Natural Numbers Set Equivalent to Ideals of Integers

Theorem
Let the mapping $$\psi: \N \to$$ the set of all ideals of $$\Z$$ be defined as $$\forall b \in \N: \psi \left({b}\right) = \left({b}\right)$$

where $$\left({b}\right)$$ is the principal ideal of $$\Z$$ generated by $$b$$.

Then $$\psi$$ is a bijection.

Proof

 * First we show that $$\psi$$ is injective.

Suppose $$0 < b < c$$.

Then $$b \in \left({b}\right)$$, but $$b \notin \left({c}\right)$$, because from Principal Ideals of Integers‎, $$c$$ is the smallest positive integer in $$\left({c}\right)$$.

Thus $$\left({b}\right) \ne \left({c}\right)$$.

It is also apparent that $$b > 0 \Longrightarrow \left({b}\right) \ne \left({0}\right)$$ as $$\left({0}\right) = \left\{{0}\right\}$$.

Thus $$\psi$$ is injective.


 * Surjectivity follows from Principal Ideals of Integers: every integer is the smallest strictly positive element of a principal ideal of $$\Z$$.