Let $f : X → X$ be an injective map such that $f ( X ) \neq X$ . Show that: a. The set $X$ is infinite.
b. For $x\in X-f(X)$, the elements $x, f(x), f(f(x)), \ldots$ are pairwise distinct.
I came across this problem while looking online for some abstract math problems to do and I have no clue how to solve it. 
For a, I said:
Suppose that $X$ is finite, meaning that there is a finite number of elements and is injective, then $f(x):X \rightarrow X$ is also contains $n$ distinct elements and is contained in $X$ based on the definition of a finite set. However, the prompt says that $f ( X ) \neq X$, therefore $X$ cannot be finite.
For b, I have no clue.
 A: For b, use a proof by induction.  Define $f^0(x)=x$ and define $f^i(x)$ to mean $f$ composed with itself $i$ times.  Our base case is easy because, by assumption, we know that $x$ and $f^1(x)$ are pairwise distinct.
Now assume as our inductive hypothesis that $\{ x, f^1(x), f^2(x), \ldots f^n(x) \}$ are pairwise distinct.  Assume $f^{i+1}(x)=f(f^i(x))=f^{n+1}(x)=f(f^n(x))$.  Then because $f$ is injective, we also know that $f^i(x) = f^n(x)$, which can't happen because our inductive hypothesis is that those elements are part of a pairwise distinct set.  Alternatively, if $x=f^{n+1}(x)=f(f^n(x))$, then $x \in \operatorname{range}(f)$, contradicting our assumption that $x \in X \setminus f(X)$.
A: Suppose that, say $f(f(x))=f(f(f(f(x))))$. Since $f$ is injective, it follows that $f(x)=f(f(f(x)))$. Now, again because $f$ is injective, $x=f(f(x))$. But this is impossible, since $x\in X\setminus f(X)$.
A: Suppose $f^{i}(x)=f^{j}(x)$ with $i <j$. [ $f^{i}$'s are the function obtained by composing $f$ with itself repeatedly]. Use injectivity of $f$ to conclude that $x=f^{j-i}(x)$. But then  $x \in f(X)$ which is  a contradiction. 
