subset under continuous functions

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Sep 17, 2016 at 14:40	comment	added	Olivier Moschetta		(2)=>(3) An open set is not necessarily the union of open balls. Rather, a set U is open if for every x in U there is an open ball B around x so that $B\subset U$.
Sep 17, 2016 at 14:33	comment	added	Olivier Moschetta		(3)=>(1) It would be far easier to show that f is continuous by showing that every open set U in $\mathbb{R}^n$ is such that the inverse image of U is open (the two things are known to be equivalent). This would work as in part 1. Let U be open in R^n. Then S=R^n\U is closed. According to (3) the inverse image of S is also closed, hence the inverse image of U is open and f is continuous.
Sep 17, 2016 at 14:32	review	Low quality posts
Sep 17, 2016 at 15:02
Sep 17, 2016 at 14:13	history	answered	Maria	CC BY-SA 3.0