Possible Duplicate:
Perfect set without rationals

Does there exist a nonempty perfect set in $\mathbb{R}$ which contains no rational number?

This problem is on p.44 PMA - Rudin

I found a proof of this on google but the proof is not 'suitable' for me, since the proof uses the concept of 'measure', which is in chapter 11, while this problem is on chapter 2.

Can one show this by direct construction?


marked as duplicate by t.b., Henning Makholm, Gerry Myerson, user940, David Mitra Aug 4 '12 at 5:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    $\begingroup$ Proceed as in the construction of the Cantor set on set $I=[\alpha,\beta]$ with $\alpha$, $\beta$ irrational. But when removing intervals, select intervals with irrational endpoints that contain rationals not already removed from a fixed enumeration of the rationals in $I$. $\endgroup$ – David Mitra Aug 4 '12 at 1:50
  • $\begingroup$ @t.b. it is.. Sorry for that, btw is anon's answer on that link correct? $\endgroup$ – Katlus Aug 4 '12 at 2:02
  • $\begingroup$ No need to apologize :) Re the answer you ask about: I didn't look closely, but it looks somewhat ... incomplete to me. By the way: if you follow the Linked column on the right hand side of the page (in the duplicate), you'll find a few more threads containing interesting answers, e.g. this one $\endgroup$ – t.b. Aug 4 '12 at 2:16
  • $\begingroup$ @David Mitra I see AC is inevitable in your argument am i right? $\endgroup$ – Katlus Aug 4 '12 at 3:19

Since the rational are countable, let $a_n$ enumerate all the rationals. Let $B_{2^{-n}}(a_n)$ be the open interval centered at $a_n$ of radius $2^{-n}$. Let $A = \bigcup_{n \in \mathbb{N}} B_{2^{-n}}(a_n)$. $A$ is open since it is it is a union of open sets. Note that $\mathbb{Q} \subset A$. Clearly $A$ is not all of $\mathbb{R}$ because "length" of $A$ is less than or equal to $2\sum_{n = 1}^\infty 2^{-n} < \infty$. (This is essentially the measure idea.) Because of this and fact that $\mathbb{R}$ is uncountable, you have that $C:= \mathbb{R} - A$ is an uncountable set. $C$ closed since it is the complement of an open set. Also $C \cap \mathbb{Q} = \emptyset$ since $\mathbb{Q} \subset A$.

Finally, by the Cantor Bendixson Theorem (Exercise 28 on page 45) which states that every uncountable closed set is (uniquely) the union of a perfec set and a countable. Hence there exists a perfect set $C'$ and a countable set $F$ such that $C = C' \cup F$. Since $C \cap \mathbb{Q} = \emptyset$, you also have $C' \cap \mathbb{Q} = \emptyset$. Hence $C'$ is a nonempty perfect set that does not contain any rational numbers.

By the way $C'$ can be constructed, look at the proof of the Cantor Bendixson Theorem to see exactly how it is made.

  • $\begingroup$ @william I don't understand why $C$ uncountable while $|A|=2^{\aleph_0}$. Help $\endgroup$ – Katlus Aug 4 '12 at 2:25
  • $\begingroup$ @Katlus Why does $|A| = 2^{\aleph_0}$ mean that $C = \mathbb{R} - A$ can't have cardinality $2^{\aleph_0}$? For instance, $(0,1)$ has cardinality $2^{\aleph_0}$, but $\mathbb{R} - (0,1)$ definitely still has cardinality $2^{\aleph_0}$. $\endgroup$ – William Aug 4 '12 at 2:28
  • $\begingroup$ @Katlus To show that $C$ is uncountable, note that $A$ takes up length less than $2\sum_{i = 0}^\infty 2^{-n} = 4$ since it a union of a sequence of open intervals of length $2(2^{-n})$. Of course $\mathbb{R}$ has infinite length. So there is alot that is missing from $A$. $\endgroup$ – William Aug 4 '12 at 2:33
  • $\begingroup$ so i see AC is used here. Am i right? $\endgroup$ – Katlus Aug 4 '12 at 2:50
  • $\begingroup$ I misunderstood. I got it now $\endgroup$ – Katlus Aug 4 '12 at 3:25

Not the answer you're looking for? Browse other questions tagged or ask your own question.