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 $\rho(Ax,Ay)&lt;\rho(x,y)$, $x\neq y,  \Rightarrow A $ have only one fix point on X.
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.
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 $Ax=ax$ $(0\leq a&lt;1)$ 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 $|Ax-Ay|&lt;|x-y|$.