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A problem (at least, I think it is) by saying that $\infty - \infty = 0$ is the fact that there are numerous examples of (un)countably infinite sets that can be subtracted from each other, and produces sets with an (un)countably infinite, or finite cardinality; and the finite ones can be any number $\in \Bbb N_0$.

So, if we define subtraction to be a binary function that takes $a$ and $b$ as its inputs, and that outputs the cardinality of the set that results from $|A| -|B|, \ |A| = a \land |B| = b$, then it is clear that the minus symbol is simply too ambiguous when $|A| = |B| = \infty$. Infinity is not a number, but it is a cardinality; that means that if we define subtraction via cardinality, then we must make the operation more precise, in order to deal with the ambiguities inherent in infinity.

So, under this new model (not sure if I can call it that), $-$ becomes $-_{f(x)}$, where $f(x)$ is an arbitrary function. When $f(x) = x$, then we write $-_{f(x)}$ as $-$, given that this denotes the typical subtraction.

All numbers correspond to their roster sets in this model; so $a$ corresponds to $\Bbb N_1^a$.

First, let's deal with two finite cases.

$$10-_{f(x)=x} 5 = 10-5 = 5$$

Then we have:

$$10-_{f(x) = 2x} 5 = |\{1,2,\dots,9,10\}| - |\{n \in \Bbb N_1^5 \mid 2n\}| = 10 - 5 = 5$$

As we can see, any one-to-one function subscripted to the minus sign will never change the answer when subtracting finite numbers. There is a difference however; the sets whose cardinalities are being subtracted from each other are different.

Thus, if $a,b$ are finite, and $f(x)$ is a one-to-one function, $a-_{f(x)} b$ is simply $a-b$. If the function is many-to-one, or one-to-many, it will be like decreasing the negative number, or increasing the negative number, respectively.

Now, for the infinite cases:

Firstly, in this model, $\infty$ without further specification is meaningless. An infinity is defined as the cardinality of a set, and with differing sets, there'll be differing infinities.

Subtraction on infinities is only defined if one set is a subset of the other after the function has acted on its elements. That allows the movement from $|A| - |B|$ to $|A-B|$, which allows for the evaluation of the cardinality, and thus the subtraction of infinities.

In this section, take $\infty = |N_1|$.

$$\infty -_{f(x)=x} \infty = \infty - \infty =|\{1,2,3, \dots \}| - |\{1,2,3, \dots \}| = |\{1,2,3, \dots \} - \{1,2,3, \dots \}| = 0$$

This is normally not defined, due to the absurd consequences defining it has. It is however my belief that in this model, those absurd consequences aren't there. That allows for this quite sensible interpretation to be used.

$$\infty -_{f(x)=2x} \infty = |\{1,2,3 \dots \} - \{2,4,6,\dots\}| = |\{ 1,3,5, \dots \}| = \infty$$

With this example, we're able to resurrect the intuitive idea that the even numbers are indeed less than all the naturals.

$$\infty -_{f(x) =\begin{cases} 1 & x\le 5 \\ x & x>5 \end{cases}} \infty = |\{1,2,3,\dots\} - \{1,6,7,8,\dots\}| = |\{2,3,4,5\}| = 4$$

This is basically just equivalent to saying:

$$\infty - \infty +4 = 0 + 4 = 4$$

I am not going to touch on the subtraction of uncountable infinities, since this post is already long and I think I've gotten the gist of my idea across.

So, here are my questions:

  1. Does this model produce absurdities?
  2. Has this been thought of before?
user110391
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    I do believe that this is not the correct sense of model, and such the model theory tag should not be used. A model is an interpretation of a logical system which satisfies all the axioms of the system. – Logan M Sep 30 '22 at 06:08
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    With finite cases... what if we consider $f(x)=3x$. Then don't we get $10-_f5={1,\cdots,10}\setminus{3,6,9,12,15}={1,2,4,5,7,8,10}=7$? – Logan M Sep 30 '22 at 06:22
  • @LoganM Yeah I wonder, if not model, what could the right word (and thus tag) be? Hopefully more people will chime in. My understanding of a model is far too vague, but it felt like it could be seen as a model. As for your example: $$10 -_f 5 = |{1,\ \dots, \ 10}| - |{3,6,9,12,15}| = 5$$. You're not subtracting elements, you're subtracting cardinalities :) In this "model", the difference of cardinalities only amounts to the subtraction of elements iff the corresponding set is a subset of the other set. This is a very helpful fact for infinite sets. – user110391 Sep 30 '22 at 07:04
  • @LoganM Also, just for the sake of learning, I'll tell you why I suspected it to be a model. What I've done is reinterpreted the traditional subtraction $a-b$ to be a function that takes the difference of the cardinalities of the roster set of $a$ and one changed set, where the changed set is the output of putting the elements of the roster set of $b$ under some arbitrary one-to-one function. Obviously, changes nothing, meaning it is a valid interpretation for all defined cases. However, this reinterpretation allows to expand the set of defined cases, so that subtraction can act on infinities. – user110391 Sep 30 '22 at 07:22
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    $\infty-\infty$ does not really make sense. If we interprete it as a limit consider $n-n$ and $n^2-n$ , the first expression has limit $0$ , but the second $\infty$ although both summands tend to $\infty$. In fact , we can have every real number $r$ as a limit by subtracting $n+r$ and $n$ where $n$ tends to $\infty$. – Peter Sep 30 '22 at 08:17
  • @Peter It is that lack of sensibility my "model" tries to fix. It ascribes size to infinity (not in the sense of countability), but in a far more basic, arithmetic way. – user110391 Sep 30 '22 at 10:20
  • Not sure what you mean. We can of course start with an infinite set and remove an infinite set of elements , for example we could create $\mathbb N\P$ (where $P$ is the set of the primes). Do you mean something like that ? – Peter Sep 30 '22 at 10:34
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    @user110391 Cardinal subtraction can only be defined for infinite cardinals when there is a strict inequality between them, so I feel this subtraction you're defining (even ignoring any other potential issues) isn't going to work they way you want in any case. – Logan M Sep 30 '22 at 13:27
  • @user110391 I guess it could be interpreted in a similar manner to a model in a very loose sense, but this most certainly will not satisfy the axioms that subtraction has on the natural numbers so the system it's a model of is likely quite useless here. – Logan M Sep 30 '22 at 13:28

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I don't think this subtraction is well-defined: Generically there can be no subtraction for $\infty$ that satisfies $1+\infty=\infty$, $\infty-\infty=r$ for some $r\in\mathbb R$ and associativity, since then we have $$r=\infty-\infty=(1+\infty)-\infty=1+(\infty-\infty)=1+r\iff 0=1.$$

In your case, for example, we could write, if I understand correctly (I am not sure if I do ) $$1=\lvert\{0\}\cup\mathbb N\rvert-_{f(x)=x}\lvert\mathbb N\rvert = \infty-_{f(x)=x}\infty=0,$$ i.e. $1=0$.

Maximilian Janisch
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  • Your contradiction is derived from my imprecise explanation. In my examples, I was using $\infty = |\Bbb N_1 |$. So, your $|\Bbb N_0| = \infty + 1$. And $\infty -_{f(x) =x} \infty + 1 = \infty - \infty +1 = 0+1 = 1$. When using the symbol $\infty$ normally, it satisfies the identity $\infty + 1 = \infty$. In one way, it doesn't though, since I think arithmetic isn't normally defined on infinity. In my model however, $\infty + 1 \ne \infty$, because infinity denotes the cardinality of a specific set, which gives it arithemtically manipulatable content. – user110391 Sep 30 '22 at 10:08
  • In my model, $\infty$ is sort of like a bounded variable. It is a variable with a value of a size such that if you were to count to it, you would never reach it. Despite this, my model ascribes greater/less-than and equality to these variables. This then allows arithmetic manipulation I think (but it may do so at the cost of sensibility xD). Whenever you write $\infty -_f \infty$, it is a given that those infinities are initially equal (though one of them is via the $f$-function possibly made unequal). In my example, I specify that my $\infty$ is denoting $|\Bbb N|$. – user110391 Sep 30 '22 at 10:14
  • Also, note that I am using "model" in lieu of what's probably a far more correct term. I am not familiar enough with the terminology to know of a different know, nor to know how incorrect model is in this context. – user110391 Sep 30 '22 at 10:16
  • If we treat "$\infty$" as a number, $1+\infty=\infty$ already implies $0=1$ – Peter Sep 30 '22 at 10:31
  • In fact, in this case we even have for every real number $r$ : $\infty+r=\infty$ implying $r=0$ , so $0$ is the only real number. – Peter Sep 30 '22 at 10:41
  • @Peter that assumes that all 'numbers' behave like real numbers, and clearly this doesn't behave like a real number in any case so I see no reason to assume that. – Logan M Sep 30 '22 at 13:23
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    @user110391 As I understand it, your $\infty$ is not well-defined. If you write $\infty=\lvert \mathbb N\rvert$, then one must have by definition of $\lvert\cdot\rvert$ that $\infty=\lvert\mathbb N_0\rvert$ also, because $\lvert\cdot\rvert$ denotes cardinality and $\mathbb N$ has the same cardinality as $\mathbb N_0$. I think you would be much better off by completely discarding the symbol "$\infty$" in your undertakings, and instead directly working on sets. However, I am afraid that your ideas simply reduce to set subtraction in this case. – Maximilian Janisch Sep 30 '22 at 15:26
  • @user110391 Actually, you might be interested in learning about ordinal numbers. There are many infinite ordinal numbers and there is a difference between $\omega$ and $\omega+1$, even though they both correspond to the cardinality $\lvert\mathbb N\rvert$. It is not about measuring the size of sets though but rather about distinguishing different well-orderings, quoting from Wikipedia: – Maximilian Janisch Sep 30 '22 at 15:41
  • “ Ordinal numbers are distinct from cardinal numbers, which measure the size of sets. Although the distinction between ordinals and cardinals is not always apparent on finite sets (one can go from one to the other just by counting labels), they are very different in the infinite case, where different infinite ordinals can correspond to sets having the same cardinal. Like other kinds of numbers, ordinals can be added, multiplied, and exponentiated, although none of these operations are commutative.” – Maximilian Janisch Sep 30 '22 at 15:42
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    @MaximilianJanisch Yeah I discovered ordinals earlier today after posting this question, and realized they were very relevant. Your comment confirms that, thanks :) – user110391 Sep 30 '22 at 17:34