The general approach here is a follow up of the approach outlined in a prior posting on [extending the projectively extended real line](https://mathoverflow.net/questions/434794/is-this-extension-of-the-projectively-extended-real-line-consistent). In particular arithmetic operators break down to ternery relations that may sometimes admit multivaluation. Here, it'll be applied to extend the affinely extended real line.


Working in $\sf ZFC$, We start with any model $(\mathbb R, =, <, +, -, \times, /)$ satisfying the customary sentences of arithmetic of reals. Now we take any two distinct sets $+\infty, -\infty$ that are not elements of $\mathbb R$ and adjoin them to $\mathbb R$ to get $\hat {\mathbb R}$. That is, we have: $$ \hat{\mathbb R} = \mathbb R \cup \{+\infty, -\infty\}$$


We define $=^\*$ by the restriction of $=$ to $\hat {\mathbb R}$, so it is the set $= \restriction\_\hat{\mathbb R}$


Similarly $\neq^\*$ is the set $\neq \restriction\_\hat{\mathbb R}$


Of course we have $\langle +\infty, -\infty \rangle \in \, \neq^\*$


Note: Whenever "$\times$" appears as an infix then it designates Cartesian product, otherwise if "$\times$" appears as a constant then it stands for the customary multiplicaiton relation over reals.


Now we extend $<$ to an extended strict less than relation $<^\*$ over $\hat{\mathbb R}$, as:


$<^\* = \, (< \cup \ (\hat {\mathbb R} \times \{+\infty\}) \cup (\{-\infty\} \times \hat {\mathbb R} )) \ \cap \neq^\*$


Then we extend the set $+$ to $+^\*$ defined as:


Let: $+' = + \ \cup \\ \hat{\mathbb R} \times \{+\infty\} \times \{+\infty\} \ \cup \\ \hat{\mathbb R} \times \{-\infty\} \times \{-\infty\} \ \cup \\ \{+\infty\} \times \{-\infty\} \times \hat {\mathbb R} $


Define: $+^\* = +' \cup \{\langle b,a,c \rangle \mid \langle a,b,c \rangle \in +' \}$


We define $-^\*$ as the reciprocal of $+^\*$:


$-^\* = \{ \langle a,b,c \rangle \mid \langle c,b,a \rangle \in +^\* \}$


Let $\hat{\mathbb R}^+; \hat{\mathbb R}^-$ be the sets of all extended reals (i.e., elements of $\hat{\mathbb R}$), strictly above or strictly below zero, respectively.


We extend the set $\times$ to $\times^\*$ defined as:


Let: $\times' = \times \ \cup \\ \hat{\mathbb R}^+ \times \{+\infty\} \times \{+\infty\} \ \cup \\ \hat{\mathbb R}^+ \times \{-\infty\} \times \{-\infty\} \ \cup \\ \hat{\mathbb R}^- \times \{+\infty\} \times \{-\infty\}\ \cup \\ \hat{\mathbb R}^- \times \{-\infty\} \times \{+\infty\} \ \cup \\ \{+\infty\} \times \{0\} \times \hat{\mathbb R} \ \cup \\\{-\infty\} \times \{0\} \times \hat{\mathbb R} $


Define: $\times^\* = \times' \cup \{\langle b,a,c \rangle \mid \langle a,b,c \rangle \in \times' \}$


We define $/^\*$ as the reciprocal of $\times^\*$:


$/^\* = \{ \langle a,b,c \rangle \mid \langle c,b,a \rangle \in \times^\* \}$


So $(\hat{\mathbb R}, =^\*, <^\*, +^\*, -^\*, \times^\*, /^\*)$ is our model of the affinely extended real line extended to be closed under all above-mentioned arithmetic operators.


So, we get all defined expressions mentioned in the Wikipedia page on arithmetic of [affinely extending the real line](https://en.wikipedia.org/wiki/Extended_real_number_line), and also settle all mentioned undefined expressions to:


$ 0/0 \to r \in \hat{\mathbb R}\\ \pm\infty + (\mp \infty) \to r \in \hat{\mathbb R} \\ \pm\infty - (\pm \infty) \to r \in \hat{\mathbb R} \\0 \times \pm \infty \to r \in \hat{\mathbb R} \\ \pm\infty /\pm\infty \to r \geq 0 \\ \pm \infty / \mp \infty \to r \leq 0 \\ r/0 \to \pm \infty$


Unlike [extending the projectively extended real line](https://mathoverflow.net/questions/434794/is-this-extension-of-the-projectively-extended-real-line-consistent), here the operational multi-valuation is not that trivial. And if we demand closure over more operators, like extending this method to the complex number line and demand closure under exponentiation\rooting; then one expect more different multi-valuation patterns.



> 
> Is there an effective way that governs extending this approach as to cover all iterative $+$ operators and their reciprocals?
> 
> 
> 


By iterative $+$ operators I mean multiplication, exponentiation, hyperexponentiation, etc..



Is there an effective way to generalize this approach of affinely extending the number line?
First of all, mentioning set-theoretic background is really just making this more complicated than it needs to be, so I'll ignore it. Additionally, I'll use "$\leadsto$" in place of your "$\rightarrow$" since I also want to talk about conventional limits.


It sounds like you want $t\leadsto c$ whenever $t$ "has a form" that equals or converges to $c$ (for $t,c$ appropriate terms). This can be made straightforwardly precise, but I don't think it's particularly useful. Specifically, given an $n$-ary partial function $f:\mathbb{R}^n\rightarrow\mathbb{R}$, we can consider its multivalued-and-partial extension $\hat{f}$ to $\hat{\mathbb{R}}^n$ by setting $\hat{f}(a\_1,...,a\_n)\leadsto b$ iff there are sequences of (non-extended!) reals $(\alpha^1\_i)\_{i\in\mathbb{N}}\rightarrow a\_1,..., (\alpha^n\_i)\_{i\in\mathbb{N}}\rightarrow a\_n$ such that for each $i$ the expression $\beta\_i:=f(\alpha\_i^1,...,\alpha\_i^n)$ is defined and $(\beta\_i)\_{i\in\mathbb{N}}\rightarrow b$. Here, "$\rightarrow$" refers to convergence in $\hat{\mathbb{R}}$ in the usual sense.


This gives all the examples you include, as well as things like $$\sin(\pm\infty)\leadsto x\iff x\in[-1,1]$$ and $C(x)\leadsto y$ for all $x,y\in\hat{\mathbb{R}}$ where $C$ is [Conway's base $13$ function](https://en.wikipedia.org/wiki/Conway_base_13_function). Additionally, if $f$ is total on $\mathbb{R}$ then $\hat{f}$ is total on $\hat{\mathbb{R}}$ and extends $f$ (for the latter point just take $\alpha^k\_i=a$), and all "value sets" $$\hat{f}[x]:=\{y: \hat{f}(x)\leadsto y\}$$ will always be **closed** for any $f$ whatsoever.


This last observation points to the right way to think about what's going on here, in my opinion: we're really just taking the topological closure of the graph of $f$ as a subset of (the appropriate power of) $\hat{\mathbb{R}}$ and declaring that to be the graph of $\hat{f}$. But I don't think we actually get anything interesting this way.


