TITLE: The axiom of infinity for Zermelo–Fraenkel set theory
QUESTION [6 upvotes]: 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.

REPLY [0 votes]: 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).
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