Banach Fixed Point Theorem

(Banach) Let $X$ be a nonempty complete metric space, and let $T : X→X$ be a contracting map, i.e. $d(T(x),T(y)) ≤ θ·d(x,y)$ for all $x,y∈X$ for some fixed $θ < 1$. Then $T$ has a unique fixed point.

Proof: (Banach) Uniqueness of fixed points is clear, since if $T$ fixes $x$ and $y$, then $θ·d(x,y) ≥ d(T(x),T(y))=d(x,y)$ and $θ ≥ 1$ unless $d(x,y)=0$ and $x=y$. To exhibit the fixed point, let $x_0∈X$ be any point, and define $x_{n+1}=T(x_n)$. Then when $α=d(x_0,x_1)$ it follows by induction that $d(x_n,x_{n+1}) ≤ θ^nα$. As well, by the triangle inequality it follows that $d(x_n,x_m) ≤ \sum_{k=n}^{m-1} θ^kα$ when $n≤m$. Since $θ < 1$ this forms part of a convergent geometric series, showing $(x_n)$ forms a Cauchy sequence in $X$. Let $x$ be the point of convergence of this sequence. $T$ is continuous since contraction maps are generally continuous, so $T(x)$ is the point of convergence of the sequence $(T(x_n))=(x_{n+1})$, showing $T(x)=x$.