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The general approach here is a follow up of the approach outlined in a prior posting on extending the projectively extended real line. 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, 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, 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..

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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. 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.

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  • $\begingroup$ I think determing the set of values an expression can take may be interesting, I noticed that this is more complex than with the projective extension of the reals, and also builds up in complexity with higher and higher operators. $\endgroup$ Nov 25, 2022 at 8:40
  • $\begingroup$ @ZuhairAl-Johar Do you have any concrete test questions you think might be interesting? That's generally a good way to motivate an idea as worth pursuing. $\endgroup$ Nov 25, 2022 at 21:58
  • $\begingroup$ The only questions I'd think would be interesting are those sprinning from this method itself, a simple example: what are $1^\infty, 0^{-\infty}$?, it appears that applying $\infty$ to operator identities would cause difficulties and those do not appear to be trivial, and as we use higher operators those become more and more difficult. If we extend the complex plane, things become even murkier. Anyhow I don't have a question in mind outside this method, that serve as an application. $\endgroup$ Nov 26, 2022 at 5:54
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    $\begingroup$ @ZuhairAl-Johar Those examples are easy to calculate: $1^\infty\leadsto x$ iff $x$ is nonnegative, and $0^{-\infty}\leadsto x$ iff $x\in\{-\infty, +\infty\}$. I don't think higher operators/complex numbers/etc. make things significantly harder; you just need to think about what the possible limiting behaviors are, and these are pretty tame. $\endgroup$ Nov 26, 2022 at 6:05
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    $\begingroup$ @ZuhairAl-Johar (I presume you mean "$(-2)^\infty$") The latter, since both $-\infty$ and $\infty$ can be "achieved" by appropriate sequences. $\endgroup$ Nov 26, 2022 at 6:34

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