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A Cauchy sequence in the uniform metric has a pointwise limit in the complete target, and the common Cauchy bound makes convergence uniform. The triangle inequality with one fixed continuous approximant shows that the limit is continuous. Thus the space of all continuous maps into a bounded complete metric space is complete, with no compactness assumption on the domain.

For contraction mappings $f,g$, compare their fixed points using $f$ on both arguments, then bound the change from $f$ to $g$ by the uniform metric. The resulting inequality gives the displayed estimate. Continuity at $f$ only needs its own contraction constant $q_f<1$, even if nearby maps have constants approaching one.

The contraction mapping theorem states that a map $f$ of a nonempty complete metric space into itself, satisfying $d(fx,fy)\le qd(x,y)$ for some $0\le q<1$, has a unique fixed point. Iteration from any point converges to it; for example $d(f^{n}x,\tau (f))\le q^{n}d(x,fx)/(1-q)$.

Assume $X$ is nonempty. Its boundedness makes $\delta$ finite. Symmetry, positivity and separation follow pointwise from $d$, and taking the supremum of $d(fx,hx)\le d(fx,gx)+d(gx,hx)$ proves the triangle inequality. Hence $\delta$ is a uniform metric.

If $(f_n)$ is a Cauchy sequence in this uniform metric, each $(f_n(x))$ converges by completeness of $X$; denote its limit by $f(x)$. Given $ϵ>0$, choose $N$ so that $\delta (f_n,f_m)<ϵ$ for $n,m\ge N$. Letting $m\to \infty$ gives $d(f_n(x),f(x))\le ϵ$ for all $x$, so $f_n\to f$ uniformly. To prove continuity, fix $x$ and useChoose $N$ so the outside terms are small, then use continuity of $f_N$ for the middle term. Thus $f\in \mathrm{Maps}(X,X)$, proving completeness of the continuous-map space in the uniform metric.

The contraction subspace is not necessarily complete. On $X=[0,1]$, $f_n(x)=(1-1/n)x$, $n\ge 2$, are contraction mappings and converge to the identity by uniform convergence. Their only possible uniform limit is not a contraction, so this Cauchy sequence has no limit in $\mathcal{C}$.

Nevertheless the fixed point assignment is continuous. Fix $f\in \mathcal{C}$ with contraction constant $q_f<1$, and put $u=\tau (f)$, $v=\tau (g)$. ThenConsequently $d(\tau (f),\tau (g))\le \delta (f,g)/(1-q_f)$. This proves continuity of the fixed-point assignment at $f$ without requiring one contraction constant valid for every $g$.