0

I am trying to construct an example of a perfect set that contains no rational.

I see in other places a construction that involves an enueration of rationals, but I thought that construction was a bit more involved than what I have (even though they bascially amount to the same idea).

My idea is to intersect the Cantor set with $\mathbf{R}\setminus\mathbf{Q}$. We know that there is a bijection from the Cantor set to the sequences of 0 and 1 by dividing every digit in the ternary expansion by 2 (or by following the path of the point in the Cantor set as being in the left or the right). Every time 2 points share the first $n$-digits in the corresponding sequences they are within $3^{-n}$ distance close, so given a point in the Cantor set and $\epsilon<3^{-n}$, by picking an irrational point in the Cantor set whose corresponding sequences share the first $n$ entries, we are done.

The only tricky part is how to actually choose such an irrational point, but that can be easily arranged by choosing any non-periodic sequence.

Is it accurate (minus the general lack of detail, of course)?

Thank you in advance!

user134070
  • 377
  • 2
    I don't understand your construction. Let $C$ be the Cantor set and $q$ be a rational in $C$. If you are are arguing that you can find irrationals in $C$ arbitrarily close to $q$, then doesn't that mean $q$ is a limit point of $C \setminus \mathbb Q$? In which case $C \setminus \mathbb Q$ is not closed in $\mathbb R$, hence not perfect. –  Feb 15 '21 at 23:03
  • @Bungo You are right, thanks, I did not see that. – user134070 Feb 15 '21 at 23:11

3 Answers3

5

There is a nice non-constructive proof that such Cantor sets exist in large numbers:

The Cantor set is up to homeomorphism the unique compact zero-dimensional metric space without isolated points.

From this it quite quickly follows that $C \times C \simeq C$, which can also easily be seen from the fact that $C \simeq \{0,1\}^{\Bbb N}$, e.g.

So a Cantor set $C$ is a disjoint union of uncountably many homeomorphic copies of $C$ (namely the $\{c\} \times C, c \in C$ in the square) and so many of them do not contain a rational, because the rationals are only countable.

Henno Brandsma
  • 250,824
2

Let $C$ be the Cantor set. Consider the collection of translates of $C$, i.e. the collection $\{C + x : x \in \mathbb R\}$, where $C+x = \{c + x : c \in C\}$. Clearly every $C+x$ is a perfect set (the topological properties aren't affected by translation), so it suffices to find one that contains no rationals.

Fix $q \in \mathbb Q$. Let $X_q = \{x \in \mathbb R: q \in C + x\}$, i.e. $X_q$ is the set of offsets $x$ such that the translate $C + x$ contains $q$.

Observe that a translate $C + x$ contains $q$ if and only if $x \in q - C$, so $X_q = q - C$. In particular, the Lebesgue measure of $X_q$ equals the measure of $q - C$, which is the same as the measure of $C$, which is zero.

If we define $X = \bigcup_{q \in \mathbb Q}X_q$, then $X$ also has measure zero, as it is a countable union of sets of measure zero. But $X$ is the set of offsets $x$ such that $C+x$ contains at least one rational. Since $X$ has measure zero, it follows that almost every choice of $x \in \mathbb R$ yields a translate $C+x$ that contains no rationals.

  • Btw, the above assumes that $C$ is the middle-thirds Cantor set, so $C$ and $X$ will have measure zero. The argument won't work for fat Cantor sets (those with strictly positive measure), because then all we can conclude from $X = \bigcup_{q \in \mathbb Q}X_q$ is that $m(X) \leq \sum_{q \in \mathbb Q}m(X_q) = \infty$, which tells us nothing; $X$ could conceivably be all of $\mathbb R$. –  Feb 16 '21 at 00:27
1

If you know that the irrationals are precisely the real numbers represented by infinite continued fractions, and the map

$$h:\Bbb Z\times{(\Bbb Z^+)}^{\Bbb Z^+}\to\Bbb R\setminus\Bbb Q:\langle a_n:n\ge 0\rangle\mapsto[a_0;a_1,a_2,\ldots]$$

is a homeomorphism, and that the Cantor set is homeomorphic to any product of infinitely many finite spaces each having at least two points, you can exhibit a great variety of explicit Cantor sets in the irrationals. For each $n\in\Bbb Z^+$ let $F_n\subseteq\Bbb Z^+$ be finite with at least two elements; then

$$\{[0;a_1,a_2,\ldots]:\forall n\in\Bbb Z^+(a_n\in F_n)\}$$

is a Cantor set in the irrationals.

The first answer to this question introduces the basics of continued fractions and proves in detail that $h$ is a homeomorphism; the second answer gives an easier proof that the spaces are homeomorphic.

Brian M. Scott
  • 631,399