Banach fixed-point theorem (contraction mapping theorem)

Step 1 — The iterates form a Cauchy sequence. Fix $x_0 \in X$ and set $x_n = T^n(x_0)$. By the contraction property applied repeatedly, $d(x_{n+1},x_n) = d(T(x_n),T(x_{n-1})) \le q\,d(x_n,x_{n-1}) \le \dots \le q^n\,d(x_1,x_0)$. For $m > n$, the triangle inequality chains this along the path $x_n, x_{n+1}, \dots, x_m$: $d(x_m,x_n) \le \sum_{k=n}^{m-1} d(x_{k+1},x_k) \le \sum_{k=n}^{m-1} q^k\,d(x_1,x_0) \le d(x_1,x_0)\sum_{k=n}^{\infty} q^k = d(x_1,x_0)\,\dfrac{q^n}{1-q}$, using the geometric series formula since $0 \le q < 1$. As $n \to \infty$, $q^n \to 0$, so this tail bound goes to 0: $(x_n)$ is Cauchy.

Step 2 — Completeness gives a limit, and continuity makes it a fixed point. Since $X$ is complete, the Cauchy sequence $(x_n)$ converges to some $x^* \in X$. The contraction inequality $d(T(x),T(y)) \le q\,d(x,y)$ shows $T$ is (Lipschitz) continuous, so $T(x^*) = T(\lim_n x_n) = \lim_n T(x_n) = \lim_n x_{n+1} = x^*$. Thus $x^*$ is a fixed point.

Step 3 — Uniqueness. Suppose $x^*$ and $y^*$ are both fixed points. Then $d(x^*,y^*) = d(T(x^*),T(y^*)) \le q\,d(x^*,y^*)$, so $(1-q)\,d(x^*,y^*) \le 0$. Since $1-q>0$, this forces $d(x^*,y^*)=0$, i.e. $x^*=y^*$.

Step 4 — The error bound. Letting $m \to \infty$ in the Step 1 estimate $d(x_m,x_n) \le \dfrac{q^n}{1-q}\,d(x_1,x_0)$ and using continuity of $d$ gives exactly $d(x_n,x^*) \le \dfrac{q^n}{1-q}\,d(x_1,x_0)$: the distance to the true fixed point after $n$ steps shrinks geometrically, and the bound can be computed before running a single iteration.