Fixed Point of Mappings is Fixed Point of Composition/General Result

From ProofWiki
Jump to navigation Jump to search

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

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 $g$.

$\blacksquare$