Injective function, $f:X\to X$ with $f(X)\subset X$, but $T\subseteq X$ is not inductive set.

Question

I'm looking for an example of the following manner:

Suppose that $f:X\to X$ is a injective function(where $X$ some set), such that the following property not holds:

If $T$ is subset of $X$, with the following two features: 1. There is an element in $T$ that is not in the range of $f$ 2. $t\in T$ implies that $f(t)\in T$. Then $T=X$.

Edit:

I'm trying to show that system that include three of Peano's axioms is independent, in the following sense: that there exist set $X\neq\emptyset$ and function $f:X\to X$ such that the system with $X$ and $f$ satisfies two, and do nod holds for the third one. The set of axioms here is:

1. $f$ is a injective function;
2. $f(X)\subset X$;
3. The first principle of induction.

Answer 1

How about this?

• $X=\{\langle a, b\rangle: a\in\mathbb{N}, b\in\{1, 2\}\}$ (so $X$ is basically two disjoint copies of $\mathbb{N}$)

• $f(\langle a, 1\rangle)=\langle 2a, 1\rangle$,

• $f(\langle a, 2\rangle)=\langle 2a+1, 1\rangle$.

Then $f$ is an injection from $X$ to itself, and $f(X)\subsetneq X$.

However, let $T=\{\langle a, 1\rangle: a\in\mathbb{N}\}\cup\{\langle 17, 2\rangle\}.$ Then $t\in T\implies f(t)\in T$, and $\exists t\in T\forall x\in X(f(x)\not=t)$ (set $t=\langle 17, 2\rangle$); but $T\not=X$.