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

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 :-(

• Well, such a set will need to have empty interior at the very least. At the same time it can't be closed, in fact it needs to fail to be closed quite spectacularly. This necessarily makes it rather weird, because it is Cantor-like in that it has empty interior but highly non-Cantor-like by virtue of being "extremely non-closed". A dense union of Cantor sets (e.g. take a union of fat Cantor sets of measure $1-1/n$ in $[0,1]$) will have empty interior and not be closed, but then the failure is similar to the failure for $\mathbb{R} \setminus \mathbb{Q}$. – Ian Jun 14 '18 at 18:17
• @Asaf I am not sure I see how this question is a duplicate of the other one. It appears to me all the other one requires (and all the answer shows) is that there is an accumulation point of $A$, not that there is such a point in $A$. – Andrés E. Caicedo Jun 14 '18 at 18:29
• (The right question to close this one as a duplicate of is here.) – Andrés E. Caicedo Jun 14 '18 at 18:32
• @Ian Sir, so such set would not exists? – Akash Patalwanshi Jun 14 '18 at 18:46
• @AkashPatalwanshi My thought process does not prove that, it merely says that any such set has to be quite different from most sets we work with. – Ian Jun 14 '18 at 18:48

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 $A$...

• And if these neighborhoods are disjoint, we are done. – Andrés E. Caicedo Jun 14 '18 at 18:31
• @AndrésE.Caicedo I had a different way to proceed in mind - we can take the neighbourhoods to have rational endpoints, and that lets us inject $A$ into a countable set. – Wojowu Jun 14 '18 at 18:33
• Yes, that's simpler. :-) – Andrés E. Caicedo Jun 14 '18 at 18:35
• Sir, Very nice approach. So finally we get contradiction, as $A$ is uncountable. So our assumption that, $A∩A'$ is empty is wrong. Hence, such a set $A$ does not exists? Am I correct, Sir? – Akash Patalwanshi Jun 14 '18 at 18:51
• @AkashPatalwanshi That's exactly right. – Wojowu Jun 14 '18 at 18:55

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.

• Sir, why such a point belongs to $A$ and $A'$ both? Yes indeed point must belong to $A'$ by definiition but, why it must belong to $A$? – Akash Patalwanshi Jun 14 '18 at 18:11
• I don't understand: why does a limit point of $A$ necessarily belong to $A$? Consider for instance $\mathbb{Q}$, this has many limit points not in itself. It also has limit points inside itself, and is countable anyway, so it doesn't answer the question, but it definitely contradicts your line of reasoning. – Ian Jun 14 '18 at 18:12
• Because $A$ is uncountable, isn't it? Or did I misunderstand your question? – Javi Jun 14 '18 at 18:48
• @Javi No, a limit point does not necessarily belong to $A$ by definition. – Ian Jun 15 '18 at 0:34
• @Javi sir, yes $A$ is uncountable but this does not imply limit point must belong to $A$. Consider an example of $\mathbb{Q^c}$, ,which is uncountable and every rational number is limit point of $\mathbb{Q^c}$ but none of them belongs to it :-) – Akash Patalwanshi Jun 15 '18 at 1:04

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.

• Yes, it is obvious that the problem is to show the theorem for which all you say is "look it up". – Andrés E. Caicedo Jun 14 '18 at 18:26