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Consider a real algebraic set $Z(f) = \{x \in \Bbb{R}^n\,|\, f(x) = 0\} \subset \Bbb{R}^n$ (not necessarily irreducible). I'm thinking about wether the (Euclidean) closure of a connected component of the complement $\Bbb{R}^n\setminus Z(f)$ has vanishing $n$:th homology or not. It feels like this should not be a hard question, but I keep finding myself thinking about an intuitive picture in $n=2$ or $n=3$ (where I think it is vanishing), unable to find a more strict approach.

Raclette
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    I think this will follow from Alexander duality. –  Dec 22 '14 at 22:02
  • That is, one does a one point compactification, embedding $\Bbb{R}^n$ in the $n$-sphere in $\Bbb{R}^{n+1}$. Then, for a connected components $C$ of the complement $\Bbb{R}^n\setminus Z(f)$, Alexander duality says that $H_n(C, \Bbb{Z}) \simeq H^{-1}(C^c, \Bbb{Z})$, but the latter is trivial by definition? – Raclette Dec 22 '14 at 22:21
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    The trick is that you need $C$ to satisfy the requirements of the theorem (and this inclusion into $S^n$ to induce an injection on $n$th homology, but this is easy). I was just being careful not to say $C$ is locally contractible, since even though it seems likely, I'm not 100% sure. –  Dec 22 '14 at 23:13
  • (where $C$ is the closure of a component etc). –  Dec 22 '14 at 23:13
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    @MikeMiller: Real semialgebraic sets are locally contractible since each is homeomorphic to a simplicial complex (Lojasiewicz). – Moishe Kohan Dec 22 '14 at 23:33

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