MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A quick reminder of the definition of Schmidt's game:

Let ${X}$ be a metric space and ${S\subset X}$ be a subset. Let ${0<\alpha,\beta<1}$ be constants. Bob chooses any open ball ${B_0\subset X}$ with radius ${\rho_0}$. Then Alice chooses a ball ${B_1\subset B_0}$ with radius ${\rho_1=\alpha\rho_0}$. Then Bob chooses a ball ${B_2\subset B_1}$ with radius ${\rho_2=\beta\rho_1}$, then Alice chooses a ball ${B_3\subset B_2}$ with radius ${\rho_3=\alpha\rho_2}$ and so on. Let ${x}$ be the (single) point in the intersection of all balls ${B_n}$. If ${x\in S}$ then Alice wins the game. Otherwise Bob wins. If Alice can force a victory, then the set ${S}$ is called ${(\alpha,\beta)}$-winning. $S$ called $\alpha$-winning if it's $(\alpha,\beta)$-winning for all $0 < \beta < 1$. One can also define $windim(S)$ to be the least upper bound on all $\alpha$ such that $S$ is $\alpha$-winning.

It's easy to see that $S$ need to be dense to be $(\alpha,\beta)$-winning. It's suprising though, that some sets of lebesgue measure $0$ are $\alpha$-winning (badly approximated numbers for $\alpha < \frac{1}{2}$).

Does some criterion exist for the inverse claim: sets of full measure (in a sense that $\mu(S^c)=0$) that are not $\alpha$-winning for some $\alpha$'s? It seems like the complement of the ternary Cantor set might be an example of that, but I couldn't find a good reasoning.

share|cite|improve this question
up vote 3 down vote accepted

I'm not sure if you're more interested in general conditions for this or an example with a simple proof. I can give a simple example and explanation. Let $S$ be the set of numbers normal in base $b$. $S$ is a set of full measure (follows by SLLN, Birkhoff ergodic theorem, etc.) W. Schmidt proved that $S^c$ is $1/2$-winning. So $S^c$ is $\alpha$-winning for all $\alpha \in (0,1/2]$. But the property of being $\alpha$-winning is preserved under countable intersections. So if $S$ were $\alpha$-winning, then $S \cap S^c=\emptyset$ would also be $\alpha$-winning. Does that answer your question? There are many other examples along these lines in some of the more recent literature on Schmidt games that I can give if you're interested.

share|cite|improve this answer
It seems like a good start, thanks. I would still appreciate some references to literature on those topics. – user35222 Jul 22 '13 at 16:33
I can give a partial list, but I want to state upfront that I'm not an expert in the area and this list will be highly skewed in favor of my own papers and results relating to normal numbers. I will say that Schmidt's result is a case of something like if you have a transformation $T$ on some space $X$, then the set of all numbers $x \in X$ where the forward orbits of $x$ under $T$ is winning. – Bill Mance Jul 23 '13 at 1:31
C. McMullen… Many of the papers of David Färm Paper with Jimmy Tseng The following paper of Moshchevitin captures this well N. G. Moshchevitin. On sublacunary sequences and winning sets (english). Math. Notes, (3–4):592–596, 2005. An application of the previous papers to Cantor series And some of the papers of Jimmy Tseng are on this topic – Bill Mance Jul 23 '13 at 1:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.