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I came up with a problem when learning real analysis.

We have known that since $Q$ is a countable set and $Q'=R$, the cardinality of $Q'$ is $c$.

Then, can we construct a countable set $A$, such that the cardinality of $A'$ is $c$, and $A\cap A'=\emptyset$ ?

Some textbook on real analysis provides this conclusion, but I wonder how to construct it.

Your help will be appreciated.

Note:

  • $c$ denotes the cardinality of $\mathbb{R}$;
  • $A^\prime$ denotes the derived set of $A$.
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  • $\begingroup$ What do $Q'$ and $A'$ mean here? $\endgroup$ – Clive Newstead Oct 3 '13 at 18:48
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    $\begingroup$ @PedroTamaroff I'll bet it's a typo and the OP meant "the cardinality of $Q'$ is $c$." $\endgroup$ – Trevor Wilson Oct 3 '13 at 19:03
  • $\begingroup$ Assuming that was a typo, it seems like a good question to me, but I'll wait for the OP to confirm this. $\endgroup$ – Trevor Wilson Oct 3 '13 at 19:05
  • $\begingroup$ It seems to be I live by an anachronic definition of accumulation point, which causes $\;A\subset A'\;$ ...but re-reading the definition this is nonsense. Deleting on its way... $\endgroup$ – DonAntonio Oct 3 '13 at 19:10
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    $\begingroup$ @CliveNewstead Probably the derived sets of $Q'$ and $A'$ respectively (as proposed in a now-deleted comment.) $\endgroup$ – Trevor Wilson Oct 3 '13 at 19:17
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Here's one way, but it might not be the simplest. Let $C$ be any Cantor set in the reals ($C$ is a set that is nonempty, nowhere dense, and perfect) and let $A$ be the set of all midpoints of the bounded complementary intervals of $C.$ Then $A' = C,$ so $A'$ has cardinality $c.$ Also, $A \cap A' = A \cap C = \emptyset,$ since all points in $A$ belong to the complement of $C.$

ADDED NEXT DAY $\;$You asked (in a comment) for me to explain in detail why $A' = C.$ I'll do this by proving these two inclusions: (1) $\,A' \subseteq C\;$ and$\;$ (2) $\,C \subseteq A'.$

(1) To prove $A' \subseteq C,$ it suffices to prove ${\mathbb R} - C \subseteq {\mathbb R} - A'.$ To this end, choose $x \in {\mathbb R} - C.$ Then $x$ belongs to at most one of the bounded complementary intervals of $C.$ Therefore, all sufficiently small open intervals containing $x$ contain at most one point of $A$ (indeed, no points of $A$ unless $x$ happens to belong to $A).$ It follows that $x$ cannot be a limit point of $A,$ and hence we have $x \in {\mathbb R} - A'.$ This completes the proof that $A' \subseteq C.$

(2) We are to prove $C \subseteq A'.$ To this end, choose $x \in C.$ Since every point of $C$ is a limit point of $C,$ there exists a sequence $\{x_n\}$ of points in $C$ such that $x_n \rightarrow x.$ By passing to a subsequence if necessary, we can assume that $\{x_n\}$ approaches $x$ monotonically from one side. That is, we have $x_n \nearrow x$ or we have $x_n \searrow x.$ Between $x_1$ and $x_2,$ there will be at least one bounded complementary interval of $C.$ Pick one of them and let $a_1$ be the point we chose from that bounded complementary interval when we formed the set $A.$ Between $x_2$ and $x_3,$ there will be at least one bounded complementary interval of $C.$ Pick one of them and let $a_2$ be the point we chose from that bounded complementary interval when we formed the set $A.$ By continuing in this manner, we get a sequence $\{a_n\}$ of points in $A$ such that $a_n \rightarrow x.$ Hence, $x \in A',$ which completes the proof that $C \subseteq A'.$

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  • $\begingroup$ I thought about finite sums $\sum_1^N \frac{e_n}{3^n}$ where $e_n$ belong to $\{0,2\}$ except for last two, which I would put $1,1$. $\endgroup$ – Sungjin Kim Oct 3 '13 at 19:31
  • $\begingroup$ @Dave L. Renfro Could you please explain in detail why $A'=C$ here? I am still kind of puzzled. $\endgroup$ – ranky Oct 4 '13 at 5:30
  • $\begingroup$ @Dave L. Renfro Thank you so much for providing this brilliant construction and its relevant explanations. But I have been still considering whether a simpler solution can be found. Can you provide some other ideas, if any, without utilizing the Cantor set? Just ideas, I mean. $\endgroup$ – ranky Oct 5 '13 at 15:32
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    $\begingroup$ @Renke Cai: I don't know of a construction not involving a Cantor set, but if I think of something later I'll post it. My feeling is that this is not possible. Indeed, unless I'm overlooking something (I'm in a hurry now, so I might be overlooking something), the impossibility follows from these two results: (1) If the closure of $A$ (the set $A \cup A')$ contains an interval, then $A \cap A' \neq \emptyset.$ (2) If the closure of $A$ contains no intervals and $A$ is uncountable, then $A \cup A'$ is an uncountable closed nowhere dense set, and so the union of a perfect set and a countable set. $\endgroup$ – Dave L. Renfro Oct 5 '13 at 18:34

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