The question:
Let $A$ be a mapping of a metric space $R$ into itself. Prove that the condition $$\rho(Ax,Ay)<\rho(x,y),\quad (x\neq y)$$ is insufficient for the existence of a fixed point of $A$.
My response:
Let $$A(x)=\sqrt{1+x^2}$$
First observe $A'(x)=\dfrac{x}{\sqrt{1+x^2}}< 1, \forall x\in \mathbb R^+$
Now using mean value theorem from calculus, $\exists c\in \mathbb R^+$ such that
$$\rho(Ax,Ay)=|A(x)-A(y)|=|A'(c)||x-y|<|x-y|=\rho(x,y)$$
However $A(x)$ has no fixed point because if it had then $A(x)=x$ would implied $x^2=1+x^2$ which is impossible.
My Actual Question: In a logical way, is not that counter example for the given question? I mean, am I missing some point that maybe I should have proved this for a generic $A$? I think my example is enough for the question but the lecturer says that "You should show this for general $A$", Why?