Set Notation (Axiom of Infinity)

Question

I'm having trouble understanding the notation used in describing the axiom of infinity (which is number 6 in the Wolfram MathWorld page). I understand what the axiom is saying, but I'm trying to understand how the following sentence:

$$\exists S \space [ \varnothing \in S \wedge (\forall x \in S) [x \cup \{x\} \in S]]$$

describes the existence of an infinity set. If someone could just break this apart for me and describe its pieces and how it all works that would be greatly appreciated.

Answer 1

Let's begin by understanding the letters and their meanings, then we'll give context to everything.

• $\exists$ is a quantifier. It means that the following symbol is a variable (a set, in the case of set theory) and we assert there is an object which the properties which we require that symbol to have.
• $S$ is that symbol. It is a placeholder that will be used to refer to some object in the universe, and describe some properties of that object.
• $\varnothing$ is the empty set.
• $\in$ is the membership relation, so when we say $\varnothing\in S$ we say that the empty set is a member of $S$.
• $\land$ is the symbol for conjunction, it means that we want both the part in the left of the symbol to be true, and the part on its right.
• $(\forall x\in S)$ is a bounded quantification. $\forall$ is the quantifier for "for all", so it says that we want that all the members of $S$ will have a certain property.
• $x\cup\{x\}$ is the union of the set $x$ with the set $\{x\}$. Remember that in set theory all the variables refer to sets.

So all in all what do we have? The axiom of infinity says the following thing:

There exists a set $S$, such that the empty set is a member of $S$, and whenever $x$ is a member of $S$, so is $x\cup\{x\}$.

Then what do we have? $\varnothing\in S$, and therefore $\varnothing\cup\{\varnothing\}=\{\varnothing\}$ is a member of $S$. Therefore $\{\varnothing\}\cup\{\{\varnothing\}\}$ is a member of $S$. Therefore $\{\varnothing,\{\varnothing\}\}\cup\{\{\varnothing,\{\varnothing\}\}\}$ is a member of $S$. And so on ad infinitum.

If we think about $\varnothing=0$, and $x\cup\{x\}$ as $x+1$ we have that $0\in S$, $1\in S$, $2\in S$, and so on. So $S$ corresponds to a set which contains the natural numbers, and so it is infinite.

Of course $S$ may include other objects, but we can conclude with the other axioms that there is some $S$ which includes only the natural numbers in the way we represent them with sets. This set is commonly known as $\omega$.

Answer 2

Using the second part of the definition recursively, we see $S$ must contain

• $\varnothing=0$
• $0\cup\{0\}=1$
• $1\cup\{1\}=2$
• $2\cup\{2\}=3$
• $\cdots$

which are all distinct. Note that when we look at the set-theoretic construction of the naturals, we have $0=\varnothing$ and $n=\{k:k<n\}=\{k<n-1\}\cup\{n-1\}=n-1\cup\{n-1\}$.