Let f : V mapped to W be a continuous function between two topological spaces V...

Let f : V mapped to W be a continuous function between two topological spaces V and W, so that (by definition) the preimage under f of every open set in W is open in V : Y is open in W implies f^−1(Y ) = {x in V | f(x) in Y } is open in V. Prove that the preimage under f of every closed set in W is closed in V . Feel free to take V = W = R^n to simplify things. Hint: show that the “preimage of” operation plays nice with set-complements, and then use the fact that every closed set is the complement of some open set. Note that R^n is both open and closed as a subset of itself.

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Colby Messinger answered 1 year ago

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