closed function with uncountable set in the domain

Question

First of all, I do apologize for the horrible title, but I really couldn't find a proper one :(.

I have to solve the following exercise:

Let $F:\mathbb{R}\rightarrow\wp(\mathbb{R})$ be a function such that $\forall x.(F(x)$ is closed and $x\notin F(x))$. Then there exists $A\subseteq\mathbb{R}$, card($A)=2^{\aleph_0}$ such that if $x\neq y$ in $A$ then $x\neq F(y)$ and $y\neq F(x)$.

Apart from the fact that I can't understand fully the reasons why it should be so, and neither am I convinced that the statement of the exercise should be true, I don't really know how to proceed.

I have also the following hint: $\forall x$ there exists an interval $I=(q_0,q_1)$ with $q_0,q_1\in \mathbb{Q}$, such that $x\in I$ and $I\cap F(x)=\emptyset$. But this is nothing really enlightening, I mean, it is just the topological definition of "a point not being in a closed set'', plus the fact that the rationals are dense in the real numbers.

I would be really grateful to anyone who can give my any sort of help.

Answer 1

In view of the hint, I am assuming that the conclusion of the theorem was supposed to be:

Then there exists $A\subseteq\Bbb R$, $|A|=2^{\aleph_0}$, such that if $x,y\in A$ and $x\ne y$, then $x\notin F(y)$ and $y\notin F(x)$.

Further HINT: For each $x\in\Bbb R$ let $I_x$ be an open interval with rational endpoints such that $x\in I_x$ and $I_x\cap F(x)=\varnothing$.

• If $I_x=I_y$, is $y$ in $F(x)$?

There are only countably many open intervals with rational endpoints, and $\Bbb R$ is not the union of countably many sets of cardinality less than $|\Bbb R|$.