The axiom of infinity for Zermelo–Fraenkel set theory

Question

The axiom of infinity for Zermelo–Fraenkel set theory is stated as follows in the wikipedia page:

Let $S(w)$ abbreviate $ w \cup \{w\} $, where $ w $ is some set (We can see that $\{w\}$ is a valid set by applying the Axiom of Pairing with $ x=y=w $ so that the set $z$ is $\{w\} $). Then there exists a set $X$ such that the empty set $\varnothing$ is a member of $X$ and, whenever a set $y$ is a member of $X$, then $S(y)$ is also a member of $X$.

I don't understand the intuition behind the set $ w \cup \{w\} $. Why this causes the set $X$ to be infinite ? I think $ \{w\} $ is enough.

Answer 1

The intuition comes from the definition of the von Neumann ordinals. We define an ordinal as the set of its predecessors, which means that the successor ordinal of $\alpha$ is exactly $\alpha\cup\{\alpha\}$.

The finite ordinals corresponds to the natural numbers, so we can easily define $0=\varnothing$ and $n+1=n\cup\{n\}$. The axiom of infinity is equivalent to saying that there is a set whose members are exactly all the finite ordinals.

Answer 2

Morally speaking[1] the Axiom of Foundation we can aid us in showing that for each positive natural number $m$ (existing in the meta-theory) and every set $x$ the set $$S^{(m)} ( x ) = \overbrace{S ( \cdots ( S}^{m\text{ times}} ( x ) ) \cdots )$$ is distinct from $x$. We first show by induction (I guess in the meta-theory) that $x \in S^{(m)}(x)$ for all $m \geq 1$, and then note that by Foundation $x \notin x$, and so each $S^{(m)}(x)$ must differ from $x$.

It then follows that the sets $\varnothing$, $S(\varnothing)$, $S ( S ( \varnothing ) )$, $\ldots$ are pairwise distinct. Since your inductive set $X$ must contain all of these, that set must be infinite (at least from the point-of-view of the meta-theory).

Added: An alternative approach, done through the exercises in Ch.1 of Jech's Set Theory, is to define a set $N$ to be the smallest inductive set (i.e., the intersection of all inductive subsets of some given inductive set), and then show that the elements of $N$ have certain properties that then imply the infiniteness (non-finiteness?) of $N$. It is somewhat tedious (since it avoids overtly mentioning ordinals, and in particular $\omega$), but is also clean and without circularity. Once ordinals are defined, it turns out that $N$ is actually $\omega$ (what a surprise!).

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[1] By which I mean that there may be inaccuracies in what I am about to say, but the gist is pretty close to the actual truth. I thank Prof. Sy David Friedman for putting this phrasing into my mathematical vocabulary.

Answer 3

With Extensionality, Power Set, and Separation, a signleton exists. All {∅}, {{∅}}, {{{∅}}}, ..., are of cardinality 1. You cannot generate an infinite set/class this way.

Class A is ordinal (written as Ord A) iff A is transitive and is well-ordered by the epsilon relation. Define the class of all ordinal numbers On = { x | Ord x }.

With neither Infinity nor Regularity, \eqref{ordom} and \eqref{ordeleqon} are provable. $$ \operatorname{Ord} \omega \tag{1} \label{ordom} $$ $$ \operatorname{Ord} A \leftrightarrow \left( A \in \mathrm{On} \lor A = \mathrm{On} \right) \tag{2} \label{ordeleqon} $$

Consequently, ω ∈ On ∨ ω = On. If ω is not a member of On, ω = On, which does not exist. Converse this proposition. If ω exists, it is an ordinal number. $$ \omega \in \mathrm V \to \omega \in \mathrm{On} $$

The Axiom of Infinity asserts the existence of ω. $$ \omega \in \mathrm V $$

A finitist denies existence of ω. This leads to ω = On. A finitist can develop natural number, integer, and rational number arithmetic but be denied the real numbers (as well as much of the rest of mathematics).