$f:U \rightarrow \mathbb{C} $ is continuous on $U$ if and only if {$z \in U| f(z)\in V$} is open for every open set V in $\mathbb{C}$

Question

I want to prove that if $f:U \rightarrow \mathbb{C} $ is continuous on $U$ if and only if {$z \in U| f(z)\in V$} is open for every open set V in $\mathbb{C}$. This is my rather incomplete approach to solve this and I hope I get some assistance on this.

$f^{-1}(V)=${$z \in U| f(z)\in V$}. Let $z\in f^{-1}(V)$. Then $f(z)\in V$. Also f is continuous at the point z. Let $\epsilon >0$. Then there exists a $\delta>0$ such that if $$ 0<|w-z|<\delta\implies |f(w)-f(z)|<\epsilon$$. $$\implies f(w)\in D_{\epsilon}(f(z))$$.

This also can be written as $f(D_{\delta}(z)\subseteq D_{\epsilon}(f(z))$

I am totally stuck here and need some help. Thanks

Answer 1

Take a $z$ in $f^{-1}(V)$. As $V$ is open in $\mathbb{C}$ and $f(z) \in V$, Choose $\epsilon > 0$ so that $D_{\epsilon} (f(z)) \subset V$. Now by your argument you have a $\delta >0$, so that $f(D_{\delta} (z)) \subset V$ i.e. $D_{\delta} (z) \subset f^{-1}(V)$.

Hence $z$ is an interior point of $f^{-1}(V)$. As $z$ is arbitrary, $f^{-1}(V)$ is open.

Answer 2

Suppose that f is continuous, you need to prove that for any open set $V$ in $\mathbb{C}$, $f^{-1}(V)$ is also an open set for the first implication.

So you can take $V$ an arbitrary open set in $\mathbb{C}$ and $y \in V$ such that $f^{-1}(y)$ it exists, and then prove that $f^{-1}(y)$ is an interior point. Notice that if $f^{-1}(y)$ does not exist for every $y \in V$ then it's inverse is empty, which is an open set and the result holds. Let's call $x=f^{-1}(y)$ (that is, one of its inverse images, since the function not necessarily is injective, it can have more than 1), so $f(x)=y$

Since $V$ is open you can find $\epsilon >0$ such that $D_{\epsilon}(f(x))\subset V$. For that $\epsilon$ you use the definition of a continuous function you said before and with that you found a $D_{\delta}(x) \subset f^{-1}(V)$, and this proves that x is an interior point, and since it was arbitrary, $f^{-1}(V)=\{z \in U|f(z) \in V\} $ is an open set.

For the other implication, suppose that $f^{-1}(V)=\{z∈U|f(z)∈V\}$ is open for every open set $V$ in $\mathbb{C}$. Take $y\in \mathbb{C}$ such that there exist $x$ such that $f(x)=y$ (once again, if we don't have such $x$ the result is trivial). For $\epsilon >0$ the disk $D_{\epsilon}(f(x))$ is an open set in $\mathbb{C}$, therefore, $f^{-1}(D_{\epsilon}(f(x)))$ is also an open set, so for $x \in f^{-1}(D_{\epsilon}(f(x))) $ there exists $\delta >0$ such that $D_\delta (x) \subset f^{-1}(D_{\epsilon}(f(x)))$, this means that $f(D_\delta (x))\subset D_{\epsilon}(f(x)) $, which means that for every $w,z\in U$, if $0<|w−z|<δ$ then $|f(w)−f(z)|<ϵ$, in other words, f is continuous.