=== Uniqueness === Let $f$ have two fixed points $p_1, p_2 \in M$. It is to be proved that $p_1 = p_2$. {{begin-eqn}} {{eqn | l = \map d {p_1, p_2}       | r = \map d {\map f {p_1}, \map f {p_2} }       | c = {{Defof|Fixed Point}} }} {{eqn | o = \le       | r = q \, \map d {p_1, p_2}       | c = {{Defof|Contraction Mapping}} }} {{eqn | ll= \leadsto       | l = \paren {1 - q} \map d {p_1, p_2}       | o = \le       | r = 0       | c = subtracting $q \, \map d {p_1, p_2}$ from both sides }} {{end-eqn}} Then from $\paren {1 - q} > 0$ and $\map d {p_1, p_2} \ge 0$: :$\map d {p_1, p_2} = 0$ From Metric Space Property $(\text M 4)$: :$p_1 = p_2$ === Existence === The fixed point $p$ will be found from an arbitrary member $a_0$ of $M$ by iteration. The plan is to obtain $\displaystyle p = \lim_{n \mathop \to \infty} a_n$ with definition $a_{n + 1} = \map f {a_n}$. The sequence of iterates converges in complete metric space $M$ because it is a Cauchy sequence in $M$, as is proved in the following. Induction on $n$ applies to obtain the '''contractive estimate''': :$\map d {a_{n + 1}, a_n} \le q^n \, \map d {a_1, a_0}$ Induction details $n = 1$: {{begin-eqn}} {{eqn | l = \map d {a_2, a_1}       | r = \map d {\map f {a_1}, \map f {a_0} }       | c = Definition of $\sequence {a_n}$ }} {{eqn | o = \le       | r = q \, \map d {a_1, a_0}       | c = {{Defof|Contraction Mapping}} }} {{end-eqn}} Assume the contractive estimate for $n = k$. Induction details for $n = k + 1$: {{begin-eqn}} {{eqn | l = \map d {a_{k + 2}, a_{k + 1} }       | r = \map d {\map f {a_{k + 1} }, \map f {a_k} } }} {{eqn | o = \le        | r = q \, \map d {a_{k + 1}, a_k}       | c = {{Defof|Contraction Mapping}} }} {{eqn | o = \le        | r = q \, q^k \, \map d {a_1, a_0}       | c = Induction hypothesis applied. Induction complete. }} {{end-eqn}} Next it is to be proved that $\sequence {a_n}$ is a Cauchy sequence in $M$ by showing that $\displaystyle \lim_{m \mathop \to \infty} \map d {a_{n + m}, a_n} = 0$ for $n$ large. {{begin-eqn}} {{eqn | l = \map d {a_{n + m}, a_n}       | o = \le       | r = \sum_{j \mathop = n}^{n + m - 1} \map d {a_{j + 1}, a_j}       | c = Metric Space Property $(\text M 2)$: Triangle Inequality and telescoping sum ideas }} {{eqn | o = \le       | r = \sum_{j \mathop = n}^{n + m - 1} q^j \, \map d {a_1, a_0}       | c = Apply the '''contractive estimate'''. }} {{eqn | o = \le       | r = q^n \paren {\dfrac {1 - q^m} {1 - q} } \, \map d {a_1, a_0}       | c = Sum of Geometric Sequence on {{RHS}} }} {{end-eqn}} We have that: :$\displaystyle \lim_{n \mathop \to \infty} q^n = 0$ and: :$\dfrac {1 -q^m} {1 - q} \le \dfrac 1 {1 - q}$ Then $\sequence {\map d {a_{n + m}, a_n} }$ has limit zero at $m = \infty$ for large $n$. Sequence $\sequence{a_n}$ is a Cauchy sequence convergent to some $p$ in $M$. Then: {{begin-eqn}} {{eqn | l = \map d {\map f p, p}       | o = \le       | r = \map d {\map f p, \map f {a_n} } + \map d {\map f {a_n}, p}       | c = Metric Space Property $(\text M 2)$: Triangle Inequality }} {{eqn | o = \le       | r = q \, \map d {p, a_n} + \map d {a_{n + 1}, p}       | c = {{Defof|Contraction Mapping}}, and $a_{n + 1} = \map f {a_n}$ }} {{end-eqn}} The {{RHS}} has limit zero at $n = \infty$. Then: :$\map d {\map f p, p} = 0$ Then by Metric Space Property $(\text M 4)$: :$\map f p = p$. {{qed}}