Continuous functions in a complete metric space I am trying to prove a problem that looks like a contraction mapping but with a different condition. Here it is:
Let $(X,d)$ be a complete metric space. Suppose $f: X \to X$ is continuous and for some $C>0$, $d(f(x),f(y)) \geq Cd(x,y)$ for all $x,y \in X$. Prove that $f(X)$ is closed. 
Thanks you. My first approached was to prove that $f(X)$ contains all its limit points. 
 A: We can actually prove more with no real difficulty: we can show that the map $f$ is a closed map, meaning that $f[F]$ is closed for all closed $F\subseteq X$.
HINT: Let $F$ be an arbitrary closed subset of $X$, and let $y\in\operatorname{cl}f[F]$ be arbitrary. $X$ is a metric space, so there is a sequence $\langle y_n:n\in\Bbb N\rangle$ in $f[F]$ converging to $y$. For each $n\in\Bbb N$ there is an $x_n\in F$ such that $f(x_n)=y_n$. (Why?)


*

*Use the fact that $d(y_m,y_n)\ge Cd(x_m,x_n)$ for all $m,n\in\Bbb N$ to show that the sequence $\langle x_n:n\in\Bbb N\rangle$ is a Cauchy sequence.  

*Now use the fact that $X$ is complete to find an $x\in F$ such that $y=f(x)$.

A: Hint: Assume $y_n=f(x_n)$ is a sequence in $f(X)$ converging to $y\in X$. You need to show that $y=f(x)$ for some $x\in X$.
If you show that $x_n$ converges to some $x$ then $y=f(x)$ by continuity of $f$. And since $X$ is complete it is enough to show that $x_n$ is a Cauchy sequence. Try using the condition on $f$ to show that, it should be swift.
