By Dedekind's definition:
Definition: We say that some set $S$ is infinite if there exists an injection $f:S\rightarrow S$ such that $\operatorname{Im}(f)\neq S$ (or $f(S)\neq S)$.
Now, the book says the following: Infinite set exists if an only if a set $N$ with following properties exists:
There is an element $1\in N$.
There is a transformation $\nu:N\rightarrow N$ (the successor function) for which the following holds: \begin{align*} &\nu \text{ is injective}\tag{i}\\ &1\notin\nu(N)\tag{ii}\\ &\text{If a subset } X\subseteq N \text{ contains $1$ and } \nu(X)\subseteq X \text{ then } X=N.\tag{iii} \end{align*}
Now, my question is, why there would be no infinite set without the existence of theset $N$? Or is this something simmilar to the axiom of infinity (e.g. the inductive set) https://en.wikipedia.org/wiki/Axiom_of_infinity? Is this definition of an infinite set equivallent to the axiom? I'm quite unsure how I should interpret this.
The point is that each infinite set must contain a set isomorphic to the natural numbers (which basically is $N$ since those are the Peano axioms).
This can be seen by the given definition as follows: Let $S$ be an infinite set, and $f:S\to S$ the corresponding injective map for which $f(S) \neq S$. Then there exists at least one element which we call, for simplicity, $1 \in S$, for which $1 \notin f(S)$. Now define the set $$ N:= \{1, f(1), f(f(1)), f(f(f(1))), ... \}.$$ Can you see why this set $N$ satisfies the three properties, with $f$ being the successor function?