Let $A \subset \mathbb{R}$ uncountable meaning $|A| \not \leq \aleph_0$ , prove that there is $B \subset A$ countable $|B| \leq \aleph_0$ such that :
$C = A- B$ is dense meaning $\forall x \in C \forall \epsilon >0 |C \cap (x-\epsilon,x+\epsilon)|\not \leq \aleph_0$
For instance $A = [0,1] \cup \{2\}$ and $B= \{2\}$ so $C = [0,1]$ is dense in every point.
I need a hint on how to construct $B$ given $A$ , or at least how to prove that such $B$ must exist
Thanks
First of all, it's worth noting that your definition of "dense" is not the usual one.
The most obvious difference is the cardinality restriction; normally we just demand nonempty intersection. Note that e.g. under your definition, no countable set can be dense. The second difference is that really you're talking about $C$'s "density" with respect to itself rather than all of the ambient space $\mathbb{R}$. I think a better name for your property would be "uncountably dense-in-itself."
So we want to show:
(Note that we could drop the cardinality restriction on $A$: if $A$ is countable, take $B=A$ and note that the emptyset is vacuously uncountably dense-in-itself.)
The key point is that we want to build $C$ from $A$ by throwing away the "bad points." Your example follows this: $A$ isn't uncountably dense-in-itself yet, because of the isolated point $2$, so we "throw it away" to get $C$. To solve the problem, we need to do two things:
Figure out precisely what "bad point" means.
Show that there are only countably many "bad points."
The right notion of "bad point" isn't hard to come up with: we say that $a\in A$ is bad if there is some open interval $I$ containing $a$ such that $I\cap A$ is countable. The interesting part is showing that there are only countably many bad points. For this, I have two hints:
Show that we can restrict attention, in the definition of "bad point," to intervals $I$ of the form $(p, q)$ with $p, q$ rational. Note that there are only countably many rationals, hence only countably many intervals of this form.
Call an interval $I$ with rational endpoints bad if $I\cap A$ is countable. Then every bad point is contained in a bad interval. Now, how many bad points can a given bad interval contain? And since there are only countably many bad intervals, what does this say about the total number of bad points?
There is one slight subtlety in the above approach. We've focused, in the definition of "bad point," on intervals with rational endpoints rather than intervals of the form $(a-\epsilon, a+\epsilon)$. This let us count the relevant intervals. It's a good exercise to convince yourself that this was an important thing to do.
It's worth noting that a related problem poses an interesting new difficulty. Suppose we drop the uncountability requirement: a set $A\subseteq\mathbb{R}$ is dense-in-itself if for every $a\in A$ and open interval $I\ni a$, we have $I\cap (A\setminus\{a\})\not=\emptyset$. Then the claim is:
(As before we can drop the requirement that $A$ be uncountable.)
At first glance this seems easier: we just want to throw away the isolated points of $A$, and it's easy to show that there are only countably many of these (use the rational interval trick again). However, we run into a problem: there may be non-isolated points of $A$ which become isolated after throwing away the isolated points! E.g. take $$A=\{0\}\cup\{{1\over 2},{1\over 3},{1\over 4},...\}.$$ $0$ is not a limit point of $A$, but once we throw away all the non-limit points of $A$ we are left with just $\{0\}$, which is now a limit point.
(If you like, add "$[17,200]$" to the $A$ defined above to make it uncountable. My point isn't about cardinality, though, it's about the way points can be arranged in a set.)
We need to keep throwing away isolated points over and over again - in fact, infinitely often. This process is the Cantor-Bendixson derivative, and to solve this problem you'll wind up going through the Cantor-Bendixson theorem that every closed set can be written as the union of a countable set and a perfect (= closed with no isolated points) set. (Note that this doesn't immediately solve the problem above, since there is no requirement that $A$ be closed!)