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Give an example of a perfect set in $\mathbb R^n$ that does not contain any of the rationals.

(Or prove that it does not exist).

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closed as off-topic by user21820, Did, mechanodroid, Paramanand Singh, Simply Beautiful Art Oct 16 '17 at 15:44

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An easy example comes from the fact that a number with an infinite continued fraction expansion is irrational (and conversely). The set of all irrationals with continued fractions consisting only of 1's and 2's in any arrangement is a perfect set of irrational numbers.

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    $\begingroup$ This is a really nice answer! $\endgroup$ – Akhil Mathew Jul 29 '10 at 4:52
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    $\begingroup$ Beautiful answer! $\endgroup$ – BBischof Jul 30 '10 at 7:16
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Consider the set of reals x whose binary expansion, if you look only at the even digit places, is some fixed non-eventually-repeating pattern z. This is perfect, since we have branching at the odd digits, but they are all irrational, since z is not eventually repeating.

You can draw a picture of this set, and it looks something like the Cantor middle third set, except that you divide into four pieces, and take either first+third or second+fourth, depending on the digits of z.

Another solution: Begin with an interval having irrational endpoints, and perform the usual Cantor middle-third construction, except that at stage n, be sure to exclude the n-th rational number (with respect to some fixed enumeration), using a subinterval having irrational endpoints. By systematically excluding all rational numbers, you have the desired perfect set of irrationals.

(Hi François!)

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    $\begingroup$ Hi Joel! Nice answer! $\endgroup$ – François G. Dorais Jul 30 '10 at 18:30
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It is well-known that $C$ is homeomorphic to $C \times C$, where $C$ is the Cantor set, as both are zero-dimensional compact metric spaces without isolated points. So $C$ contains uncountably many disjoint homeomorphic copies of $C$ and at most countably many of them can contain rationals...

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    $\begingroup$ This is now my favorite example of a nonconstructive proof. $\endgroup$ – Robert Wolfe Sep 9 '14 at 20:36
  • $\begingroup$ @Bryan Thx. It shows the power of the characterisation theorem nicely, I think. It's a folklore argument, don't know whose it is originally. $\endgroup$ – Henno Brandsma Sep 9 '14 at 20:43
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Just consider a translation of Cantor set $C$, denote as $E=C+\{x_0\}$. The perfectness of $E$ is trivial due to the perfectness of $C$. To make $E\cap\mathbb{Q}=\varnothing$, we need to choose an $x_0\notin \mathbb{Q}-C$. The only thing left is to show $\mathbb{Q}-C\neq\mathbb{R}$, i.e. $\mathbb{Q}+C\neq\mathbb{R}$. By Baire Category theorem $$\mathbb{Q}+C=\bigcup_{r\in\mathbb{Q}}\{r\}+C$$ can't have any interior point, since $\{r\}+C$ don't have any interior point, for any $r\in\mathbb{Q}$. The conclusion follows.

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It can be proven that the Cantor set is perfect. Certainly, this contains infinitely many rationals. How about modifying the construction of the Cantor set by defining: $I_1 = [\sqrt{2},\sqrt{2}+1/3] \cup [\sqrt{2}+2/3,\sqrt{2}+1]$, $I_2 = [\sqrt{2},\sqrt{2}+1/9] \cup [\sqrt{2}+2/9,\sqrt{2}+1/3]\cup[\sqrt{2}+2/3,\sqrt{2}+7/9]\cup[\sqrt{2}+8/9,\sqrt{2}+1]$, etc and setting $P = \cap_{i=1}^\infty I_i$? Each of end points of any interval that appears in the construction is a member of $P$ and is irrational. However, is it true that all the members of $P$ must be an end point of a certain interval? I am tempted to think so because we can prove that $P$ does not contain any interval.

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    $\begingroup$ There are only countably many endpoints, but a nontrivial perfect set is uncountable. $\endgroup$ – JDH Jul 30 '10 at 1:45
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    $\begingroup$ Is this correct, then? $\endgroup$ – Lelouch Jun 10 '17 at 10:49
  • $\begingroup$ @Lelouch: JDH's comment is a sufficient reason for why the last two sentences in the post are false. There are uncountably many points in $P$ but only countably many end-points. However JDH's comment does not address whether the rest of the post is a correct answer to the question. Even if the construction works, this answer does nothing to justify it. One would have to prove that the ternary expansion of $\sqrt{2}$ has infinitely many of each digit, otherwise $P$ would have a rational member. $\endgroup$ – user21820 Oct 16 '17 at 11:26
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    $\begingroup$ Indeed, 6 upvoters might want to explain how this addresses the question at all. $\endgroup$ – Did Oct 16 '17 at 11:55
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Let $A$ be an open subset of $R$ of finite measure and containing $Q$. This is possible because $Q$ is countable. Let $B=R$ \ $A$. Now $B$ is closed, and uncountable (because it has infinite measure). Let $ C$ be the family of open real intervals that, each, have countable intersection with $B$. Then $\cup C$ is equal to $\cup D$ where $ D $ is a countable subset of $ C$, so $B$ has countable intersection with $\cup C$. The uncountable closed set $E= B$ \ $\cup C$ is perfect. Indeed, if $p \in E$ and $V$ is an open interval containing $p$, then $E \cap V$ is uncountable.

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