Infinity of every ZF model

Question

Let's define $S(x) = x \cup \{x\}$.

1. Prove that axioms of ZF (semantically) imply that all sets $\emptyset, S(\emptyset), S(S(\emptyset)), \dots$ are pairwise distinct.

2. Prove (without axiom of infinity) that every model of ZF theory is infinite.

I don't really understand what does it mean that axioms semantically imply. I can write a formula $\Phi(x)$ that will be true for $S^{(n)}(\emptyset)$ and will be false for $S^{(m)}(\emptyset)$. Is it sufficient?

For the second part it seems that it is an implication of 1. since all models of ZF have to contain all $S^{(n)}(\emptyset)$ so our model has at least infinite subset.

Answer 1

Obviousely in each model there are such elements as: $\emptyset = S^{(0)}(\emptyset), S(\emptyset) = S^{(1)}(\emptyset), \dots$. If we prove that they are pairwise distinct it will imply that every ZF model have to be infinite.

Let's write down some $S^{(i)}$. One can gain:

$$ \emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}, \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}, \dots $$

Everyone can see that if $n < m$ then $S^{(n)}(\emptyset) \in S^{(m)}(\emptyset)$. Let's prove it by induction:

1. Base. $S^{(n)}(\emptyset) \in S^{(n+1)}(\emptyset)$ by definition.
2. Step. Assume $S^{(n)}(\emptyset) \in S^{(m)}(\emptyset)$, but $S^{(m+1)}(\emptyset) = \{S^{(m)}(\emptyset), \{S^{(m)}(\emptyset)\}\}$, hence each element of $S^{(m)}(\emptyset)$ we can also find in $S^{(m+1)}(\emptyset)$, therefore in $S^{(n)}(\emptyset)$ too.

Assume the contrary. Let $\exists n, m, n < m: S^{(n)}(\emptyset) = S^{(m)}(\emptyset)$. But we proved that $S^{(n)}(\emptyset) \in S^{(m)}(\emptyset) = S^{(n)}(\emptyset)$. This contradicts axiom of regularity. The end.