2

Could someone please state and explain the axiom of infinity in ZF set theory? This isn't homework, it's just something that has interested me for awhile.

Asaf Karagila
  • 393,674
Jen.
  • 61
  • did you read the wikipedia page (or part of it) to get started ? http://en.wikipedia.org/wiki/Axiom_of_infinity – Denis Jan 30 '14 at 22:52
  • Yes, I did. I still don't understand it. Why is the union of two empty sets different from just the empty set itself? – Jen. Jan 30 '14 at 22:53
  • 2
    Beware, ${\emptyset}$ has one element, so it is not the emptyset. – Denis Jan 30 '14 at 22:56
  • The purpose of the ZF Axiom of Infinity is to provide a basis for the set of natural numbers. Readers may be interested in my own proposed Axiom Infinity based on the discussion here. It might even give some insight into the ZF version. See my formal proof at http://www.dcproof.com/AxiomOfInfinity.htm – Dan Christensen Feb 02 '14 at 06:43

2 Answers2

3

The axiom of infinity says that there exists a set $A$ such that $\varnothing\in A$, that is the empty set is an element of $A$, and for every $x\in A$ the set $x\cup\{x\}$ is also an element of $A$.

The definable function $f(x)=x\cup\{x\}$ is an injection from $A$ into itself, and since $f(x)\neq\varnothing$ for every $x$, it follows that $f$ is not surjective. Therefore $A$ must be infinite.

Do note that $\{\varnothing\}$ is not the empty set, and so $\varnothing\cup\{\varnothing\}=\{\varnothing\}\neq\varnothing$.

Asaf Karagila
  • 393,674
  • it is a little misleading to write "definitely not" as if it was "even less" the empty set as before, although it is the same set as before. – Denis Jan 30 '14 at 22:59
  • Why do you suppose Zermelo and company used this instead of more intuitive Peano-like axioms for the natural numbers? – Dan Christensen Jan 31 '14 at 04:04
  • @Dan, I will write you a proper answer later on, when I'm at the keyboard again. The general idea is that PA deals with natural numbers and set theory deals with sets. – Asaf Karagila Jan 31 '14 at 04:08
  • @AsafKaragila Isn't Inf just an attempt to deal with natural numbers within set theory? In my own set theory, I leave out anything like it, and simply state Peano's Axioms at the beginning of proofs on number theory. It avoid embarrassing results like $2\subset 5.$ – Dan Christensen Jan 31 '14 at 13:58
  • @Dan: If you want to speak about the "set of natural numbers" and "the set of real numbers", which is really one of the basic uses for sets - turning a collection of mathematical objects into a mathematical object on its own - the axiom of infinity simply guarantees that there is an infinite set, and for that the rest is given. If you don't think that infinite sets should exist, then don't assume it. But then the natural numbers are not a set. There's nothing wrong with that, but it's something to keep in mind. You can no longer speak about the set of even numbers, or so on. [...] – Asaf Karagila Jan 31 '14 at 14:09
  • [...] Moreover, the axioms of $\sf ZFC$ where you replace the infinity axiom with its negation are bi-interpretable with $\sf PA$. And you still get something like $2\subseteq 5$ there. But why does that bother you? Does it bother you that when you write a program both arrays of strings and integers are both saved in the memory as binary data? Does it bother you that both the thing you love and the thing you hate are made of energy? If you decide that everything mathematical is made of sets, or can be represented by using sets, then it's a valid question whether or not $2\subseteq 5$. [...] – Asaf Karagila Jan 31 '14 at 14:12
  • [...] This question was discussed, at length, on MathOverflow. At least twice. Andreas Blass posted a very nice reason why assuming everything is a set is positive: if you want to understand the mathematical universe, and it is made of infinitely many different types, you need to understand them all. If you assume it's just made of sets, then you just need to understand sets. If all you do is number theory, then fine. Work in $\sf PA$. Nobody is forcing you to work in set theory. But if you want to work with sets, then you need to have a reasonable axiomatic framework for sets. [...] – Asaf Karagila Jan 31 '14 at 14:14
  • [...] And if you analyze the axioms of $\sf ZFC$ then they are reasonable. Including the axiom of infinity, which is reasonable because we want to work with infinite sets. If you disagree with some of the axioms, you can remove them, again, nobody is forcing you to do anything. The question is what do you want to do with your sets, and what axioms do you need for that. If you look at the properties of naive set theory, then only axiom [schema] in $\sf ZFC$ which is objectionable is the Replacement schema, and if you remove it (but keep the Separation schema) you end up with $\sf ETCS$. – Asaf Karagila Jan 31 '14 at 14:16
  • @AsafKaragila Thank you for your long and thoughtful response, Asaf. You say Inf is just to postulate the existence of at least one infinite set. But it looks remarkably like the set of natural numbers described by Peano's axioms, but with embarrassing "junk" like $2\subset 5$ thrown in. I still don't see any real benefit to this approach. – Dan Christensen Jan 31 '14 at 14:43
  • @Dan: From the other axioms of set theory we can prove that if there exists one infinite set, then there exist a Dedekind-infinite set, and therefore the set of the natural numbers exist. However it is mighty convenient to use inductive sets instead, if only for the sake of simplicity and avoiding the need to define finiteness (there shouldn't be any argument that inductive sets are infinite). The point of set theory is not to cushion number theory, but to deal with sets. Again, if your focus is just number theory, use a different foundation that don't use set theory instead. – Asaf Karagila Jan 31 '14 at 15:01
  • @AsafKaragila It's really a kind of leap of faith that infinite sets exist -- whether or not it is built into the axioms of set theory. But not a very big leap. If you admit the existence -- within set theory or not -- of a function $f$ on a set $X$ that is injective but not surjective, you can extract at least one infinite subset of $X$ that identical in structure to the natural numbers as defined by Peano's axioms, using $f$ as the successor function. See "What are numbers again?" at my math blog http://dcproof.wordpress.com/ – Dan Christensen Jan 31 '14 at 15:34
  • @Dan: It's an infinitely stronger leap of faith to assume that induction "works out". Because induction proves that there exists a mathematical object which cannot be represented by any physical quantity. If you have that, infinite sets make a vastly smaller leap. If you want "existence" of mathematical object to mean physical existence, then you probably have to focus your effort on numbers well below $10^{100}$. – Asaf Karagila Jan 31 '14 at 15:39
  • @AsafKaragila Actually the principle of induction can be derived from the process I describe above with $f$ and $X$, as can the other Peano axioms. – Dan Christensen Jan 31 '14 at 15:44
  • @Dan: I'm not arguing that adding the axiom of infinity is not adding more power to the system. It is. I said so a few comments earlier. I'm saying that if you are concerned with leaps of faith, then the believe that induction "works" is greater than the belief that infinite sets "exist". Because induction implies that mathematics is not something which can necessarily be described using our physical reality. Therefore the "existence" of infinite sets is not a physical existence (as the existence of $10^{10000}$ is not a physical existence). – Asaf Karagila Jan 31 '14 at 15:46
  • @AsafKaragila I think you missed my point. If you admit the existence of functions on a set that are injective but not surjective (not a great leap), you don't have to assume induction as an axiom, you can derive it. And you don't have to deal with nonsense like $2\subset 5.$ – Dan Christensen Jan 31 '14 at 15:55
  • @Dan: If you work in set theory and assume that everything in the universe is a set, then the statement $2\subseteq 5$ is a syntactically legal statement. Because the two of those are sets. The point of the von Neumann ordinals is that they are very nice and well-behaved with the syntax (their order is the only relation we really have in the language, $\in$). But no one is forcing you to use this representation of the natural numbers. If you don't want $2\subseteq 5$ to even be a legal expression then you need to allow non-sets into the world, but that doesn't add anything to the theory. [...] – Asaf Karagila Jan 31 '14 at 15:58
  • [...] Set theories with atoms, or urelements, are known and they have been used in the past. But they also add a hassle into everything, because now you have two sorts of objects and you need to be careful when mixing them. So people stopped using those theories for the better part. So everything, including the integers, is a set in a universe of set theory. If you can't accept that, try type theory instead. – Asaf Karagila Jan 31 '14 at 16:00
  • @AsafKaragila It would bother me if I could derive $2\subset 5$ in my system. What other nonsense might crop up? $0^0=1$??? – Dan Christensen Jan 31 '14 at 16:02
  • @Dan: I suppose that it also bothers you that the computer you're using at this very moment is using integers to represent characters. I suppose that the whole idea of a computer and a universal language in which we can encode pretty much anything is fundamentally flawed. Shame, we had such a successful run with it. And please, we had a nice and friendly discussion. Don't bring that increasingly annoying $0^0=1$ thing again. – Asaf Karagila Jan 31 '14 at 16:04
  • @AsafKaragila Our brains are computers. This may all be a dream and logic only one possibility among many. Enough said. – Dan Christensen Jan 31 '14 at 16:11
  • 1
    @Dan: You believe whatever you want to believe. If you have a problem with integers being sets, don't use set theory. Use type theory, use second-order logic as a foundation (you can avoid sets if you're very careful). Become an ultrafinitist. I don't really care. I think this discussion has reached its conclusive end now. – Asaf Karagila Jan 31 '14 at 16:13
  • @AsafKaragila My own proposed Axiom of Infinity based on my thoughts above: http://dcproof.com/AxiomOfInfinity.htm – Dan Christensen Feb 02 '14 at 06:00
0

In ZF-Infinity, $\omega=\{\alpha\mid\alpha\text{ and its elements are either }0\text{ or successor}\}=\mathbb{N}$ can be either a set or not a set (proper class). The Axiom of Infinity basically postulate that $\omega$ is indeed a set, which enable set theory to deal with all sorts of infinite sets. Without it, one can only prove finite sets exist.

Kaa1el
  • 2,058