3.4 Continuity and topology

Let $f:S\to T$ be a function, and let $A$ be a subset of the codomain $T.$ Recall that $f^{-1}\left(A\right)$ denotes the subset of the domain $S$ consisting of all those $x\in S$ for which $f\left(x\right)\in A.$

Our original definition of continuity was in terms of $ϵ$ 's and $\delta$ 's. [link] established an equivalent form of continuity, often called “sequential continuity,” that involves convergence of sequences.The next result shows a connection between continuity and topology, i.e., open and closed sets.

1. Suppose $S$ is a closed subset of $C$ and that $f:S\to C$ is a complex-valued function on $S.$ Then $f$ is continuous on $S$ if and only if $f^{-1}\left(A\right)$ is a closed set whenever $A$ is a closed subset of $C.$ That is, $f$ is continuous on a closed set $S$ if and only if the inverse image of every closed set is closed.
2. Suppose $U$ is an open subset of $C$ and that $f:U\to C$ is a complex-valued function on $U.$ Then $f$ is continuous on $U$ if and only if $f^{-1}\left(A\right)$ is an open set whenever $A$ is an open subset of $C.$ That is, $f$ is continuous on an open set $U$ if and only if the inverse image of every open set is open.

Suppose first that $f$ is continuous on a closed set $S$ and that $A$ is a closed subset of $C.$ We wish to show that $f^{-1}\left(A\right)$ is closed. Thus, let $\left\{x_n\right\}$ be a sequence of points in $f^{-1}\left(A\right)$ that converges to a point $c.$ Because $S$ is a closed set, we know that $c\in S,$ but in order to see that $f^{-1}\left(A\right)$ is closed, we need to show that $c\in f^{-1}\left(A\right).$ That is, we need to show that $f\left(c\right)\in A.$ Now, $f\left(x_n\right)\in A$ for every $n,$ and, because $f$ is continuous at $c,$ we have by [link] that $f\left(c\right)=limf\left(x_n\right).$ Hence, $f\left(c\right)$ is a limit point of $A,$ and so $f\left(c\right)\in A$ because $A$ is a closed set. Therefore, $c\in f^{-1}\left(A\right),$ and $f^{-1}\left(A\right)$ is closed.

Conversely, still supposing that $S$ is a closed set, suppose $f$ is not continuous on $S,$ and let $c$ be a point of $S$ at which $f$ fails to be continuous. Then, there exists an $ϵ>0$ and a sequence $\left\{x_n\right\}$ of elements of $S$ such that $c=lim{x}_n$ but such that $|f\left(c\right)-f\left(x_n\right)|\ge ϵ$ for all $n.$ (Why? See the proof of Theorem 3.4.) Let $A$ be the complement of the open disk $B_{ϵ}\left(f\left(c\right)\right).$ Then $A$ is a closed subset of $C.$ We have that $f\left(x_n\right)\in A$ for all $n,$ but $f\left(c\right)$ is not in $A.$ So, $x_n\in f^{-1}\left(A\right)$ for all $n,$ but $c=lim{x}_n$ is not in $f^{-1}\left(A\right).$ Hence, $f^{-1}\left(A\right)$ does not contain all of its limit points, and so $f^{-1}\left(A\right)$ is not closed. Hence, if $f$ is not continuous on $S,$ then there exists a closed set $A$ such that $f^{-1}\left(A\right)$ is not closed. This completes the proof of the second half of part (1).

Next, suppose $U$ is an open set, and assume that $f$ is continuous on $U.$ Let $A$ be an open set in $C,$ and let $c$ be an element of $f^{-1}\left(A\right).$ In order to prove that $f^{-1}\left(A\right)$ is open, we need to show that $c$ belongs to the interior of $f^{-1}\left(A\right).$ Now, $f\left(c\right)\in A,$ $A$ is open, and so there exists an $ϵ>0$ such that the entire disk $B_{ϵ}\left(f\left(c\right)\right)\subseteq A.$ Then, because $f$ is continuous at the point $c,$ there exists a $\delta >0$ such that if $|x-c|<\delta$ then $\left|f\right(x)-f(c\left)\right|<ϵ.$ In other words, if $x\in B_{\delta }\left(c\right),$ then $f\left(x\right)\in B_{ϵ}\left(f\left(c\right)\right)\subseteq A.$ This means that $B_{\delta }\left(c\right)$ is contained in $f^{-1}\left(A\right),$ and hence $c$ belongs to the interior of $f^{-1}\left(A\right).$ Hence, if $f$ is continuous on an open set $U,$ then $f^{-1}\left(A\right)$ is open whenever $A$ is open. This proves half of part (2).

Finally, still assuming that $U$ is open, suppose $f^{-1}\left(A\right)$ is open whenever $A$ is open, let $c$ be a point of $S,$ and let us prove that $f$ is continuous at $c.$ Thus, let $ϵ>0$ be given, and let $A$ be the open set $A=B_{ϵ}\left(f\left(c\right)\right).$ Then, by our assumption, $f^{-1}\left(A\right)$ is an open set. Also, $c$ belongs to this open set $f^{-1}\left(A\right),$ and hence $c$ belongs to the interior of $f^{-1}\left(A\right).$ Therefore, there exists a $\delta >0$ such that the entire disk $b_{\delta }\left(c\right)\subseteq f^{-1}\left(A\right).$ But this means that if $\in S$ satisfies $|x-c|<\delta ,$ then $x\in B_{\delta }\left(c\right)\subseteq f^{-1}\left(A\right),$ and so $f\left(x\right)\in A=B_{ϵ}\left(f\left(c\right)\right).$ Therefore, if $|x-c|<\delta ,$ then $\left|f\right(x)-f(c\left)\right|<ϵ,$ which proves that $f$ is continuous at $c,$ and the theorem is completely proved.