Perfect set without rationals

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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What is a perfect set? Also, this looks like a homework problem. – Kevin H. Lin Jul 28 '10 at 21:36
Why are you asking if you apparently know the answer? – Mariano Suárez-Alvarez Jul 28 '10 at 21:37
@Kevin: Line Bundle is not asking homework questions - no-one would have homework on so many different areas at once – Casebash Jul 28 '10 at 21:37
@Kevin, a perfect set is one which is equal to its derived set, as in en.wikipedia.org/wiki/Perfect_set – Mariano Suárez-Alvarez Jul 28 '10 at 21:38
Mariano has started a discussion at meta: meta.math.stackexchange.com/questions/313/… – Kevin H. Lin Jul 28 '10 at 21:40

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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This is a really nice answer! – Akhil Mathew Jul 29 '10 at 4:52
Beautiful answer! – BBischof Jul 30 '10 at 7:16

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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Hi Joel! Nice answer! – François G. Dorais Jul 30 '10 at 18:30

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 all but countably many of them can contain rationals...

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This is now my favorite example of a nonconstructive proof. – Bryan Sep 9 '14 at 20:36
@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. – Henno Brandsma Sep 9 '14 at 20:43

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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There are only countably many endpoints, but a nontrivial perfect set is uncountable. – JDH Jul 30 '10 at 1:45

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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Let $C$ represent the Cantor set. Consider the set $C+\alpha$, where $\alpha=\sum_{n=1}^{\infty}10^{2^n}$.

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And this works because...? – Andrés E. Caicedo Feb 4 '11 at 6:22
See this: math.stackexchange.com/q/381690/462 I do not know of an explicit $\alpha$ that works. – Andrés E. Caicedo May 5 '13 at 19:37

protected by Arctic CharOct 27 '15 at 8:22

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