Fixed Point of Mappings is Fixed Point of Composition/General Result
Contents
Theorem
Let $S$ be a set.
Let $n \in \N$ be a strictly positive integer.
Let $\N_n$ be the initial segment of $n$ in $\N$.
That is, let $\N_n = \left\{{0, 1, \dots, n-1}\right\}$.
For each $i \in \N_n$, let $f_i: S \to S$ be a mapping.
Let $x \in S$ be a fixed point of $f_i$ for each $i \in \N_n$.
Let $g = f_0 \circ f_1 \circ \dots \circ f_{n-1}$ be the composition of all the $f_i$s.
Then $x$ is a fixed point of $g$.
Proof
The proof proceeds by mathematical induction on $n$, the number of mappings.
Base Case
Flawed. The base case needs to be the two-element case for obvious reasons.
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If $n = 1$, then $g = f_0$.
Since $x$ is a fixed point of $f_0$, it is also a fixed point of $g$.
$\Box$
Inductive Step
Suppose that the theorem holds for $n$. We will show that it holds for $n+1$.
Let $x \in S$ be a fixed point of $f_i$ for each $i \in \N_{n+1}$.
Let $g = f_0 \circ f_1 \circ \dots \circ f_{n-1} \circ f_n$ be the composition of all the $f_i$s.
Since the theorem holds for $n$, $x$ is a fixed point of $f_0 \circ f_1 \circ \dots \circ f_{n-1}$.
By Composition of Mappings is Associative:
- $g = \left({ f_0 \circ f_1 \circ \dots \circ f_{n-1} }\right) \circ f_n$
Thus by Fixed Point of Mappings is Fixed Point of Composition (for two mappings), $x$ is a fixed point of $g$.
$\blacksquare$
