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Mar 12, 2017 at 19:57 comment added M. Rahmat You are right, this is why I accepted you partial answer. How about the general case, where $A$ is just closed?
Mar 12, 2017 at 19:11 comment added Robert Israel @M.Rahmat Unless I misunderstand your notation, that example is not compact.
Mar 12, 2017 at 18:31 comment added M. Rahmat Correct, but this is not necessary: the union of the spheres $S(n,1/n)$ may have a neighborhood $V$, as explained but the your $\varepsilon$ does not exist. What is the necessary condition?
Mar 12, 2017 at 8:40 history answered Robert Israel CC BY-SA 3.0