Is it an axiom of set theory that $|\mathbb N| = \omega$? $\omega$ is the first infinite cardinal. $|\mathbb N| = \omega$. is this an axiom of set theory? or it has a proof based on preceding axioms?
thanks.
 A: It's not an axiom, because no axiom of $\sf ZFC$ is talking about $\Bbb N$.
But regardless of set theory, we know that $\Bbb N$ is the unique second-order model of Peano axioms; or the unique (non-empty) well-ordered set which has no maximal element, and every proper initial segment is finite.
This is why we can prove, from the axioms of $\sf ZF$, that $\omega$ satisfy the second-order version of Peano axioms: all the arithmetic (cardinal or ordinal) and the second-order induction axiom, stating that every inductive subset is equal to the entire set of natural numbers.
Now, we can prove that every two models of second-order Peano are isomorphic. Therefore we can prove that if $A$ is a model of second-order Peano, then $A$ has a bijection with $\omega$. So if we know that every set "worthy of being called $\Bbb N$" is equipotent with $\omega$, it means that $\sf ZF$ proves that the natural numbers have the same cardinality as $\omega$.
A: Well, it's common to define $\mathbb N := \omega$ and since $|\omega | = \omega$, i.e. there is a bijection $f \colon \omega \to \omega$ (the identity), the result follows.
