Infinite Set of Natural Numbers is Countably Infinite

Theorem
Let $\N$ be the set of natural numbers.

Let $S$ be an infinite subset of $\N$.

Then $S$ is countably infinite.

That is, there is a bijection $f: S \to \N$.

Proof
We will demonstrate that there is a bijection $f': \N \to S$.

By Inverse of Bijection is Bijection, the inverse of $f'$ will be a bijection from $S$ to $\N$.

Let $P$ be the set of infinite subsets of $S$.

Define a mapping $g:P \to P$ by letting $g(T) = T \setminus \{{ \min T }\}$.

Note that $T$ actually has a smallest element because $T$ is infinite and thus non-empty, so the Well-Ordering Principle applies.

Note that $g(T) \in P$ because an infinite set with one element taken away is infinite.

Recursively define the mapping $h: \N \to P$ thus:


 * $h(0) = S$
 * $h(n^+) = g(h(n))$

Let $f': \N \to S$ be defined by letting $f'(n) = \min h(n)$.

$f'$ is strictly increasing:

For any $n \in \N$, $f'(n) = \min h(n)$ and $f'(n^+) = \min h(n^+) = \min \left({h(n) \setminus \left\{{ \min h(n) }\right\} }\right)$.

But $\min h(n)$ precedes each element of $h(n)$, so strictly precedes each element of $h(n^+)$.

Thus $f'(n) = \min h(n) < \min h(n^+) = f'(n^+)$.

Therefore $f'$ is strictly increasing.

Thus $f'$ is an injection by Strictly Increasing Mapping on Toset is Order Embedding.

$f'$ is a surjection:

Suppose for the sake of contradiction that $f'$ is not surjective.

By the Well-Ordering Principle, there is a smallest $x \in S$ that is not in the image of $f$.

By the construction of $f'$, $x ≠ \min S$, so $\min S < x$.

Thus there is a largest element, $y$ of $S$ that strictly precedes $x$.

But then $y = f'(n)$ for some $n$, so $f'(n^+)= x$.

Note: it is possible to use the Cantor-Bernstein-Schröder Theorem instead of proving $f'$ is surjective.