Prove the following result known as the contractive application theorem: if $X$ is a compact metric space and $f: X \rightarrow X$ a contractive application (that is, there exists $K <1$ such that $d (f (x), f (y)) <Kd (x, y)$ for all $x \in X$ ) then there exists a single point $x\in X$ such that $f (x) = x$. (hint: if $a \in X$, consider the sequence $x_1 = a, x_n = f (x_ {n-1})$).
I have seen similar results, but with the difference that in the statement they speak of completeness not compactness. And I can't understand or demonstrate how to go from completeness to compactness. The idea that I use for the test is the following: If I can go from compact to complete, it would be enough for me to try cauchy, that would imply convergence and with the continuity of the function $f$ I obtain the fixed point. Uniqueness is much calmer.
The question is, how to go from compactness to completeness? ... And if this is not possible, how else could this statement prove.
Let us know what you've tried so far!
Using the hint, we have for example that $d(x_1,x_2)=d(f(x_0),f(x_1))\leq Kd(x_0,x_1).$
Similarly, using the triangle inequality we have for example that $$d(x_1,x_3)\leq d(x_1,x_2)+d(x_2,x_3)\leq[K+K^2]\cdot d(x_0,x_1).\hspace{0.5cm}{\text{(Why?)}}$$
What can you say in general, then, about $d(x_n,x_m)$? (HINT: create a Cauchy sequence, which you know will converge by the completeness of $X.$)