MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm interested in knowing classes of topological spaces $X$ which admit a basis of open sets $\{U_i\}_{i\in I}$ such that $U_i\cap U_j$ is connected for all $i,j\in I$. Do manifolds have this property? Riemannian ones maybe? If not, what if we relax the condition by saying that $U_i\cap U_j$ has a finite number of connected components for $i,j\in I$?

share|cite|improve this question
up vote 4 down vote accepted

I reckon you consider the empty subspace to be connected (since for a Hausdorff space, at least one such intersection must be empty). In that case, for a Riemannian manifold, you can get not just connectedness, but contractibility for all (nonempty) finite intersections of basis elements, by taking a neighborhood basis to consist of geodesically convex neighborhoods. (An open set $U$ is geodesically convex if any two points in $U$ are connected by a unique geodesic in $U$, one whose length is the distance between the points). For it is immediate that any (nonempty) intersection of geodesically convex neighborhoods is also geodesically convex, hence contractible. Details are spelled out in this nLab article.

Of course, a (paracompact Hausdorff) smooth manifold admits a Riemannian metric, so you have the result for smooth manifolds.

I believe CW complexes have the same property (a proof of the existence of a good open cover is sketched in the same nLab article), but I haven't checked that carefully.

share|cite|improve this answer
@Todd, that's great, thanks! – Fernando Muro Mar 8 '13 at 15:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.