Consider a theory of sets in which there is no infinite set. One may argue that a robust theory of this kind would mimic the real world more accurately and would be more applicable. Why are such theories not prevalent?
Asked
Active
Viewed 510 times
2 Answers
7
I don't find the examples in the comments persuasive. Maybe we can do without the real numbers; I never liked them anyway. But here is a different sort of example that I think might be more compelling.
Consider a finite computer program whose input is a finite string of symbols and whose output, when it has one, is "yes" or "no" depending on whether the string has some property. Just which property isn't important right now. But I am not thinking of anything tricky. It could be something like "the string has even length" or "the string is a number that is a multiple of 23" or "the string describes a graph that has a Hamiltonian cycle", where we would normally imagine that the property was perfectly well-defined, and clear at least in principle.
The set of strings for which the algorithm says "yes" is infinite, and I cannot think of any way in which a finite set could be said to describe the situation more accurately or realistically. What finite set would you use? You would be in the position of saying that there was some finite constant $\mathcal L$ such that the algorithm would fail to say "yes" for every string longer than $\mathcal L$. But what $\mathcal L$ would you pick?
I suppose you could argue that any real implementation of the algorithm on a real computer would take more than the lifetime of the universe to finish if given a string bigger than, say, $10^{10^{10}}$ symbols, which is certainly true, but what does that add to the understanding of the algorithm? All it does is drag in a lot of contingent facts about current engineering practice and currently-accepted theories of physics. But the whole point of studying algorithms is to abstract away exactly such contingent facts, and to model computation in a simpler way. We are only trying to model a certain computation, not the entire universe, so why do we need to include the speed of light and the cosmological constant in our model?
That said, there is a legitimate school of thought in mathematics that holds that idea of arbitrarily large integers is incoherent. You should read about ultrafinitism. You may find it interesting.
J. W. Tanner
- 63,683
- 4
- 43
- 88
MJD
- 67,568
- 43
- 308
- 617
-
2
-
Another possibility would be to say that: it is impossible to construct the complete set of strings for which the algorithm says "yes". That would be the most accurate and realistic way to describe this situation in the real world. Just like Euclid didn't say that the set of prime numbers is infinite, he said "No matter how many primes you have already found, you can always find a new prime number." – Kasper May 31 '18 at 07:45
5
If you take the theory ZF and replace the Axiom of Infinity with its negation, you get a theory of sets in which there must not be any infinite sets.
It turns out that this theory is equivalent to first-order Peano Arithmetic! The paper here has a good explanation.
Obviously PA is very well studied!
Oscar Cunningham
- 16,939
-
It seems to me that negating the axiom of infinity doesn't get rid of all infinite sets, just the inductive ones. And that seems consistent with the idea that the resulting system is PA, since the nonstandard models of PA do have infinite objects. – MJD Feb 21 '18 at 12:49
-
@MJD I do not understand what you mean. Negating the axiom of infinity gets rid of all infinite sets. If you then study models of PA in a setting where infinite sets exist and you have tools such as compactness, then obviously you can get nonstandard models with externally infinite objects. But that is because of the metatheory you are using, not because of the theory you are studying. – Andrés E. Caicedo Feb 21 '18 at 15:49
-
This is really not my area of expertise, but I am sure that the axiom of infinity doesn't directly assert the existence of arbitrary infinite sets. It only asserts the existence of one specific infinite set, namely $\omega$, with the property that $\varnothing\in\omega$ and $x\in\omega\implies x\cup{x}\in\omega$. Its negation rules out sets with this particular property, but not infinite sets per se. It may be that one can use replacement or something to derive from ¬Inf a proof that there are no infinite sets, but in my great ignorance it is not obvious to me how this would work. – MJD Feb 21 '18 at 16:14
-
@MJD it is a straightforward fact that if $X$ is infinite then its double power set is Dedekind infinite, which implies (by replacement and separation) that $\omega$ exists. – Andrés E. Caicedo Feb 21 '18 at 16:20
-
-
1@MJD I actually wrote that as an answer a while back, see here. – Andrés E. Caicedo Feb 21 '18 at 16:36
-
1@MJD: Another way to see this is that an infinite set cannot have a finite rank, so an infinite ordinal must exist, so $\omega$ exists. – Asaf Karagila Feb 22 '18 at 06:32
Also just to point out, large natural numbers are not very natural despite their name. They don't really exist in nature any more than the real numbers do.
– Feb 21 '18 at 02:39