# 高微的問題~請高手幫忙解

Let U and V be open sets in Rn and let f be a one-to-one mapping from U onto V (so that there is an inverse mapping f-1 : V → U ). Suppose that f and f-1 are both continuous. Show that for any set S whose closure is contained in U we have f(boundary of S)= boundary of (f(S)).

### 2 個解答

• 最佳解答

Note that x in the boundary of S <=> f(x) in the f(boundary of S)Now given y in the f(boundary of S), then there is x in the boundary of S such that y = f(x)case1. x in Swe can choose {x_n} in U-S s.t. x_n -> x. By continuity we have f(x_n) -> f(x) and f(x_n) in V-f(S), hence we have1. B_r(f(x))∩(V-f(S)) ≠ empty set for all r>02. B_r(f(x))∩f(S) at least contains f(x)So f(x) = y in the boundary of f(S)Case2 for x in U-S is similar.Thus say f(boundary of S) is contained in boundary of (f(S)).Since those hypotheses are symmetric, hence the thing "boundary of (f(S)) is contained in f(boundary of S)" holds automatically by the preceding argument.Hence  f(boundary of S) = boundary of (f(S))