Is this extension of the projectively extended real line, consistent?

Question

This posting has been Edited. The edited material shall be noted.

The projectively extended real line $\hat {\mathbb R}= \mathbb R \cup \{\infty\}$ is one system which allows division by zero! Yet it has many undefined arithmetic operations.

If we weaken the requirement of the operators being binary functions to being ternary relations, and so can relate many values to the same arguments; then those undefined operations can be settled in the sense of getting some kind of closure under them.

I'll adopt the notation "$a * b \to c$" to mean: the operator $*$ is a ternary relation that sends the ordered pair $(a,b)$ to $c$; for convience, it can be read as: $c$ is a result of $a * b$.

Now we may coin a closure notion over $\hat {\mathbb R}$ under an operator $*$ as:

$ \hat {\mathbb R} \text { closed under } * \iff \forall x,y \in \hat{\mathbb R} \exists z \in \hat {\mathbb R}: x * y \to z$

I'll revert to the functional expression (i.e. $x * y = z$) only when $z$ is unique per $(x,y)$

The following is a possible extension:

$\infty + \infty = \infty \\ \infty - \infty \to r \\ \infty \times 0 \to r \\ 0 \times \infty \to r \\ \infty / \infty \to r \\ 0 /0 \to r$

Where $r$ is any element of $\hat{\mathbb R}$.

[EDIT]: the above was the original suggestion, which was motivated by defining subtraction and division as the reciprocal relation of addition and multiplication respectively, i.e.:

$a - b \to c \iff c + b \to a \\ a / b \to c \iff c \times b \to a$

But, ($\infty + \infty \to \infty$) doesn't conform to that! We have $r - \infty = \infty$ so this results in $\infty + \infty \to r$, so this to be corrected to ($\infty + \infty \to r$); for all $r \in \hat{\mathbb R}$.

Additional motivation is to have multiplication by zero be equivalent with self subtraction.

Accordingly $\hat{\mathbb R}$ would be closed under operators "$+,-,\times, /$" in the sense defined here.

Perhaps a similar move can be done for the Riemanian sphere $\mathbb C^*$, to gain closure under more operators.

Is this system consistent relative to $\sf ZFC$? More specifically, what are the particulars of what could be viewed to constitute somehwat natural interpretation of this system in $\sf ZFC$?

Answer 1

This is an account on the particulars of an interpretation of the original system [before the edit] presented in this question in set theory.

First we define an extended kind of rationals to suit adding a rational that is higher than all other rationals, the latter would correspond to $\infty$. The set of all those rationals shall be denoted by $\mathbb Q^*$

$\text {Define}: r \in \mathbb Q^* \iff r \subseteq \mathbb R \times \mathbb R \land [\operatorname {image}(r) = \mathbb R \lor \operatorname {preimage}(r) = \mathbb R] \land \\\exists a,b \in \mathbb Z : r= \{\langle a \times x, b \times x \rangle \mid x \in \mathbb R \}$

In the above definition $r$ is meant to represent the rational number $a/b$ for integers $a,b$.

A strict smaller than relation $<$ can be defined on $\mathbb Q^*$ as:

$r < s \iff \\ r \neq s \land \exists x \exists a \, \exists b \,( a < b \land \langle a, x \rangle \in r \land \langle b, x \rangle \in s)$

As usual elements of $\mathbb Q^*$ strictly below $0/1$ are negatively signed, those above except $\infty$ are positively signed, while the rest (i.e.; $0, \infty$) are unsigned, this can also be captured in terms of sets as the signed rationals being those which have both their images and preimages being $\mathbb R$.

So we define: $\infty = \{ \langle x,0 \rangle \mid x \in \mathbb R \}$, i.e. the reciprocal of $0/1 = \{\langle 0,x\rangle \mid x \in \mathbb R \}$

Addition of any two extended rationals $r,s$ is given by:

$r + s =\{\langle a+b,c \rangle \mid \langle a,c \rangle \in r\land \langle b,c \rangle \in s\}$

This gives: $r + \infty = \infty + r= \infty$ for all $r \in \mathbb Q$

Subtraction over $\mathbb Q^*$:

$r - s \to q \iff \\ q+s = r \lor q= \{\langle a-b,c \rangle \mid \langle a,c \rangle \in r\land \langle b,c \rangle \in s\} $

This gives: $r - \infty = \infty-r = \infty$, for all $r \neq \infty$

and $ \infty - \infty \to r $, forall $r \in \mathbb Q^*$.

Multiplication over $\mathbb Q^*$:

$r \neq 0, s \neq 0 \\ r \times s = \{\langle k \times x,a \rangle , \langle t \times h, m^2 \rangle \mid (\langle t,m \rangle \in r [s] \land \langle h,m \rangle \in s [r]) \land (\langle k,1 \rangle \in r [s] \land \langle x,a \rangle \in s [r]) \} $

Where: $(z \in r[s] \land u \in s[r])$, is short for: $(z \in r \land u \in s) \lor (z \in s \land u \in r)$

if $r =0 \lor s=0 \implies r \times s \to (r-r)+(s-s) $

So, we get: $r \times \infty = \infty \times r= \infty$, for all $r \neq 0$

and, $0 \times \infty \to r; \infty \times 0 \to r $, forall $r \in \mathbb Q^*$

Now, that we defined addition and multiplication of the extended rationals, we can define extended reals as Dedekind cuts over $\mathbb Q^*$

A Dedekind cut shall be defined here as a binary partition on $\mathbb Q^*$, with one block being an initial segment (i.e.; closed under $<$) of $\mathbb Q^*$ that is open upwardly. So, for example $\{\mathbb Q^* \setminus \{\infty\}, \{\infty\}\}$ is a Dedekind cut, and it is taken to represent the real number that corresponds to the extended rational number $\infty$, and so it'll be denoted by "$[\infty]$". The elements of a Dedekind cut are to be termed initial, terminal abbreviated as init,term, the former is the one closed dowardly, the latter is the one closed upwardly.

We can define a total order $<$ on Dedekind cuts themselves, this is given by:

$K < L \iff \operatorname {init}(K) \subsetneq \operatorname {init}(L)$

That said, then clearly $[\infty]$ is strictly greater than all other cuts. Formally this is: $$ r \neq [\infty] \implies r < [\infty]$$, for every extended real $r$.

The set of extended reals to be designated by $\hat {\mathbb R}$

As a terminology if $S$ and $C$ are nonempty subsets of $\mathbb Q^*$, then:

$S \ * \ C = \{a \ * \ b \mid a \in S \land b \in C \} $

Where "$*$" is some arithmetic operator.

Define: $X=\{S,-\} \iff X=\{S, \mathbb Q^* \setminus S \}$

Addition of extended reals:

$K+L = \{ \operatorname {init}(K) + \operatorname {init}(L), - \} $

This yields: $r + [\infty] = [\infty]$, for all $r \in \hat{\mathbb R}$

Subtraction of extended reals:

$ K-L \to X \iff \\ X + L=K \lor \\K < [\infty] \land L=[\infty] \land X=[\infty]$

Another definition is: $K-L \to X \iff \\ X + L=K \lor \\ X = \{\operatorname {init} (K) - \{x \in \operatorname {term}(L) \mid L \neq [\infty] \to x \neq \infty \}, - \}$

This gives: $ r - [\infty]= [\infty]-r = [\infty] $, for all $r \neq [\infty]$,

and: $[\infty] -[\infty] \to r $, for all $ r \in \hat {\mathbb R}$.

Define: $\operatorname {Comp} (S) = S \cup \{\operatorname {Inf}(S)\} $

$\operatorname {Comp}$ is read as the completion set of.

Multiplication of extended reals:

$K > 0 \land L > 0: \\K \times L= \{\operatorname {Comp} (\operatorname {term}(K) \times \operatorname {term}(L)) , -\}$

$ K > 0 \land L < 0: K \times L = 0 - (K \times (0 - L))$

$ K < 0 \land L < 0: K \times L = (0-K) \times (0-L)$

$ K = 0 \lor L = 0:\\ K \times L \to X \iff (K-K) + (L-L) \to X$

Accordingly: $r \times [\infty] = [\infty]$, if $r \neq 0$;

and: $0 \times [\infty] \to r, [\infty] \times 0 \to r $; for all $r \in \hat {\mathbb R}$

Division of extended reals:

$ K/L \to X \iff X \times L \to K $

Yielding: $[\infty]/r = [\infty], r/[\infty]=0$, for all $r \neq [\infty] $;

and: $r/ 0 = [\infty], [\infty]/[\infty] \to r, 0/0 \to r $, for all $r \in \hat{\mathbb R}$.

This would establish the interpretation of the projectively extended real line in $\sf ZFC$, and also provides an answer to the undetermined expressions mentioned in the Wikipedia page, along the lines mentioned in this question. So, it proves the consistency of this system.