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Dividing open domains in $\mathbb R^2$ in parts of equal area From this question, what can go wrong if $U,V$ are not connected?

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In the accepted answer to the linked question $f(x)$ was shown to be a monotone function which was strictly increasing on the open interval of those $x$ such that $0<f(x)<|A|$. Remove connectedness and $f$ may not be strictly increasing on a given open subset of such $x$. In particular, there can be infinitely many values with $f(x)=|A|/2$. A simple way to see this is to imagine two open circular discs with disjoint closures as being your set $A$. Any line which contains exactly one disc in each of the half-planes it determines will cut the area of the combination "in half", but you can shift any one of those lines by a small amount to get a parallel line with the same property.

However, the function remains continuous and the Intermediate Value Theorem still guarantees that a line with prescribed direction $d$ exists which cuts $A$ in half. The difference is that the line is no longer guaranteed to be uniquely specified by its orientation. However, you can still pick a specific such line for any given $d$ by taking the midpoint of those $x$ with $f(x)=|A|/2$ (note that $f$ monotone guarantees this is a closed interval, possibly a single point), which allows you to use much the same argument to simultaneously bisect both $A$ and $B$.

zibadawa timmy
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