finding bijection such that $|\{ x\in A : x \neq f(x)\}| =\mathfrak{c} $

Question

Let $|A| = 2^{\mathfrak{c}}$. I am finding function $f$ is bijection from $A$ to $A$ such that $|\{ x\in A : x \neq f(x)\}| =\mathfrak{c} $. Any ideas? I will try to prove it later.

Answer 1

I assume that by $\mathfrak{c}$ you mean the cardinality of the reals, but this argument works for any cardinality with a slight adjustment and choice (if we use the reals then we get a lot of functions etc that otherwise we would have to prove or assume the existence of).

We may take it that $A$ is the power set of the reals, so the sets of the form $B_x = \{x\}$ for some $x \in \mathbb{R}$ have cardinality $\mathfrak{c}$. If we define $f$ such that $f(B_x) = B_{x+1}$ and $F(C) = C$ for all other sets, then we are done.