30
$\begingroup$

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).

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Did, mechanodroid, Paramanand Singh, Simply Beautiful Art
If this question can be reworded to fit the rules in the help center, please edit the question.

41
$\begingroup$

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.

$\endgroup$
  • 5
    $\begingroup$ This is a really nice answer! $\endgroup$ – Akhil Mathew Jul 29 '10 at 4:52
  • 3
    $\begingroup$ Beautiful answer! $\endgroup$ – BBischof Jul 30 '10 at 7:16
25
$\begingroup$

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!)

$\endgroup$
  • 2
    $\begingroup$ Hi Joel! Nice answer! $\endgroup$ – François G. Dorais Jul 30 '10 at 18:30
20
$\begingroup$

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...

$\endgroup$
  • 2
    $\begingroup$ 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
7
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ There are only countably many endpoints, but a nontrivial perfect set is uncountable. $\endgroup$ – JDH Jul 30 '10 at 1:45
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Indeed, 6 upvoters might want to explain how this addresses the question at all. $\endgroup$ – Did Oct 16 '17 at 11:55
0
$\begingroup$

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.

$\endgroup$