and some kind of lattice be determined on it from which we can recover the lattice of functions. It will probably dictate the structure of the function sets (rings, groups, etc).

It seems the topological equivalence I am looking for is just a homeomorphism. Dense sets will remain dense, sets with limit points preserve those. Am I thinking correctly?

What do we know about the lattice of subsets of $\mathbb{R}$ upto homeomorphic equivalence?

$\endgroup$
1
  • $\begingroup$ Could you please state your questions a bit more precisely? Also, the polynomials do not form a lattice. $\endgroup$
    – erz
    Jan 28 at 9:40
1
$\begingroup$

from which we can recover the lattice of functions

There are a lot of ideas here, but if this is the main question the answer is no. The following rings (indeed, algebras) all have exactly the same separating sets, namely the dense subsets of $\mathbb{R}$:

  1. all continuous functions

  2. all bounded continuous functions

  3. all continuous functions which go to zero at infinity

  4. all $C^1$ functions

  5. all smooth functions

etc.

The notion of a separating set as defined here seems interesting. An equivalent definition is that $S$ is separating for $\mathcal{R}$ if $0$ is the only function in $\mathcal{R}$ which vanishes on $S$. But there isn't this correspondence that you're asking for.

$\endgroup$
4
  • $\begingroup$ I am not sure those two definitions are equivalent. For example the constant function f(x)=1 and g(x)=cos(x) will not be separated by the set $\{2*pi*n | n \in \mathbb{Z}\}$. Ah, wait then the Ring property here would give h(x)=cos(x)-1. Which then satisfies the property you are speaking about. Am I thinking correctly? $\endgroup$
    – Heraiwa
    Jan 28 at 22:41
  • $\begingroup$ On the set of functions you described as being different, that's cool. So it means at least that the separating set (upto homeomorphism) doesn't capture the entire lattice. It does up to some equivalence. Now I am considering what that equivalence translates to in the lattice of functions. $\endgroup$
    – Heraiwa
    Jan 28 at 22:44
  • $\begingroup$ If we assume the set of functions is a group, that seems to have the same effect as assuming it is a ring over $\mathbb{R}$. I can't think of separating sets that would work to separate a group vs ring of functions generated from a set of functions. Perhaps if we impose the functions to form a field there is a difference. For example considering the field of functions of the form p(z)/q(z) from the Reimann sphere to itself. Where p and q are polynomials. $\endgroup$
    – Heraiwa
    Jan 29 at 0:44
  • $\begingroup$ I'm not sure what is the right question to ask, but I feel there is a research project here ... $\endgroup$
    – Nik Weaver
    Jan 29 at 1:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.