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

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.

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.
• I had a go at this earlier and I had a bit of issue. Let $\alpha = \{a\}$ and let $A$ be the power set of $\alpha$. Then $|\alpha| = 1 = \mathfrak{c}$ and $|A| = 2^1 = 2^\mathfrak{c} = 2$. I don't see a way to have a bijection from $A$ to $A$ such that $|\{x\in A:x\neq f(x)\}|=\mathfrak{c}=1$ because to preserve bijectivity, $\emptyset$ must either map to $\emptyset$ or $\alpha$ and $\alpha$ has to map to $f(\emptyset)^c$. In the naive case that $\emptyset$ maps to $\alpha$ and $\alpha$ maps to $\emptyset$, $|\{x\in A:x\neq f(x)\}| = 2 \neq 1$... – Darius Jan 21 at 2:47