Find an injective but not surjective function to a infinite set X A set if infinite iff there exists an injective function f that maps X to X for which f(x) is a proper subset of X. I only know that set X is infinite and I do not know anything about the range of the set. Could any one give me an example of an injective but not surjective mapping of X?
Thank you 
 A: If the only thing you know about $X$ is that it is infinite, then it is not possible to describe such a mapping precisely enough that you can know what it does to each element. You don't even know what the elements are, so how would the example be able to tell what happens to them?
Using the Axiom of Choice, one can prove (see below) that every infinite $X$ has a subset $Y$ that is in bijective correspondence with $\mathbb N$. In that case you can construct $f$ as "for every element $x$ that corresponds to a natural number $n$, $f(x)$ is the element that corresponds to $n+1$; if $x$ doesn't correspond to a natural, $f(x)=x$.

Proof of above claim. Let $X$ be infinite; we want to construct a injection $\varphi:\mathbb N\to X$. Then the image of $\varphi$ will be the requested $Y$.
Fix a choice function $g:\mathcal P(X)\setminus\{\varnothing\}\to X$, and define $\varphi$ recursively:
$$ \varphi(n) = g\bigl( X\setminus \{\varphi(i) \mid i<n\} \bigr) $$
The argument $X\setminus\{\varphi(i) \mid i<n\}$ is always non-empty because $X$ is infinite and there are only finitely many $\varphi(i)$ with $i<n$. So there are elements in $X$ that don't get removed.
Then $\varphi$ is injective: If $b>a$, then $\varphi(a)\ne\varphi(b)$ because $\varphi(b)$ will have been constructed as $g$ of a set where $\varphi(a)$ has explicitly been removed.
(Actually one needs only the axiom of Countable Choice, but that's a bit more involved -- see sketch by Asaf in the comments).
A: Let $X$ be the natural numbers and let $f ( x ) = x+1$ the mapping is injective and range$(f )=\mathbb {N}\setminus \{1\}$ which is infinite.
