# Decompose a unit ball into 3 convex disjoint parts parts with common boundary

I guess that for arbitrary $$n\in\mathbb{N}$$ it is impossible to decompose the open unit ball $$B(0,1)$$ of $$R^n$$ in four disjoint sets $$A$$,$$B$$,$$C$$,$$D$$, such that $$A$$,$$B$$,$$C$$ be convex open subsets and the set $$D$$ the common boundary for the three, i.e $$D=\partial A=\partial B=\partial C$$.

I think it is true, but I can not catch the key idea to prove this.

If we do not care about convexity, it is known that it is possible to decompose an open ball of $$R^2$$ in three disjoint connected parts with common boundary (i.e.Wada Lakes), similar is possible in higher dimensions.

If my guess is not true please provide me with reference or example, If someone see how to prove the claim please provide me with proof or a clue at least.

• For n=3, why you think it's impossible? You may think of dividing 🍊 into 3 even parts. – GNUSupporter 8964民主女神 地下教會 Feb 15 at 10:46
• If you use the square/cube instead of the balls (which shouldn't change very much) you can make $3$-d Wada lakes by taking the $2$-d Wada lake and crossing it with the interval. Edit: I just noted that you have the additional constraint of convexity in higher dimensions? No Wada lake like constructions can give convex sets. – quarague Feb 15 at 10:52

The boundary of the unit ball must be part of the boundary of the pieces. Hence, if all pieces have the same boundary and are convex, then they must all contain the interior of the unit ball.

If you discard the boundary of the unit ball itself, there is actually a solution. You can partition the unit ball in the points with positive, zero and negative first coordinate. All three parts will have the points with zero first coordinate as boundary.

• Thanks, I changed the question little bit. I also do not see your point completely. $\{(x_i)\mid x_1 =0\}$ is not open set. Please be more specific if any. – Guard Feb 16 at 4:51