Can a number be normal in an arbitrary set of bases? Suppose we have a subset $S$ of the natural numbers greater than or equal to $2$. Is there a real number $x$ such that $x$ is $k$-normal for all $k$ in $S$, and not $k$-normal for all $k$ in the complement of $S$, no matter what $S$ we choose?
 A: If $r$ and $s$ are powers of the same integer, then $x$ is normal in base $r$ if and only if it is normal in base $s$. Schmidt proved that if $r$ and $s$ are not powers of the same integer then there is an uncountable infinity of numbers that are normal to base $r$ but not even simply normal to base $s$. 
The reference is W Schmidt, On normal numbers, Pacific J. Math. 10 (1960) 661–672, MR0117212 (22 #7994). 
A lot of work has been done since then to generalize this result. For example, Brown, Moran, and Pearce proved in 1985 that, given a base $s$, every real number can be expressed as a sum of four numbers, each of which is non-normal to base $s$, but normal to every base $r$ that isn't a power of the same integer as $s$. 
A more recent paper is Becher and Slaman, On the normality of numbers to different bases, J. Lond. Math. Soc. (2) 90 (2014), no. 2, 472–494. I haven't seen this paper, nor a review, so I can't report on its contents. 
A: If $x$ is normal in base $2$ it will be normal in base $4$.  So if $2$ is in $S$ but $4$ is not, there will be no such $x$.  Many similar examples are possible.
