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Let $C$ be the standard Cantor middle-third set. As a consequence of the Baire category theorem, there are numbers $r$ such that $C+r$ consists solely of irrational numbers, see here.

What would be an explicit example of a number $r$ with this property?

Short of an explicit example, are there any references addressing this question?

A natural approach would be to see that all irrationals in $C$ are transcendental, so it would suffice to take $r=\sqrt2$. But this is open, see here.

Many thanks for the answers. (It would be interesting to know whether $\sqrt2$ works.)

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Alternately, you can use "measure zero". More generally, given any any measure zero set $Z$ and any countable set $Q$, there is $r$ so that $Z+r$ misses $Q$. – Gerald Edgar Feb 22 '14 at 21:05
@GeraldEdgar Yes. Conversely, Bjørn's measure-theoretic answer can be restated in terms of category and (Cohen) 1-generic reals. – Andrés E. Caicedo Feb 23 '14 at 23:00
up vote 27 down vote accepted

One way to obtain explicit examples, which combines the ideas of (weak forms of) randomness and base 3 expansions, is to use the fact that normality in a given base is preserved under rational addition, which was proved by D. D. Wall in his 1949 Berkeley PhD Dissertation. (I'm relying on D. Doty, J. H. Lutz, and S.Nandakumar [Finite-state dimension and real arithmetic, Information and Computation 205(11):1640-1651, 2007] for this reference.) Here a number $r$ is normal in base $b$ if for any finite nonempty string $\sigma$ drawn from the alphabet $\{0,...,b-1\}$, the limiting frequency of the appearances of $\sigma$ as a substring of the base $b$ expansion of $r$ is $b^{-|\sigma|}$. Since elements of $C$ are not normal in base $3$, any number $r$ that is normal in base $3$ has the desired property. Examples of such numbers can be found at, for instance.

In fact, normality is overkill. Let $r$ be disjunctive in base $3$, i.e., every finite ternary string appears as a substring of the ternary expansion of $r$ (which is both a comeager and a conull property). We need to show is that if $q$ is a positive rational then $r+q \notin C$. (Here addition is mod $1$.) If the ternary expansion of $q$ has infinitely many $1$'s then the fact that the ternary expansion of $r$ contains $0^n$ for all $n$ means that $q+r \notin C$. Otherwise, the fact that this expansion contains $0^m10^n$ for all $m,n$ does the trick.

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Great answer, also welcome Denis! – Bjørn Kjos-Hanssen Feb 22 '14 at 22:12
very nice answer... – Anthony Quas Feb 22 '14 at 22:24
The normals former a meager set, can we make a comeager answer? – Bjørn Kjos-Hanssen Feb 22 '14 at 22:53
Great, thank you! Much nicer and general than I had hoped for. You may want to add your latest comment to the answer, to increase its visibility, since it is a stronger result. – Andrés E. Caicedo Feb 23 '14 at 23:08
Thanks for the suggestion. I've incorporated my comment (which addressed Bjørn's question) into my answer. – Denis Hirschfeldt Feb 23 '14 at 23:23

I don't think this should be hard. Think base 3. Something is irrational if and only if it does not have a ultimately periodic base 3 expansion. Something's in $C$ if and only if it has no 1's in its base 3 expansion.

Here's an explicit $t$ that I think does the job: $$ t=\sum_{n=0}^\infty \sum_{j=2^{2n}}^{2^{2n+1}-1}3^{-j}. $$ In base 3 it's $$ t=0.10011110000000011111111111111110000000000000000000000000000000011\ldots $$ Since $t$ does not have an ultimately periodic base 3 expansion, it's irrational.

Let's call the coordinate ranges where $t$ has 0's "0-blocks" (i.e. from $2^k$ to $2^{k+1}-1$ for $k$ odd) and the coordinate ranges where $t$ has 1's "1-blocks". Now if you form $x+t$ for any $x\in C$, $x+t$ has arbitrarily long blocks with no 1's in the base 3 expansion (corresponding to the blocks of 0's in $t$) - if you're unlucky, the last digit of $x+t$ in a 0-block might be a 1, but there are no others.

Hence if assume for a contradiction that $x+t$ is rational (and so has ultimately periodic base 3 expansion), the repeating block must contain only 0's and 2's.

Now consider the expansion of $x+t$ on the 1-blocks. A calculation shows that the only way to avoid having 1's in the interior of the blocks is for $x$ to have 2's in those blocks and $x+t$ to consist of 0's in the 1-blocks.

Now since $x+t$ has arbitrarily long blocks of 0's, we deduce that the periodic block must consist of all 0's -- that is $x+t$ is a triadic rational.

Since $1-t$ has ternary expansion $$ 1-t=0.122111122222222111111111111111122222222\ldots $$ it's easy to check that there's no $x\in C$ with the property that $x+t$ is a triadic rational.

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Thank you. This is the kind of answer I was expecting, and it is nice that the calculations end up not being as messy as I feared. I myself lost patience trying to keep track of what you call the 1-blocks of $x+t$ (for the $t$ I used myself, which I see is different from yours) when I first thought about this. – Andrés E. Caicedo Feb 23 '14 at 23:05
Anthony, I have some doubts about your explicit example of $t$. Let's write it like that: $$ t = 0.1(00)^{2^0}(11)^{2^1}(00)^{2^2}(11)^{2^3}\ldots (00)^{2^{2n}}(11)^{2^{2n+1}}\ldots $$ Then take $x \in C$: $$ x = 0.0(20)^{2^0}(02)^{2^1}(20)^{2^2}(02)^{2^3}\ldots (20)^{2^{2n}}(02)^{2^{2n+1}}\ldots $$ As a result we get $$ x + t = 0.1(20)^\infty \in \mathbb Q, $$ because within each block of length $2^n$ we can calculate the sum separately. (continued) – Kajetan Janiak Jan 19 at 18:44
More precisely, for each subsequent pair of digits we use the following two formulas: $$ 0.(0)^n00 + 0.(0)^n20 = 0.(0)^n20 $$ and $$ 0.(0)^n11 + 0.(0)^n02 = 0.(0)^n20. $$ – Kajetan Janiak Jan 19 at 18:44

If $q\in x+C$, i.e., $q=x+c$ for some $c\in C$, then $x=q-c$, i.e., $x\in q-C$. Now $q-C$ is a closed set of measure zero and moreover it is a $\Pi^0_1$ class if $q\in\mathbb Q$. Thus each weakly 1-random real avoids $q-C$. In particular Chaitin's $\Omega$ does not belong to $q-C$.

There are infinitely many versions of Chaitin's number, so to get an explicit example we must choose one, coming from a canonical construction of a prefix-free universal Turing machine.

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Is it clear that there is a computable example? The Baire category argument just says that there are comeager-many examples. – Noah Schweber Feb 22 '14 at 20:25
I guess the set $q-C$ is fairly simple, so a resource-bounded random real (like a polynomial time random one) should suffice. Perhaps one of those is even sort of explicitly definable, like a "polynomial time $\Omega$"... – Bjørn Kjos-Hanssen Feb 22 '14 at 20:40
Hi Bjørn. Thanks, nice idea. – Andrés E. Caicedo Feb 23 '14 at 23:01

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