I want to show continuity of the following function by using the $\epsilon$-$\delta$-definition: \begin{align} f:\mathbb{C} \ &\times \ \mathbb{C} \rightarrow \mathbb{C} \\ (z,w) &\rightarrow z + w \end{align} We use the Euclidean norm as the measure for distance in the domain. The real problem comes down to that I can't seem to find an inequality that can help us. Since we use the Euclidean Norm, I thought we could perhaps use the following inequality: $|(z,w)| \leq \sqrt{\text{Re}(z)^2 + \text{Im}(z)^2 + \text{Re}(w)^2 + \text{Im}(w)^2}$.
We choose $(x,y) \in \mathbb{C} \ \times \ \mathbb{C}$ randomly. Also we choose $\epsilon > 0 $ randomly. We also choose $(c,d) \in \mathbb{C} \ \times \ \mathbb{C}$ randomly. Then, we need to show that there is a $\delta$ such that if $|(x,y) - (c,d)| \leq \delta$ then $|x + y - c - d| \leq \epsilon$.
If we then start with $|x+y-c-d| \leq |x-c| + |y-d|$. I feel as if we can use this to find the delta that we need, but I don't know how to exactly.
I'd appreciate a hint to try and show this continuity by using the definition.
Thanks for your time,
K. Kamal
We have that $|(x,y)-(c,d)|^2 \leq \delta^2$.
Let's expand what we have on the left side of the inequality: $|(x,y)-(c,d)|=|(x-c,y-d)| = \sqrt{(|x-c|^2 + |y-d|^2}$, just like $\lVert\cdot\rVert_2$ in $\mathbb{R}^2$. So, taking squares at both sides, you get that $|(x,y)-(c,d)|^2 = |x-c|^2 + |y-d|^2$, so:
$|x-c|^2 + |y-d|^2 \leq \delta^2$
You want to relate that with $|x-c| + |y-d|$.
First prove that $|x-c|^2 \leq \delta^2$. Then you have that $|x-c| \leq \delta$.
Do something similar for $|y-d|$. Which $\delta=\delta(\varepsilon)$ can you pick now to get the desired result?