📂Set Theory

Axiom of Infinity

Axioms

$$\exists U \left( \emptyset \in U \land \forall X ( X \in U \implies S(X) \in U) \right)$$ There exists a set $U$ that contains the empty set and $X$ as elements, and also contains $S(X)$ as an element.

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• For a set $X$, $S(X)$ is defined as a set that is equivalent to $S(X):= X \cup \left\{ X \right\}$.

Explanation

Rather than tediously explaining why this is called the infinity axiom, it’s better to look at the proof of the existence of the set of natural numbers $\mathbb{N}$.

Theorem: Existence of the Set of Natural Numbers

$\mathbb{N}$ exists.

Proof

Strategy: By using the construction method proposed by von Neumann, which corresponds natural numbers themselves to sets, the set of natural numbers $\mathbb{N}$ is directly constructed. Thereby, $\mathbb{N}$ exists and immediately possesses the properties of natural numbers.

Let’s define the empty set $\emptyset$ and its $S(n)$ as follows. $$0 : = \emptyset \\ ( n + 1 ):= S(n) = n \cup \left\{ n \right\}$$ Then, $$1 = 0+1 = S ( 0 ) = \left\{ 0 \right\} \\ 2 = 1+1 = S ( 1 ) = \left\{ 0, \left\{ 0 \right\} \right\} = \left\{ 0, 1 \right\} \\ 3 = 2+1 = S ( 2 ) = \left\{ 0, \left\{ 0 \right\}, \left\{ 0, \left\{ 0 \right\} \right\} \right\} = \left\{ 0, 1, 2 \right\} \\ \vdots$$ By the infinity axiom, $\mathbb{N} = \left\{ 1, 2, 3, \cdots \right\}$ exists and satisfies the following property. $$n_{1} \in n_{2} \iff n_{1} < n_{2} \\ n_{1} \subset n_{2} \iff n_{1} \le n_{2}$$

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The claim that there are infinitely many natural numbers is somehow true, but in fact, no one in this universe has ever seen infinitely many natural numbers. No matter how long, consistently, or many natural numbers one seeks, it’s impossible to prove inductively that an infinite set exists. The infinity axiom was introduced to explain this infinity, and intuitively, there’s absolutely no reason to reject it.