Topics Mathematics

Question Do infinite sets exist? Most mathematicians say yes, but to me it seems like infinite sets can only exist if we use inductive reasoning but not deductive reasoning. For example, in the set {1,2,3,4,...} we can't prove that the ... really means what we want it to. No one has shown that the universe doesn't implode before certain large enough "numbers" are ever glimpsed, so how can we say they exist as part of an "object" like a set. We can only do this by assuming the existence of the rest of the set since that seems logical base on our experience. But that seems like a rather weak argument.

Accepted:
March 7, 2013

Comments

We can use mathematical induction to prove that (i) infinitely many natural numbers exist from the premise that (ii) 1 is a natural number and the premise that (iii) every natural number has a successor. Although it's called mathematical "induction," it's actually deductive reasoning. I take it that (ii) is beyond dispute, and (iii) is at any rate very hard to deny! It won't do to demand proof of (ii) or (iii) before accepting this proof of (i), for if the premises in any proof must themselves have been proven, then we have an infinite regress: nothing could be proven in a finite amount of time.

We've therefore proven that infinitely many natural numbers exist. The notation "{1,2,3,4,...}" is just one way of referring to the set containing all and only those infinitely many numbers. It's perhaps a fallible way of referring to that set, because it assumes that the audience knows which number comes next in the series. A more reliable way of referring to the set is "the set whose members are the natural numbers." In any case, the existence of the members of {1,2,3,4,...} doesn't require that they be "glimpsed" before time runs out (or, indeed, glimpsed at all).

The argument here actually requires two more premises: (iv) that different numbers have different successors and (v) that 1 is not the successor of anything. If (v) failed, 1 could be its own successor and the only number. If (iv) failed, then 2 could be 1's successor and also its own.

It's perhaps also worth noting that, although (ii)-(v) do imply that there are infinitely many numbers, it does not follow from them that there are sets that have infinitely many members. This is because (ii)-(v) say nothing about sets, and we cannot simply assume that there is a set containing all the infinitely many numbers. Analogues of (ii)-(v) hold in so-called hereditarily finite set theory, in which there are no infinite sets. (Indeed, one can consistently add the axiom "there are no infinite sets" to this theory.)

Finally, the general observation that not everything can be proven does not imply that one can't reasonably ask that (ii)-(v) be proven, nor that one might not worry that, say, (iii) is to close to "there are infinitely many numbers". Efforts have been made to prove (ii)-(v) from more basic assumptions about numbers. See e.g. the SEP entry on Frege's Theorem.