# Thread: Help with understanding the infinity axiom in ZF

1. ## Help with understanding the infinity axiom in ZF

I need some help in understanding why:

$\exists{x}[\{\}\in{x}\wedge\forall{y}\in{x}((y\cup\{y\})\in{x })]$

means that there exits an infinite set. I am confused about the last part of this statment $\forall{y}\in{x}((y\cup\{y\})\in{x})]$

Thanks for any help

2. ## Re: Help with understanding the infinity axiom in ZF

In ZF, the axiom of infinity is equivalent to Ex x is infinite.

What about Ay(y in x -> yu{y} in x) don't you understand?

It just says: For all y, if y is in x, then yu{y} is in x.

In other words, x is closed under the successor operation (where the successor operation takes any y and gives yu{y}).

Now, if you want to prove that the axiom of infinity is equivalent in ZF to Ex x is infinite, then show what you have done so far, and we'll try to help you from there...

3. ## Re: Help with understanding the infinity axiom in ZF

In ZF, the axiom of infinity is equivalent to Ex x is infinite.
What definition of "infinite" do you use here?

4. ## Re: Help with understanding the infinity axiom in ZF

First, previously I should have said:

In ZF\{axiom of infinity} we have that the axiom of infinity is equivalent with Ex x is infinite,

since it is trivial that in ZF we have that the axiom of infinity is equivalent with Ex x is infinite (since both are theorems of ZF anyway).

And any ordinary set theoretical definition of 'infinite' will do.

There are a number of different approaches. Here's one (with Tarski infinite):

n is a natural number <-> (n is well ordered by epsilon on n & n is well ordered by the inverse of epsilon on n)

x is finite <-> ~En(n is a natural number & x is equinumerous with n)

x is infinite <-> ~ x is finite

But also, in ZF\{the axiom of infinity} we have that

the axiom of infinty is equivalent with Ex x is Dedekind infinite, where 'Dedekind infinite' is defined as:

x is Dedekind infinite <-> Ey(y is a proper subset of x & x is equinumerous with y)

5. ## Re: Help with understanding the infinity axiom in ZF

as i understand it, although you need not actually take as axiomatic the existence of the null set, it is often done.

once you have the null set, you can form the set of all subsets of the null set (the power set). since the null set HAS no subsets, the power set just

contains one element, the null set itself. now we have two sets. given any two sets, we can form their union,

which in this case, is typically called the natural number 1 (and the power set of the null set is called 0).

continuing, we can define 2 as 1 U {1}. then the axiom of infinity says there exists a set: {1,2,3,4,......}.

this is equivalent to saying that the natural numbers form a set (that is, that the natural numbers are a well-defined set

formed by well-defined set operations from the one set we are sure exists: the null set).

what your version of the axiom of infinity actually says, in ordinary english, is: "there exists an inductive set".

the fact that $\{\ \} \in x$ establishes the "base case", and the last part establishes the "inductive step"

if $y \in x$ then $y \cup \{y\} \in x$ (in a usual definition of the natural numbers, this is saying:

if n is in X, then so is n+1).

it is quite possible to have a model of set theory without this axiom (in which every set is finite). which although logically satisfactory, leaves us in the position of not being able to use constructions we take for granted in mathematics, such as defining real numbers as convergent sequences of rational numbers.

6. ## Re: Help with understanding the infinity axiom in ZF

Originally Posted by Deveno
as i understand it, although you need not actually take as axiomatic the existence of the null set, it is often done.
once you have the null set, you can form the set of all subsets of the null set (the power set). since the null set HAS no subsets, the power set just
contains one element, the null set itself. now we have two sets. given any two sets, we can form their union,
which in this case, is typically called the natural number 1 (and the power set of the null set is called 0).
continuing, we can define 2 as 1 U {1}. then the axiom of infinity says there exists a set: {1,2,3,4,......}.
this is equivalent to saying that the natural numbers form a set (that is, that the natural numbers are a well-defined set

formed by well-defined set operations from the one set we are sure exists: the null set).

what your version of the axiom of infinity actually says, in ordinary english, is: "there exists an inductive set".

the fact that $\{\ \} \in x$ establishes the "base case", and the last part establishes the "inductive step"

if $y \in x$ then $y \cup \{y\} \in x$ (in a usual definition of the natural numbers, this is saying:

if n is in X, then so is n+1).

it is quite possible to have a model of set theory without this axiom (in which every set is finite). which although logically satisfactory, leaves us in the position of not being able to use constructions we take for granted in mathematics, such as defining real numbers as convergent sequences of rational numbers.
Do you know that not all important mathematicians except the empty set. Moore said that a point set cannot be empty. That would be an oxymoron. Points sets contains points.

7. ## Re: Help with understanding the infinity axiom in ZF

Originally Posted by Plato
Do you know that not all important mathematicians except the empty set. Moore said that a point set cannot be empty. That would be an oxymoron. Points sets contains points.
perhaps you meant "accept"? indeed, there is a sort of duality to "everythingness" and "nothingness" which means that we can say certain things in a natural language which lead us to some tough questions in formal logic. however, many mathematicians accept (or at least study) Zermelo-Fraenkel set theory, and in that theory, the empty set exists.

Wikipedia goes so far as to say: "The axiom of empty set is generally considered uncontroversial, and it or an equivalent appears in just about any alternative axiomatisation of set theory."

if we want set theory to be in some sense meaningful, then we would want some sets to actually exist (so that we are not describing a theory with no models). if there is some set U we can agree actually exists, in ZF we can PROVE the empty set exists, by defining:

$\emptyset = \{x \in U|(x \in x) \wedge \neg (x \in x)\}$.

i do not intend for any of my posts to constitute an endoresement of ZF set theory, i regard it as somewhat of an abomination: the fact that the axioms are so convoluted strikes me as evidence we haven't hit upon the right primitive notion (membership) for basing mathematics upon.

set theory isn't really my forte, i dislike it intensely. and i make no claim to be an important mathematican, on either count. i only seek to make the axiom of infinity somewhat more intelligible for the original poster, and make no claim whatsoever for the validity or non-validity of ZF in general.

8. ## Re: Help with understanding the infinity axiom in ZF

thanks very much I think that I have a better understanding of it now, just trying to get to grips with the axioms (this post went a bit beyond what I understand at the moment) but I'm sure I'll have more questions later!

thanks very much for the help

9. ## Re: Help with understanding the infinity axiom in ZF

Originally Posted by Plato
Do you know that not all important mathematicians except the empty set. Moore said that a point set cannot be empty. That would be an oxymoron. Points sets contains points.
And we might say that my wallet contains bills, but when I run out of cash we don't say my wallet is no longer a wallet. Of course, this point hinges on, as Deveno pointed out, our primitive notions of sets. My wallet is always a wallet, even when it holds no bills, because it is a container of bills. In like manner, one might view a point set as a container of points. Sometimes that container can be empty. Such a critter should pose no intuitive difficulty. An oxymoron arises only if we perceive sets as manifestations of their content (members). However, even on this view we could say the lack of members (points) manifests in an empty set. In that respect, the difference in perspectives have a commonality.

10. ## Re: Help with understanding the infinity axiom in ZF

Originally Posted by Deveno
the null set HAS no subsets
The empty set does have a subset. The empty set has exactly one subset, viz. itself.

Originally Posted by Deveno
he power set of the null set is called 0
In the usual treatment the empty set is 0 and the power set of the empty set is 1. By the way, we don't need to refer to power sets for this, since the power set of the empty set is just the singleton whose only member is the empty set.

Originally Posted by Deveno
the axiom of infinity says there exists a set: {1,2,3,4,......}
As usually formulated (and as you alluded to elsewhere in your post), the axiom of infinity is that there is a successor inuctive set. Then with an instance of the axiom schema of separation, we prove that there is a least successor inductive set, then with the axiom of extensionality we prove that there is a unique such least successor inductive set, then we may prove also that this unique least successor set has as its members all and only the finite ordinals, viz. 0, 1, 2 ...

11. ## Re: Help with understanding the infinity axiom in ZF

Originally Posted by Deveno
in ZF we can PROVE the empty set exists, by defining:

$\emptyset = \{x \in U|(x \in x) \wedge \neg (x \in x)\}$.
Just to be clear, that definition is enabled by an instance of the axiom schema of separation along with the axiom of extensionality.

Originally Posted by Deveno
the fact that the axioms are so convoluted
I don't see any such fact. Indeed, the axioms seem quite straightforward to me. What is your basis for claiming that the axioms are convoluted?

12. ## Re: Help with understanding the infinity axiom in ZF

Originally Posted by MoeBlee
The empty set does have a subset. The empty set has exactly one subset, viz. itself.
quite correct, my bad. no elements (kicks self hard).

In the usual treatment the empty set is 0 and the power set of the empty set is 1. By the way, we don't need to refer to power sets for this, since the power set of the empty set is just the singleton whose only member is the empty set.
again, i apologize. that's what i meant, was in a hurry. but my point was, if we start from just the axioms, what sets do we have to work with? to create a non-empty set when we start with just the empty set, how shall we get an non-empty set if we don't use the power set construction?

As usually formulated (and as you alluded to elsewhere in your post), the axiom of infinity is that there is a successor inuctive set. Then with an instance of the axiom schema of separation, we prove that there is a least successor inductive set, then with the axiom of extensionality we prove that there is a unique such least successor inductive set, then we may prove also that this unique least successor set has as its members all and only the finite ordinals, viz. 0, 1, 2 ...
i don't see how we can guarantee a unique least successor set, except up to a bijection. consider the following two sets:

{Ø, {Ø}, {{Ø}, {{Ø}}}, {{Ø}, {{Ø}}, { {{Ø},{{Ø}}} },.......} and

{Ø, {Ø}, {{Ø}}, {{{Ø}}},.........} which of these two different sets shall we say is "least"?

I don't see any such fact. Indeed, the axioms seem quite straightforward to me. What is your basis for claiming that the axioms are convoluted?
'tis an opinon. just the formal statement of the axiom schema of separation is a bit involved with its introduction of free and bound variables of an (unspecified) predicate (which to my mind, begs the question what predicates are "allowable"?). that is to say, the formal formulation of ZF relies on notions of a background first-order (formal) language. i certainly see the power of such an approach, but the development of so much machinery just to be able to describe to our satisfaction a statement like: "1 is a natural number" leads me to believe we have taken a wrong turn somewhere. you are entitled to disagree, and there are certainly many adherents of ZF(C), so you're in good company.

13. ## Re: Help with understanding the infinity axiom in ZF

Originally Posted by Deveno
i don't see how we can guarantee a unique least successor set, except up to a bijection.
I have to concur, one cannot say there is a unique least inductive set. It is a particular issue in the philosophy of mathematics how we can go about saying what should be the "proper" content of the natural numbers. Your two examples are the von Neumann and Zermelo constructions, respectively. They are both adequate representations of $\omega$, yet they are structurally different beasts.

14. ## Re: Help with understanding the infinity axiom in ZF

Originally Posted by Deveno
i don't see how we can guarantee a unique least successor set, except up to a bijection.
Successor inductive. In this context, "leastness" refers to the subset relation. There is a unique least successor inductive set in the sense that there is a unique set w such that w is successor inductive and for all x, if x is successor inductive then w is a subset of x.

Originally Posted by Deveno
{Ø, {Ø}, {{Ø}}, {{{Ø}}},.........}
That set is not successor inductive. Recall that X is successor inductive if and only if 0 is in X and for all y, if y is in X then yu{y} is in X.

Of course, we could define 'successor' or 'successor inductive' in a different way, indeed so that the Zermelo natural numbers you just mentioned suit such a definition. But in context, we refer to the more ordinary definitions I'm using.

Originally Posted by Deveno
the formal statement of the axiom schema of separation is a bit involved with its introduction of free and bound variables of an (unspecified) predicate (which to my mind, begs the question what predicates are "allowable"?
The axiom schema of separation:

For all P and x, if P is a formula and x is a variable not free in P, then all closures of the following are axioms:

ExAy(y in x <-> (y in z & P))

So you see that the only technical matter about variables is that x is not free in the defining formula P. That is hardly "convoluted". Indeed it's straightforward and common sense. It's common sense that you wouldn't define a set, calling it 'x' with a condition that itself mentions x. Indeed convolultion (and contradiction) would result by allowing x to be free in P.

As to what predicates are allowable. This is rigorous and straightforward. Any formula P in which x does not occur free.

The axiom schema of separation expresses a straightforward notion: Given a set z, we can form the set x that is a subset of z and x has as members all and only those objects that are in z and that have property P. That is even common everyday reasoning. Given the set of all umbrellas we can also form the set of all umbrellas that are red. That is given the set z of all umbrellas, we have the set x that is a subset of z and such that the members of x are all and only those objects in z that are red.

Originally Posted by Deveno
the formal formulation of ZF relies on notions of a background first-order (formal) language.
Yes, as does any formal first order theory. What formal first order theory that axiomatizes mathematics for the sciences would one point to as being less complicated than ZC set theories?

Of course, one doesn't have to concern oneself with formal theories for mathematics, but if one is interested in a formal axiomatization, then I don't see that ZFC is espeically convoluted or complicated.

Originally Posted by Deveno
i certainly see the power of such an approach, but the development of so much machinery just to be able to describe to our satisfaction a statement like: "1 is a natural number"
But the machinery is not JUST to be able to say "1 is a natural number".

Originally Posted by Deveno
there are certainly many adherents of ZF(C), so you're in good company.
Just to be clear, I'm not an "adherent" of ZFC. I find that it has certain virtues and also that one may have certain philosophical objections to it. I enjoy studying it, but I don't take it as some theory that I must "adhere" to.

15. ## Re: Help with understanding the infinity axiom in ZF

Originally Posted by bryangoodrich
one cannot say there is a unique least inductive set.
In this context, 'successor inductive' is used in the sense: X is successor inductive if and only if 0 is in X and for all y, if y is in X then yu{y} is in X.

And, in this context, 'least' is used in the sense of the subset relation.

So, from the axioms we derive the theorem:

There exists a unique w such that w is successor inductive and for all X, if X is successor inductive then w is a subset of X.

Originally Posted by bryangoodrich
the von Neumann and Zermelo constructions, respectively. They are both adequate representations
Of course. That is not at issue. Rather, given some particular defintion of 'successor inductive' and taking 'least' in this sense of subsets, there is a least successor inductive set. This is just using the most ordinary definitions in textbook set theory.

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