Element of Natural Number is Natural Number

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Theorem

Let $n$ be a natural number.

Let $m \in n$.

Then $m$ is also a natural number.


Proof

The proof proceeds by induction.

For all $n \in \N$, let $\map P n$ be the proposition:

for all $m \in n$: $m$ is a natural number.


Basis for the Induction

$\map P 0$ is the case:

for all $m \in 0$: $m$ is a natural number.

This is true vacuously, as $\O$ has no elements by definition of empty class.

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 $\map P {k^+}$ is true.


So this is the induction hypothesis:

for all $m \in k$: $m$ is a natural number.


from which it is to be shown that:

for all $m \in k^+$: $m$ is a natural number.


Induction Step

This is the induction step:

Let $m \in k^+$.

Then either:

$m = k$

or:

$m \in k$

In the first case, $m$ is the natural number $k$.

In the second case, it follows by the induction hypothesis that $m$ is a natural number.


In both cases $m$ is a natural number.


So $\map P k \implies \map P {k^+}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \N: \forall m \in n$: $m$ is a natural number.

$\blacksquare$


Sources

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