# general topology – Proving the equivalence of some different definitions of continuity

I wanted to verify the correctness of my solution to a basic point-set topology exercise. Also, could the ($$2 leftrightarrow 3$$) component of the proof be made more direct and enlightening?

Proposition. Let $$f : X rightarrow Y$$ be a function from space $$X$$ to space $$Y$$. The following conditions are equivalent:

1. given any open $$U subseteq Y$$, $$f^{-1}(U)$$ is open,
2. given any closed $$F subseteq Y$$, $$f^{-1}(F)$$ is closed, and
3. given any subset $$A subseteq X$$, $$f(mathsf{Clos}(A)) subseteq mathsf{Clos}(f(A))$$.

Proof.

Let $$f : X rightarrow Y$$ be a function from space $$X$$ to space $$Y$$.

(1 $$leftrightarrow$$ 2) $$f^{-1}(U)$$ is open for every open $$U subseteq Y$$ $$leftrightarrow$$ $$f^{-1}(U)^c$$ is closed for every open $$U subseteq Y$$ $$leftrightarrow$$ $$f^{-1}(U^c)$$ is closed for every open $$U subseteq Y$$ $$leftrightarrow$$ $$f^{-1}(U^c)$$ is closed for every $$U^c subseteq Y$$ where $$U$$ is an open set of $$Y$$ $$leftrightarrow$$ $$f^{-1}(F)$$ is closed for every closed $$F subseteq Y$$ since complements of opens are exactly the closed sets.

(2 $$rightarrow$$ 3) Suppose $$f^{-1}(F)$$ is closed for every closed subset $$F subseteq Y$$. We need to show that $$f(mathsf{Clos}(A)) subseteq mathsf{Clos}(f(A))$$ for every subset $$A subseteq X$$. Let $$A subseteq X$$ be an arbitrary subset of $$X$$ and let $$x in mathsf{Clos}(A)$$. It remains to be shown that $$f(x) in mathsf{Clos}(f(A))$$ i.e. $$x in f^{-1}(mathsf{Clos}(f(A)))$$. By assumption, we have that $$f^{-1}(mathsf{Clos}(f(A)))$$ is closed so it suffices to show that $$x in mathsf{Clos}(f^{-1}(mathsf{Clos}(f(A))))$$. We are done if we can show that every open neighbourhood of $$x$$ intersects $$f^{-1}(mathsf{Clos}(f(A)))$$. Let $$U$$ be an arbitrary open neighbourhood of $$x$$. Because $$x in mathsf{Clos}(A)$$, $$U$$ must contain some $$a in A$$. Clearly, $$f(a) in f(A) subseteq mathsf{Clos}(f(A))$$ which is to say that $$a in f^{-1}(mathsf{Clos}(A))$$.

(3 $$rightarrow$$ 2) Suppose $$f(mathsf{Clos}(A)) subseteq mathsf{Clos}(f(A))$$ for every subset $$A subseteq X$$. Let $$F subseteq Y$$ be a closed subset. We need to show that $$f^{-1}(F)$$ is closed i.e. $$mathsf{Clos}(f^{-1}(F)) subseteq f^{-1}(F)$$ which is equal to $$f^{-1}(mathsf{Clos}(F))$$ because $$F$$ is closed. Let $$x in mathsf{Clos}(f^{-1}(F))$$. We have by assumption that
$$begin{equation*} f(mathsf{Clos}(f^{-1}(F))) subseteq mathsf{Clos}(f(f^{-1}(F))) = mathsf{Clos}(F) end{equation*}$$
meaning $$f(x) in mathsf{Clos}(F)$$ i.e. $$x in f^{-1}(mathsf{Clos}(F))$$ which was what we needed.