The title says it all. The motivation behind this question is that I'm stuck with a theorem in topology, which states that if $f,g:X \rightarrow \mathbb{K}$ (where $\mathbb{K}$ is either $\mathbb{R}$ or $\mathbb{C}$ - both endowed with the euclidean topology - and $X$ is some topological space) is continuous, then $f\cdot g$ is as well. Since the sketch of the proof goes like this: "We know that from that the multiplication $\textrm{mult}: \mathbb{K}^2 \rightarrow \mathbb{K}$ is continuous and $C:X \rightarrow \mathbb{K}^2, C(x)=(f(x),g(h))$ is as well, so must be $f \cdot g = C \circ \textrm{mult}$" I'm left with the problem (not having taken a full course in complex analysis yet) with which topology $\mathbb{C}^2$ traditionally is endowed, when one proves, that the multiplication of complex numbers is continuous.
(For this theorem I need the fact that this topology I'm looking for on $\mathbb{C}^2$ - viewed as the domain of $\textrm{mult}$ - is at least contained in the product topology obtained from $ \mathbb{C}^2 = \mathbb{C} \times \mathbb{C} $ , both equipped with the euclidean topology - $ \mathbb{C}^2$ viewed this time as the range of $C$ - because else the proof won't work. My guess is, that the topology I'm looking for on $\mathbb{C}^2$ - as the domain of $\textrm{mult}$ - is the product topology above described, but I couldn't quickly do a proof that using this topology $\textrm{mult}: \mathbb{C}^2 \rightarrow \mathbb{C}$ is continuous and since I've wasted enough time on this problem and also haven't taken a complex analysis course, I thought it is finally time to ask you guys about this, so that I can get a clear view of things.)
Side questions:
I'm also wondering why "euclidean topology" was mentioned, because as far as I knew, this makes sense only for $\mathbb{R}$ (I'm guessing with $\mathbb{C}$ it is meant, that the euclidean topology of $\mathbb{R}^2$ is used; could it be, that if the topology I'm looking for on $\mathbb{C}^2$ - as the domain of $\textrm{mult}$ - is the product topology, that it is the same as the euclidean one for $\mathbb{R }^4$ ?)
Does anybody know a different proof of the above theorem ? (One that especially doesn't use nets, because I've done a sketch of a proof thatis way and it is just painful - and it also doesn't adapt easily to other types of composing $f$ and $g$)