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.

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