1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 $\emptyset$, $|\{x\in A:x\neq f(x)\}| = 2 \neq 1$... $\endgroup$ – Darius Jan 21 at 2:47
  • 1
    $\begingroup$ Its a good point - I was assuming that we had infinite cardinalities. $\endgroup$ – user24142 Jan 21 at 5:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.