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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    $\begingroup$ 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 $Q$. $\endgroup$ Feb 22, 2014 at 21:05
  • $\begingroup$ @GeraldEdgar Yes. Conversely, Bjørn's measure-theoretic answer can be restated in terms of category and (Cohen) 1-generic reals. $\endgroup$ Feb 23, 2014 at 23:00

4 Answers 4


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 http://en.wikipedia.org/wiki/Normal_number, 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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    $\begingroup$ Great answer, also welcome Denis! $\endgroup$ Feb 22, 2014 at 22:12
  • $\begingroup$ very nice answer... $\endgroup$ Feb 22, 2014 at 22:24
  • $\begingroup$ The normals former a meager set, can we make a comeager answer? $\endgroup$ Feb 22, 2014 at 22:53
  • $\begingroup$ 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. $\endgroup$ Feb 23, 2014 at 23:08
  • $\begingroup$ Thanks for the suggestion. I've incorporated my comment (which addressed Bjørn's question) into my answer. $\endgroup$ Feb 23, 2014 at 23:23

EDIT: This is modified from the original version which was incorrect, as pointed out by @KajetanJaniak.

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: $$ t=\sum_{n=0}^\infty \sum_{j=2^{2n}}^{2^{2n+1}-1}3^{-j}\mathbf 1_{j\text{ odd}}. $$ In base 3 it's $$ t=0.1|00|0101|00000000|0101010101010101|00000000000000000000000000000000|01\ldots $$ (the $|$ are not part of the number, but just make it easier to read). 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 01's "01-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 01-blocks. A calculation shows that the only way to avoid having 1's in the sum in the interior of the 01-blocks is for $x$ to consist of a concatenation of 0022's and 2022's in those blocks and $x+t$ to consist of 0200's and 2200's there. Since the blocks occurring in $x+t$ have one less two than the corresponding blocks in $x$, we see that $x+t$ "looks different" on the 0-blocks than it does on the 01-blocks. Hence $x+t$ cannot have a periodic base 3 expansion, and must be irrational.

  • $\begingroup$ 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. $\endgroup$ Feb 23, 2014 at 23:05
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    $\begingroup$ 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) $\endgroup$ Jan 19, 2016 at 18:44
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    $\begingroup$ More precisely, for each subsequent pair of digits we use the following two formulas: $$ 0.(0)^n00 + 0.(0)^n20 = 0.(0)^n20 $$ and $$ 0.(0)^n11 + 0.(0)^n02 = 0.(0)^n20. $$ $\endgroup$ Jan 19, 2016 at 18:44
  • $\begingroup$ @KajetanJaniak: Thanks for the comment. It seems I was too hasty in my last but 2 paragraph. I feel hopeful there is some way to modify the example and make it work. $\endgroup$ Oct 16, 2017 at 17:47
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    $\begingroup$ Anthony: Add a mention to that effect at the beginning of the post? $\endgroup$
    – Did
    Oct 16, 2017 at 21:12

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.

  • $\begingroup$ Is it clear that there is a computable example? The Baire category argument just says that there are comeager-many examples. $\endgroup$ Feb 22, 2014 at 20:25
  • $\begingroup$ 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 $\Omega$"... $\endgroup$ Feb 22, 2014 at 20:40

Here is a simple example. Let $(f(n),g(n))$ enumerate all pairs of positive naturals over natural $n$. Let $x$ be the value of the ternary expansion starting with .1 where the $n$-th block of $0$s (between $1$s) has minimum length that is greater than the previous block length and is such that the $1$ immediately after that is at a position of the form $f(n)·k+g(n)$ for natural $k$. This is always possible because $f(n)$ is positive. Then $C−x$ does not contain any rational $r$.

Proof: Take any rational $r$. Then $r$ has ternary expansion that eventually regularly repeats some block $B$ of size $b$. If $B$ has a $0$, then it will repeat at positions of the form $f(n)·k+g(n)$ for some natural $n$ such that $f(n) = b$, and so at a position large enough adding $x$ to $r$ will add a $1$ to some repeat of that $0$ such that the next $1$ is beyond the next block, and so the result will have a $1$, and hence $x+r$ is not in $C$. If $B$ has a $1$, then at a position large enough adding $x$ to $r$ will add only $0$s to two adjacent repeats of $B$, since $x$ has increasingly longer blocks of $0$s, and the next $1$ after that will not affect the first repeat block, and hence $x+r$ is not in $C$. Since we can assume that $B$ has either a $0$ or a $1$, we are done.


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