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Can someone help me?

How can I prove that exists a number $k \in \mathbb R$ that $$A = \{x + k;\ x \in \text{Cantor set} \} \subset\text{ Irrationals}\;?$$

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  • $\begingroup$ Do you want this for the standard middle-thirds Cantor set, or is any Cantor space okay? In the latter case, digits can be played to get the result easily. $\endgroup$ – Karolis Juodelė May 5 '13 at 0:48
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    $\begingroup$ Hint (for one possible approach of many): $k$ should not be of the form $q-x$ for any rational $q$ and any $x$ in the Cantor set. What is the measure of this set? $\endgroup$ – GCD May 5 '13 at 0:49
  • $\begingroup$ I want the standard middle-thirds Cantor set. $\endgroup$ – Pedro Amorim May 5 '13 at 0:55
  • $\begingroup$ See also math.stackexchange.com/q/1064/462 $\endgroup$ – Andrés E. Caicedo May 5 '13 at 19:38
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Let $C$ be the middle-thirds Cantor set. If $x\in\Bbb R$ and $q\in\Bbb Q\cap(x+C)$, then there is a $y\in C$ such that $q=x+y$, and hence $x=q-y\in q-C$. If $\Bbb Q\cap(x+C)\ne\varnothing$ for every $x\in\Bbb R$, then

$$\Bbb R=\bigcup_{q\in\Bbb Q}(q-C)\;.$$

Verify that each of the sets $q-C$ for $q\in\Bbb Q$ is a closed, nowhere dense subset of $\Bbb R$, and then apply the Baire category theorem to get a contradiction.

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  • $\begingroup$ Do we have explicit examples? $\endgroup$ – Andrés E. Caicedo May 5 '13 at 15:03
  • $\begingroup$ @Andres: Of a translation taking $C$ into $\Bbb P$? Good question; I don’t know. $\endgroup$ – Brian M. Scott May 5 '13 at 15:10
  • $\begingroup$ Thanks. I was almost there, but I was using all the real numbers instead of only the rationals therefore, I couldn't use the Baire theorem $\endgroup$ – Pedro Amorim May 5 '13 at 21:41
  • $\begingroup$ @Pedro: You're welcome. $\endgroup$ – Brian M. Scott May 6 '13 at 1:22
  • $\begingroup$ Or you could have used the sub-additivity of measure. See below. $\endgroup$ – Kasun Fernando Sep 27 '14 at 16:04
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If you read Spanish, you may want to look at the book

Wilman Brito, El Teorema de Categoría de Baire y aplicaciones, Consejo de publicaciones, Universidad de los Andes, Mérida, Venezuela, 2011.

I found this book through this MO question. What follows comes from this reference.

Section 1.10 is about the Cantor set and the result you are asking about, that was first noted by Ludwig Scheeffer, in 1884, with the argument in Brian's answer. The reference is

Ludwig Scheeffer. Zur Theorie der stetigen Funktionen einer reellen Veränderlichen, Acta Math. 5 (1), (1884), 49–82. MR1554648.

(The article is behind a paywall.) The proof can be seen in English in The theory of sets of points, by William Henry Young and Grace Chisholm Young, recently (2006) republished as part of the Cambridge Library Collection. (See Theorem 18 in Chapter IV, and $\S67$ in Chapter V.)

More precisely, Scheeffer shows that if $E$ is countable and $N$ is nowhere dense, then there is a dense set $D$ of points $d$ such that $(E+d)\cap N$ is empty. Note that Brian's answer actually shows a bit more (for $E=\mathbb Q$ and $N=C$, the Cantor set), namely, that the set of $x$ such that $C+x$ consists entirely of irrational numbers, is comeager. This is a particular case of a 1954 result by Frederick Bagemihl, see

Frederick Bagemihl. A note on Scheeffer's theorem, Michigan Math. J. 2 (1953--54), (1955), 149–150. MR0065614 (16,455c).

Bagemihl showed that if $F$ is meager and $N$ is countable, then there is a comeager set $G$ such that $(x+N)\cap F=\emptyset$ for all $x\in G$. This is an improvement of Scheeffer's result, replacing the assumption that $N$ is nowhere dense with the weaker requirement that it is meager, and the conclusion that $D$ is dense with the stronger requirement that it is comeager.

In 1981, in

Duane Boes, Richard Darst, and Paul Erdős. Fat, symmetric, irrational Cantor sets, Amer. Math. Monthly 88 (5), (1981), 340--341. MR0611391 (83i:26003),

it is proved that there is a comeager set $G\subseteq[0,1]$ such that $(0,1)\cap C_\alpha$ consists only of irrational numbers, for all $\alpha\in G$. Here, for $0<\alpha\le 1$, $C_\alpha$ is defined as the Cantor set, starting with $[0,1]$ and at stage $n>0$ removing the middle interval of length $\alpha/3^n$ from each remaining interval. (So $C_1=C$ is the usual Cantor set.)


Returning to the question, note that all the results above are established via applications of the Baire category theorem. For explicit examples of reals $x$ such that $x+C$ consists solely of irrational numbers, see this MO question. In particular, any $x$ such that every finite sequence of $0$s, $1$s, and $2$s appears as a substring of the ternary expansion of $x$ has this property.

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  • $\begingroup$ Are you sure that's the correct reference for Scheefer's result? Bagemihl's A note on Scheefer's theorem refers to Scheefer's Zur Theorie der stetigen Funktionen einer reellen Veränderlichen, Acta Math. 5 (1884). 279-296. Maybe he published two versions of that theorem? By the way, how could he be using Baire's (1899) category theorem in 1884? $\endgroup$ – bof Feb 22 '14 at 21:35
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    $\begingroup$ @bof I've corrected the link, thanks for noticing this. Scheeffer proves directly a version of Baire's theorem in his note. $\endgroup$ – Andrés E. Caicedo Feb 26 '14 at 17:56
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We had a similar problem in one of our Real Analysis assignments. It was a generalization of the given problem.

Let $E \subset \mathbb{R}$ be a set with Lebesgue outer measure $0$. Then there exists $h \in \mathbb{R}$ such that $E+h \subseteq \mathbb{Q}^c$.

We can prove this using the identity Brian derived.

If $\Bbb Q\cap(E+h)\ne\varnothing$ for every $h\in\Bbb R$, then

$$\Bbb R=\bigcup_{h\in\Bbb Q}(h-E)\;.$$

$$m(\Bbb R) \leq \sum_{h \in \Bbb Q}{} m^*(h-E)=0$$

which gives us a contradiction. So, $\Bbb Q\cap(E+h)=\varnothing$ for some $h \in \Bbb R$.

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