TITLE: Is Lebesgue's "universal covering" problem still open?
QUESTION [46 upvotes]: The following problem has been attributed to Lebesgue. Let "set" denote any subset of the Euclidean plane. What is the greatest lower bound of the diameter of any set which contains a subset congruent to every set of diameter 1? There are a number of interesting geometric problems of this type. Is it possible that some of them may be difficult to solve because the solution is a real irrational number which (when expressed in decimal form) is not even recursive-and so cannot be approximated in the usual way?

REPLY [5 votes]: Philip Gibbs wrote:

the area of the universal cover for an angle $\sigma=0.4$ degrees can be computed to be $0.8441177$.

I believe Philip later noticed an error in this area computation.  He also later noticed that there's a constraint that $\sigma$ needs to obey, which is not obeyed by this choice: the best choice of $\sigma$ is about 
$$ \sigma = 1.294389444703601012^\circ $$
and the area of the universal covering this gives is about
$$ 0.844115297128419059\dots .$$
We wrote a paper on this together with Karine Bagdasaryan, and Greg Egan did some high-precision calculations that give the above numbers.   Our paper explains the details:

John C. Baez, Karine Bagdasaryan and Philip Gibbs, The Lebesgue universal covering problem.

Abstract: In 1914 Lebesgue defined a "universal covering" to be a convex subset of the plane that contains an isometric copy of any subset of diameter 1. His challenge of finding a universal covering with the least possible area has been addressed by various mathematicians: Pal, Sprague and Hansen have each created a smaller universal covering by removing regions from those known before. However, Hansen's last reduction was microsopic: he claimed to remove an area of $6 \cdot 10^{−18}$, but we show that he actually removed an area of just $8 \cdot 10^{-21}$. In the following, with the help of Greg Egan, we find a new, smaller universal covering with area less than $0.8441153$. This reduces the area of the previous best universal covering by a whopping $2.2 \cdot 10^{−5}$.