“Wheel Theory”, Extended Reals, Limits, and “Nullity”: Can DNE limits be made to equal the element “$0/0$”?

Question

"Wheels" are a little-known kind of algebraic structure:

They modify the concept of a field or a ring in such a way that division by any element is possible, including division by zero, while also avoiding contradictions (such as $2 = 1$) in the algebra. They do this by essentially promoting and generalizing the "inversion" operator $x^{-1} = \frac{1}{x}$ to a primary operation, and modifying the distributive laws.

Specifically, a wheel is an algebraic structure $(W, +, *, /)$ consisting of a set $W$, two binary operations $+$ and $*$, which are just addition and multiplication, and a third, unary operation $/$, which could be called "division" or "involution", satisfying:

1. $(W, +)$ and $(W, *)$ are commutative monoids.
2. $/$ is involutive, i.e. for all $a \in W$, $//a = a$.
3. A number of modified distributivity principles: for all $a, b, c \in W$, $$ac + bc = (a + b)c + 0c$$ $$(a + bc)/b = a/b + c + 0b$$ $$(a + 0b)c = ac + 0b$$ $$/(a + 0b) = /a + 0b$$
4. $0 * 0 = 0$
5. Existence of additive annihilator: for all $a \in W$, $0/0 + a = 0/0$.

Following the lead of a somewhat eccentric "computer scientist" who proposed some stuff along these lines but otherwise was kinda loopy, I call $0/0$ "nullity", and denote it $\Phi$.

We can then form a wheel from the reals by forming the set $W = \mathbb{R} \cup \{ \infty, \Phi \}$, where we take $/0 = \infty$ and $\Phi = 0/0$. This infinity is unsigned, as in the real projective line. Addition and multiplication are defined similarly, except whenever an operation is "undefined", we define it to equal $\Phi$. In particular, we have $\infty + \infty = \Phi$, $0 * \infty = \Phi$, $0^0 = \Phi$, etc.

We can define a "topological wheel" to be a wheel where the set $W$ has topological structure and the functions $+$, $*$ and $/$ are continuous functions, in a manner analogous to the definitions of topological rings and fields. The topology put on the real wheel above would be like that of the projective line plus an isolated point $\Phi$. This is the inspiration for the term "wheel": you can draw this structure on a piece of paper as a circle with a point for $\Phi$ in the center (of course, you can put in anywhere not on the circle, but this is where the term comes from), and that will look like a cart wheel with axle.

So in this space, we have that "undefined" operations like $\frac{0}{0}$ yield $\Phi$. Yet with limits, we still have that, say, $\lim_{x \rightarrow 0} \sin\left(\frac{1}{x}\right)$ DNE. So my question is:

Is it possible to put a topology on this wheel so that all functions have a limit, with those whose limit DNE in the usual topology having limit $\Phi$ and those whose limit exists in the usual topology have that same limit here?

If "no", what is the largest possible class of functions including all those whose limits exist in the usual topology for which the above can be done?

EDIT: Hmmmmmmm... I notice that the "wheel-shaped" topology actually doesn't give a topological wheel after all! In particular, the map $x \mapsto x + \infty$ is not continuous in this topology. Note that the preimage of the open set $\{ \Phi \}$ (which is open since $\Phi$ is an isolated point and is actually in fact clopen) pulled back through this map is not $\{ \Phi \}$ but $\{ \infty, \Phi \}$, since $\infty + \infty = \Phi$. Yet this set is not open, but closed, being the union of the closed sets $\{ \infty \}$ and $\{ \Phi \}$ and $\{ \infty \}$ is not a connected component, so it can't be clopen and must be only closed.

So this begs another question: is there even any topology on this wheel at all which makes it into a topological wheel and such that real limits are preserved? If so, does such a topology automatically give "DNE" (in the projective reals) limits $\Phi$ as a value?

Answer 1

First of all I'd like to say that I'm very amused by the fact that you and I independently decided to call wheel theory's $0/0$ 'nullity' after James Anderson's 'transreal arithmetic'.

I'm fairly sure that if you take the real projective line topology on $\mathbb{R}\cup\{\infty\}$ and append $\Phi$ as an open extension topology (i.e. the open sets are precisely the pre-existing open sets in $\mathbb{R}\cup \{\infty\}$ and the entire space $\mathbb{R}\cup\{\infty,\Phi\}=\odot_\mathbb{R}$) then you get a topological wheel. Furthermore I think this may be the only way to extend the ordinary real projective line to get a topological wheel (largely because of reasoning similar to that in your edit), but I haven't proved that.

This topology is somewhat reminiscent of generic points in the Zariski topology on the spectrum of a ring in that nullity is 'next to' every other number, but it's not exactly the same. Also it's somewhat natural in that it's the quotient topology of $\mathbb{R}^2$ under the equivalence relation $(a,b)\sim(c,d)$ iff $(a,b)$ and $(c,d)$ are not $(0,0)$ and $(a,b)=(e\cdot c,e \cdot d)$ for some nonzero $e$, which is just the construction of the real projective line without deleting $(0,0)$.

As far as the limits are concerned you almost get what you want. Every sequence converges to $\Phi$ and at most one other point. The non-$\Phi$ limit point exists iff the sequence converges in the real projective line topology and is equal to that limit.

Furthermore I think that this may be the best that you can do. If you consider any series $a_n\in\mathbb{R}$ that normally does not converge, but in your topological wheel converges to $\Phi$, then the series $(a_n, -a_n)$ converges in the product topology $\odot_\mathbb{R} \times \odot_\mathbb{R}$ to $(\Phi,\Phi)$, addition is a continuous map $+ : \odot_\mathbb{R} \times \odot_\mathbb{R} \rightarrow \odot_\mathbb{R}$ therefore the sequence $a_n - a_n=0$ must converge to $\Phi+\Phi=\Phi$ as well as the obvious limit of $0$.