If $X$ is a compact space and $A:X\rightarrow X$ is such that $\rho(Ax,Ay)<\rho(x,y)$, $x\neq y, \Rightarrow A $ have only one fix point on X. I have problems with this demostration can anybody help me please? 
If $X$ is a compact space and $A:X\rightarrow X$ is such that $\rho(Ax,Ay)<\rho(x,y)$, $x\neq y,  \Rightarrow A $ have only one fix point on X.
 A: Um, if x and y are two unequal fixed points then $\rho(Ax,Ay)= \rho(x,y)$.
So there can't be more than one fixed point.  
A: Assuming $X\ne \emptyset.$
Let $f(x)=\rho (x,Ax).$ We have $$f(y)=\rho (y,Ay)\leq \rho (y,x) +\rho (x,Ax)+\rho (Ax,Ay)\leq$$ $$\leq \rho (y,x)+\rho (x,Ax)+\rho (x,y)=$$ $$=2\rho(x,y)+f(x).$$ $$\text {So }\quad f(y)-f(x)\leq 2\rho (x,y).$$ Interchanging $x$ and $y$ we have  $$f(x)-f(y)\leq 2\rho (y,x).$$ From the last 2 inequlities above,   we have $|f(x)-f(y)|\leq 2\rho (x,y).$
Therefore $f$ is continuous from $X$ to $\Bbb R.$
Now $X$ is compact and not empty and $f$ is continuous, so $f(X)$ is a non-empty compact subset of $\Bbb R,$ so  $f(X)$ has a minimum value . 
If $f(x)>0$ then $f(x)\ne\min f(X).$ Because with $y=Ax,$  we have $x\ne y$ (because $\rho (x,y)=\rho (x, Ax)=f(x)\ne 0$), so $$f(x)=\rho(x,Ax)=\rho (x,y)>\rho (Ax,Ay)=\rho (y,Ay)=f(y).$$ 
Therefore  $\min f(X)=0.$ So there exists $x$ with $f(x)=0.$
Now $f(x)=0\iff \rho (x,Ax)=0 \iff x=Ax.$ But if $f(x)=f(y)=0$ with $x\ne y$ then $\rho (x,y)>\rho (Ax,Ay)=\rho (x,y),$ a paradox. So there is exactly one $x$ such that $x=Ax.$
A: Under the lack of the compactness argument you can not assure that the infimum of the function $f(x)=\rho(Ax,x)$ equals zero so that the solution of the equation $Ax=x$ is unique. 
In particular, on the metric space endowed by the interval $X=(0,1)$ and the Euclidean metric $\rho(x,y)=|x-y|$, the family of linear functions $Ax=ax$ $(0\leq a<1)$ has no fixed points. 
Recall that the compactness argument is not a necessary condition to assure that a function has a unique fixed point: on the metric space $X=(-\frac{1}{2},\frac{1}{2})$, the quadratic function $Ax=x^2$ has a unique fixed point ($x=0$) and at the same time satisfies the condition $|Ax-Ay|<|x-y|$.
