This question is from Jonathan D. H. Smith's Introduction to Abstract Algebra book.
The question basically says that for an infinite set $X$ a function $f : X \to X$ is called almost identical if the set $\left\{ x \in X \mid x \neq f\left(x\right) \right\}$ of elements $x$ of $X$, differing from their image $f(x)$ under $f$, is finite. If $F$ be a subset of $X^X$ consisting of almost identical functions, show that $F$ is a monoid of functions.
So I think I have shown that $F$ is a semi-group of functions. But, I am confused about showing that an identity exists in $F$ because for all $x$ in $X$, $\operatorname{id}(x)$ = $x$. Does that mean that the set of elements $x$ of $X$ differing from their image $\operatorname{id}(x)$ under $\operatorname{id}$ is empty and hence finite?
This is very confusing to me. Can somebody provide a hint? Thank you a lot.
A set $S$ is said to be finite if $S$ is empty or there exists a bijection $S \rightarrow I_n$ for some $n \in \mathbb{N}$, where $I_n = \{ 1, 2, ..., n \}$.