In a more detailed way,
Definition: A set $X$ is said to be Dedekind-infinite if it has a proper subset $A$ equipotent (similar/in bijection) to $X$.
Let $X$ be a Dedekind-infinite set. Does it exists $A, B$, subsets of $X$ such that, $A\sim X$, $B\sim \mathbb{N}$ and $A\cap B=\emptyset$?
If not, what conditions are needed to ensure this?
In this case $U\sim V$ means "There is a bijective function $f:U\rightarrow V$".
The question comes to me and I think it's true; I can't find a counterexample.
Also, I think, the proposition is intuitive. The set of natural numbers is the "least" inductive set, and then, the first infinite set, and it has that property.
Thanks in advance
The answer is yes. The key observation is that any countably infinite set $A$ can be partitioned into two countably infinite subsets $A=A_1\sqcup A_2$ - this is just Hilbert's Hotel - and of course we then have $A\sim A_1\sim A_2$.
So suppose $X$ is Dedekind-infinite. I claim that $X$ has a countably infinite subset $A$. Before proving this, let's see why this is helpful: if $A\subset X$ is countably infinite, write $A=A_1\sqcup A_2$ as above. Then $A_2\cup (X\setminus A)\sim X$ (why? because $A_2\sim A$!), $A_1\sim\mathbb{N}$, and $[A_2\cup (X\setminus A)]\cap A_1=\emptyset$.
So now we just have to show that a Dedekind-infinite set has a countably infinite subset. Let $f: X\rightarrow X$ be injective but not surjective, and let $x\in X\setminus ran(f)$. Now think about the sequence $x, f(x), f(f(x)), f(f(f(x)))$, ... It's a good exercise to show that this sequence never repeats - that is, for $m\not=n$ we have $f^m(x)\not=f^n(x)$. So this defines a countably infinite subset of $X$, namely $\{f^n(x): n\in\mathbb{N}\}$.