# Continuity and Uniform Continuity of inverse functions

Let $$f : X → Y$$ be a given function, and suppose that $$f^{-1}(C)$$ is an open subset of $$X$$ whenever C is an open subset of $$Y$$ .

(a) Prove that f is continuous on $$X$$.

(b) Prove that $$f^{-1}(B)$$ is a closed subset of $$X$$ whenever B is a closed subset of $$Y$$

(c) If $$Y = R$$, and $$f$$ is continuous, and a $$\epsilon$$ $$R$$, what kind of set is A = {$$x$$ $$\epsilon$$ X : $$f(x)$$ <= a}? Justify your answer

I already solved part a, and my attempt for part b is:

$$f^{−1}(B)$$ = $$(f^{−1}(B^c))^c$$ ⋯ (1) ($$E^c$$ denoting the complement of $$E$$).

So if B is closed, then $$B^c$$ is open, $$f^{−1}{(B^c)}$$ is open and its complement is closed. This means $$f^{−1}(B)$$ is closed by (1).

But I'm finding trouble in solving part c. Any help please?

• Hint: When $B$ is closed, then $Y\setminus B$ is open. – Severin Schraven Mar 28 at 9:12

$$f^{-1}(B)=(f^{-1}(B^{c}))^{c}$$ $$\cdots\, \,\,\,$$ (1) ($$E^{c}$$ denoting the complement of $$E$$). So if $$B$$ is closed, then $$B^{c}$$ is open, $$f^{-1}(B^{c})$$ is open and its complement is closed. This means $$f^{-1}(B)$$ is closed by (1).