Today, in my class of set theory, my professor said that replacement axiom doesn't true in the structure $(\omega, \in)$. However, I think that he was wrong and here is my proof.
First the axiom form that I use is the following: Let $\phi(x,y,A,w_1,\dots,w_n)$ be a formula in the language of set theory where $x,y,A,w_1,\dots,w_n$ are parameters. So, replacement axiom says:
$$
\forall A,w_1,\dots,w_n(\forall x\in A\exists! y\phi\rightarrow\exists Y\forall x\in A\exists y\in Y\phi).
$$
Now, let $\phi$ a formula as above and take $A,w_1,\dots,w_n\in \omega$. Also, suppose that $\forall x\in A\exists!\phi$. As $A$ is a natural number, $A=\{0,1,\dots,A-1\}$. Now, define $y_k\in\omega$ as the unique element such that $\phi(k,y_k,w_1,\dots,w_n)$ where $k\in A$.
If we take $Y=\max\{y_1,\dots,y_{A-1}\}+1$, then $\forall x\in A\exists y\in Y\phi$. So, the axiom holds.
Is there some mistake in my argument?
Best Answer
The axiom that you called "replacement" in the question is true in $(\omega\in)$, but its usual name is "collection", not "replacement". The axiom usually called "replacement" is false in $(\omega,\in)$, as shown in Robert Shore's answer.