# Continuity of a function on a metric space and its consequences

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 $$Y = \mathbb{R}$$, and $$f$$ is continuous, and $$a \in \mathbb{R}$$, what kind of set is $$A = \{x \in X : f(x) \leq a\}$$? 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?

• Observe that $A=f^{-1}((-\infty,a])$ and that $(-\infty,a]$ is closed in $R$. Then you will need the converse of (a). – user647486 Apr 6 at 11:41
• Your part B is good. Maybe state explicitly that $f^{-1}(B^c)^c = f^{-1}(B)$, but yes. – Kaj Hansen Apr 6 at 12:18