# Property of the Plane

I was reading the blog topological musings, and the category talk reminded me of this neat thing I posted on AoPS awhile ago.

By using the Mayer-Vietoris sequence in cohomology, we determine whether or not $\mathbb{R}^2$ can be written as the union of two open connected sets, U and V, such that $U\cap V$ is disconnected.

Well, it seems that we are concerned with $H^0$ stuff, since that tells us the number of connected components.

The sequence gives $0\to H^0(\mathbb{R}^2)\to H^0(U)\oplus H^0(V)\to H^0(U\cap V)\to 0$, a nice short exact sequence in which we mostly know they are connected, so

$0\to \mathbb{R}\to \mathbb{R}\oplus \mathbb{R}\to H^0(U\cap V)\to 0$, basically the question reduces to: Is it possible for the second to last term to have dimension greater than 1, but the sequence to remain exact?

Well, no, since the exactness tells us the dimension of $H^0(U\cap V)$ is the dimension of $\frac{\mathbb{R}\oplus\mathbb{R}}{\mathbb{R}}\cong \mathbb{R}$.

So we have a nifty result: the plane cannot be broken into two connected parts in which the intersection of these two parts is disconnected.

Challenge for the readers: This seems extremely simple and obvious. Find a proof that does not require advanced techniques such as cohomology.