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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.

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I believe the symbol used there, instead of $\phi$ is $\varnothing$, i.e., the empty set. –  amWhy May 17 '13 at 3:36
    
Changed it anon, and thanks amWhy, still a little confused though. –  Amateur Math Guy May 17 '13 at 3:37
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I took the liberty of replacing the \bigcup with a \cup, which is the more common symbol for binary unions. –  Trevor Wilson May 17 '13 at 6:12

2 Answers 2

up vote 3 down vote accepted

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 $\omega$.

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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 $n=\{k:k<n\}=\{k<n-1\}\cup\{n-1\}=n-1\cup\{n-1\}$.

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