Contraction Mapping Principle

First take $q \in [0, 1)$: distances are always $\ge 0$, so the contraction constant $q$ — a ratio of distances — may be taken with $q \ge 0$ (if $q < 0$ then $d(Tx,Ty) \le q\,d(x,y) \le 0$ would force every $Tx$ to coincide, and in any case the induction multiplying by $q$ and the geometric tail formula below would break over the sign). Squeeze one step at a time: $d(x_{k+1},x_{k+2}) = d(Tx_k,Tx_{k+1}) \le q\,d(x_k,x_{k+1})$, and since $q \ge 0$ the inequality direction is preserved, so by induction $d(x_k,x_{k+1}) \le q^k\,d(x_0,x_1)$ for every $k$.

For $m>n$, chaining the triangle inequality gives $d(x_n,x_m) \le \sum_{k=n}^{m-1} d(x_k,x_{k+1}) \le \left(\sum_{k=n}^{m-1} q^k\right) d(x_0,x_1)$. Extending the sum to $\infty$ yields the geometric tail $\sum_{k=n}^{\infty} q^k = \frac{q^n}{1-q}$, so $d(x_n,x_m) \le \frac{q^n}{1-q}\,d(x_0,x_1)$.

The right side tends to $0$ as $n \to \infty$, independently of $m$, so $(x_n)$ is Cauchy. With $q = 1/2$ and $x_0 = 1$, $d(x_1,x_0) = 1/2$ makes the bound $2^{-n}$; at $n = 8$ it is $\frac{1}{256} \approx 0.0039$ — matching the caption's figure.

Since $|T(x) - T(y)| = \left|\frac{x}{2} - \frac{y}{2}\right| = \frac12|x - y|$, $T$ is a contraction with $q = 1/2$. And $x \in (0,1]$ gives $T(x) = x/2 \in (0, 1/2] \subset (0,1]$, so $T$ maps $(0,1]$ into itself.

A fixed point would satisfy $x = x/2$, i.e. $x = 0$, but $0 \notin (0,1]$. Hence $T$ has no fixed point in $(0,1]$.

Starting from $x_0 = 1$ gives $x_n = 2^{-n}$: $1, \frac12, \frac14, \dots$.

This sequence satisfies the geometric bound from the first problem, so it is Cauchy — but its limit $0$ lies outside the space.

Without completeness a Cauchy sequence loses its "place to land" and the whole theorem collapses; at $n = 8$, $x_8 = \frac{1}{256} \approx 0.0039$, matching the caption's figure.

Completeness makes the Cauchy sequence $(x_n)$ converge to a point $x^*$ inside the space. A contraction is continuous since $d(Tx,Ty) \le q\,d(x,y)$, so $Tx^* = T(\lim x_n) = \lim Tx_n = \lim x_{n+1} = x^*$ — $x^*$ is a fixed point.

Justification for passing to the limit: the metric is continuous — by the reverse triangle inequality, $\lvert d(x_n,x_m) - d(x_n,x^*)\rvert \le d(x_m, x^*) \to 0$ as $m \to \infty$, so $d(x_n,x_m) \to d(x_n,x^*)$.

Hence letting $m \to \infty$ in the first problem's estimate (and $d(x_1,x_0) = d(x_0,x_1)$) gives the error bound $d(x_n,x^*) \le \frac{q^n}{1-q}\,d(x_1,x_0)$.

For $T(x) = x/2 + 1$ with $q = 1/2$ and $x_0 = 1$, the bound is $\frac{(1/2)^8}{1/2}\cdot\frac12 = \frac{1}{256} \approx 0.0039$, and the actual iterate $x_8 = 2 - 2^{-8} = 1.99609375$ has error $|x_8 - 2| = \frac{1}{256}$, matching the bound exactly.

Uniqueness: if $x^*$ and $y^*$ are both fixed points, $d(x^*, y^*) = d(Tx^*, Ty^*) \le q\,d(x^*, y^*)$. Since $q < 1$, this forces $d(x^*, y^*) = 0$, i.e. $x^* = y^*$.