Given a manifold which Hausdorff, second-countable. Does there exist a countable open cover $\{U_i\}_{i=1}^\infty$ such that the closure $\bar{U_i}$ of each $U_i$ is compact?
Does every manifold can be "covered" by compact sets?
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$\begingroup$
general-topology
manifolds
1 Answers
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It can be done with $\Bbb R^n$, and you can cover your manifold with countably many open sets homeomorphic to $\Bbb R^n$. What can you conclude from this?
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0It is true that only works on the topological space? – 2017-01-10