show that $\{x \in X | F(x) \neq x\}$ is countable. let $X$ be a set and $F: X \to X$ is a function such that for every  $A \subseteq X$ ,  $A- F(A)$ is  countable.

show that $\{x \in X | F(x) \neq x\}$ is countable.


i can show for every $x \in X$ , $F^{-1}(x)$ is countable.

 A: I can do it using the axiom of choice, I don't know if it's really needed.
At any rate, the result is still true under the weaker hypothesis that
$\lbrace x | F(x) \neq x\rbrace$ has an uncountable well-ordered subset,
as will become clear from the proof.
As you already realized, $F^{-1}(\lbrace x \rbrace)$ is countable for every
$x\in X$. It follows that $F^{-1}(A)$ is countable for every countable $A\subseteq X$.
Assume the axiome of choice, so that $X$ has a well-ordering which we denote by $\prec$. Suppose, by contradiction, that $X'=\lbrace x | F(x) \neq x\rbrace$ is not countable. Then, by the remark concluding the above paragraph, if $A$ is any countable subset of $X'$, $A\cup F^{-1}(A)$ (which is countable) does not exhaust all of $X'$. Denote by
$m(A)$ the $\prec$-smallest element in $X'$ outside $A\cup F^{-1}(A)$.
We then define a sequence $(y_{\alpha})$ for all countable ordinals $\alpha$ (in other words, $\alpha \lt {\aleph}_1$) by transfinite induction, as follows : let $y_0\in X'$ be arbitrary. Then, for any countable ordinal $\alpha$, $A_{\alpha}=\lbrace y_\beta,F(y_\beta)\rbrace_{\beta \lt \alpha}$ is countable and we then set $y_\alpha=m(A_\alpha)$.
Now, consider the set $Y=\lbrace y_\alpha | \alpha \lt {\aleph}_1\rbrace$. It has cardinality ${\aleph}_1$. Let $u\in Y$, so that $u=y_{\alpha}$ for some $\alpha\lt {\aleph}_1$.
I claim that $Fu\not\in Y$. Otherwise, we would have $Fu=y_{\beta}$ for some $\beta\lt {\aleph}_1$. If $\beta \lt \alpha$, we have $y_{\alpha} \in F^{-1}(A_{\alpha})$ which is impossible by construction. If  $\beta \gt \alpha$, we have $y_{\beta} \in A_{\beta}$ which is again impossible by construction. Finally, if $\beta=\alpha$ we contradict $u\in X'$.
We have thus shown that $FY$ is disjoint from $Y$, so $Y\setminus FY=Y$. But then
$Y$ must be countable by the initial hypothesis and this is absurd. This finishes the proof.
