Let $M = (X, d)$ be a D1107: Metric space such that

(i) | $f : X \to X$ is a D49: Proper contraction on $M$ |

Then
\begin{equation}
\# \{ x \in X : f(x) = x \}
\leq 1
\end{equation}

Result R4174
on D49: Proper contraction

Proper contraction has at most a single fixed point

Formulation 0

Let $M = (X, d)$ be a D1107: Metric space such that

(i) | $f : X \to X$ is a D49: Proper contraction on $M$ |

Then
\begin{equation}
\# \{ x \in X : f(x) = x \}
\leq 1
\end{equation}

Proofs

Let $M = (X, d)$ be a D1107: Metric space such that

(i) | $f : X \to X$ is a D49: Proper contraction on $M$ |

If $f$ has no fixed points, then the claim clearly holds. Assume thus that $f$ has at least one fixed point and let $x, y \in X$ be fixed points of $f$. Since $f$ is a proper contraction, there is a constant $0 \leq C < 1$ such that
\begin{equation}
d(x, y) = d(f(x), f(y)) \leq C d(x, y)
\end{equation}
Since $d(x, y) \geq 0$ is an unsigned quantity, this can only happen when either $C$ or $d(x, y)$ is zero. If $d(x, y) = 0$, then $x = y$ by definition of a D58: Metric. Since $x$ and $y$ were arbitrary fixed points of $f$, result R2833: Nonempty set is singleton iff all elements equal each other guarantees that the set $\{ x \in X : f(x) = x \}$ is a singleton.

If $C = 0$, then $0 \leq d(x, y) \leq 0$ and thus $d(x, y) = 0$. Metric axioms imply $x = y$ and the claim again becomes a consequence of R2833: Nonempty set is singleton iff all elements equal each other. $\square$

If $C = 0$, then $0 \leq d(x, y) \leq 0$ and thus $d(x, y) = 0$. Metric axioms imply $x = y$ and the claim again becomes a consequence of R2833: Nonempty set is singleton iff all elements equal each other. $\square$