Basic Universe has Infinite Number of Elements

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Theorem

Let $V$ be a basic universe.

Then $V$ has an infinite number of elements.


Proof

Let $a_n$ be the class defined as:

$\forall n \in \N: a_n = \begin{cases} \O & : n = 0 \\ \set {a_{n - 1} } & : n > 0 \end {cases}$

It is shown by the Principle of Mathematical Induction that $a_n$ is a set for all $n \in \N$.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

$a_n$ is a set.


Basis for the Induction

$\map P 0$ is the case:

$a_0$ is a set.

But $a_0 = \O$ by definition.

By the Axiom of the Empty Set:

$\O$ is a set.
$\O \in V$

Thus $\map P 0$ is seen to hold.

This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$a_k$ is a set.


from which it is to be shown that:

$a_{k + 1}$ is a set.


Induction Step

This is the induction step:


By the induction hypothesis we have that $a_k$ is a set.

From Singleton Class of Set is Set, its singleton class $\set {a_k}$ is also a set.

But by definition:

$a_{k + 1} = \set {a_k}$

Hence $a_{k + 1}$ has been shown to be a set.


So $\map P k \implies \map P {k + 1}$ and so by the Principle of Mathematical Induction, $a_n$ is a set for all $n \in \N$.

$\Box$


It is now proved that $a_m \ne a_n$ when $m \ne n$.

Without loss of generality, let $m > n$.

Aiming for a contradiction, suppose $a_n = a_{m + n}$ for some $m > 0, n \ge 0$.

Let $n = 0$.

Then we would have:

$a_m = \O$

But as $m > 0$ we have that $a_m = \set {a_{m - 1} }$ for some $m - 1 \in \N$.

But $\O \ne \set x$ for any set $x$.

Now suppose $n > 0$.

Then by Singleton Classes are Equal iff Sets are Equal:

$a_{n - 1} = a_{n - 1 + m}$

We can repeat this step a total of $n$ times to reach:

$a_0 = a_{0 + m}$

from which again it follows that $a_m = \O$ for some $m > 0$.

This has been shown to be false for all $m$.


Hence it cannot be the case that $a_n = a_{m + n}$ for some $m > 0, n \ge 0$.

Thus there exists an injection from $\N$ to $\set {a_n: n \in \N}$.

That is, $\set {a_n: n \in \N}$ is countably infinite by definition.


By definition of basic universe, $V$ is a universal class.

By definition of universal class, every set is an element of $V$.

Hence:

$\forall n \in \N: a_n \in V$

and it follows that $V$ is infinite.

$\blacksquare$


Sources

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