Can an uncountable subset of $\mathbb{R}$ have empty intersection with its derived set?

Question

Question: Is there an uncountable subset $A$ of $\mathbb{R}$ such that $A\cap A'=\emptyset$, where $A'$ denotes the derived set of $A$?

I just know that an uncountable set $A $ must have limit points (that is, $A'\ne\emptyset $) and that, if $A$ is countable, then it is easy to find such an example, say, we can take $A=\mathbb{N}$ (in which case $A'=\emptyset $ and therefore $A\cap A'=\emptyset $). But, here $A$ is an uncountable set :-(

Please help me, stuck on this

Answer 1

Hint: suppose no point of $A$ is a limit point of $A$. Then every point has an open neighbourhood in which it is the only element of $A$...

Answer 2

You said that you know that every uncountable subset of $\mathbb{R}$ has a limit point, and that's true indeed. Then such point belongs to $A$ and to $A'$ by definition. Hence $A\cap A'$ cannot be empty.

Answer 3

Not only can $A\cap A'$ NOT be empty when $A$ is uncountable, it must in fact be itself uncountable. Consider: Partition $A$ into two sets: Those points which are also in $A'$, and those which are not. The points in the second set are by definition NOT limit points of $A$, therefore they are isolated points of $A$. There's a theorem [look it up] that any set of isolated points must be countable. So if the first subset were countable, their (disjoint!) union - $A$ itself - would have to be countable, contrary to assumption. And of course, if the first subset is empty it is definitely countable.

NOTE: I got this analysis from an answer to a question posed in either American Mathematical Monthly or Mathematics Magazine a few years ago.