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I have a question about infinity axiom in ZF and maybe, it has nonsense. So I apologize in advance if it is the case.

In ZF, the infiny axiom can be state as $\exists X(X\neq\emptyset\wedge\forall x\in X(S(x)\in X))$ and it says that exists an infinite set (moreover, it is equivalent to say that $\omega$ exists) more exactly, it says that exists an inductive set. In this last point is where arise my question, is it possible to build an infinite set without infinite axiom?

Note that the question depends of that we means by "infinite". A usual way to say that a set $A$ is infinite is through of $\preceq$-relation: $A$ is a infinite set iff $\omega\preceq A$. And with this definition it is possible to show that ZF-Inf+$\forall x(x\ \text{is finite})$ is consistent.

Now, there exists another "definition" of infinite and it is "Dedekind-infinite": A set $A$ is Dedekind-infinite iff exists $X\subsetneq A$ such that $X\approx A.$

So, can we state the existence of a infinite set (with another "reasonable" definition of infinite) that it is not be infinite in the "usual way"?

For example, can we get that ZF+$\neg\mathbf{Inf}$+$\exists x(x\ \text{is Dedekind-infinite})$ is consistent?

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marked as duplicate by Asaf Karagila set-theory Nov 28 '18 at 13:41

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