Let the kernel-like operator, $K$, be defined as
$K(f,R) = \text{Closure}({\{x; \lvert f(x)\rvert < \frac{1}{R}\})}$
where $f:\mathbb{R} \to \mathbb{R}$ is a continuous function on the complex numbers with the usual topology. Essentially, the kernel operator finds the preimage of the function on an open disk centered around 0 and takes the closure of it. The purpose of wanting the closure is so I can compare different outputs using the Hausdorff metric which involves using compact sets. Some examples of using this operator are,
$K(z^2, 1) = \{z; \Re(z)^2+\Im(z)^2 < 1\}$
For the next one, let $a = \Re(z)$ and $b = \Im(z)$.
$K(z^2 - 2, 1) = \{z; 1<a<\sqrt{3}, -\sqrt{-a^2+\sqrt{8 a^2+1}-2}<b<\sqrt{-a^2+\sqrt{8 a^2+1}-2}$
$-\sqrt{3}<a<-1, -\sqrt{-a^2+\sqrt{8 a^2+1}-2}<b<\sqrt{-a^2+\sqrt{8 a^2+1}-2}\}$
The last one was done with some assistance from wolfram alpha to actually deal with the resulting inequality.
Now, the essence of my question is the kernel operator continuous? As continuity of a function depends on the topology you have on its domain and range, I'll clarify what the domain and range I'm investigating look like.
The domain of $K$ is any continuous complex-valued function defined on $A$ where $A$ is a closed ball in the usual topology of $\mathbb{C}$ and the topology is the one induced using the uniform metric. The uniform metric refers to taking the maximum of the difference of the two functions (and this is well-defined due to the extreme value theorem). The usual uniform metric involves a supremum, but as the max will always exist here which one you use doesn't make a difference in this setting. I also chose to restrict the domain of $K$ to continuous functions defined on a compact set because $K$ would not be continuous if you allowed functions defined on a domain that isn't compact. If the domain isn't complete you could have issues come from missing points. If the domain isn't bounded you can have problems by having your sequence of functions smallest place where it intersects your open disk always increase so that the functions converge to one that has not touching the disk while each function in the sequence intersected the disk.
The range will be the closed subsets of $A$ (which makes them compact sets). The topology I'm interested in here is the one that arises from the Hausdorff metric.
The purpose of all this is that choosing large values of $R$ let's you approximate the zero's of the function as you are taking the mainly taking the preimage of a small disk about 0. In fact,
$\bigcap\limits_{i=1}^{\infty} K(f,R) = f^{-1}(\{0\})$
which is what I mean by bigger $R$ getting you closer to the 0 points. Using this instead of finding the 0's directly is the preimage of the set $\{0\}$ is problematic because the zero's of a sequence of functions that converge to another function may be very different than the final function ($f_n = 1/n$ as a good example). Taking the preimage of an open disk avoids this type of issue since it avoids functions almost reaching the boundary.
Now, is this kernel-like operator continuous with respect to f. Specifically, if $\displaystyle\lim_{n \to \infty} f_n = f$ uniformly, then does $\displaystyle\lim_{n \to \infty} K(f_n,R) = K(f,R)$ using the Hausdorff metric.
Edit: Given the current difficulties involving the boundary I have one question related specifically to the issues that occur on the boundary. The question is (still assuming $f_n \to f$ uniformly),
$\displaystyle\lim_{n \to \infty} K(f_n,R) \Delta K(f,R) \subset E(R)$
where $E(R) = \partial\{x; \lvert x\rvert < \frac{1}{R}\}$ and $\Delta$ is for symmetric difference of those two sets and $\partial$ means the boundary of that set. Essentially are the points where those two sets differ only the boundary of the open disk that they are taking the preimage of.