Imaginary $\delta$ in proof of continuity

Question

Define $f: \mathbb R^2 \to \mathbb R$ as $$f(x_1,x_2)=x_1+x_2$$ Prove that $f$ is continuous where the distance function on $\mathbb R^2$ is $$d'(x,a)=\sqrt{(x_1-a_1)^2+(x_2-a_2)^2}$$

I went about proving this using the $\delta$ - $\epsilon$ definition of continuity. I chose $\delta=\sqrt{\epsilon-1}$. Then, given that $$d'(x,a)\lt \delta$$ I need to derive $$|f(x_1,x_2)-f(a_1,a_2)|\lt \epsilon$$ My derivation went like this: $$d'(x,a)\lt \delta$$ $$\sqrt{(x_1-a_1)^2+(x_2-a_2)^2}\lt \sqrt{\epsilon-1}$$ $$(x_1-a_1)^2+(x_2-a_2)^2\lt \epsilon-1$$ Then, by a variation of the triangle inequality, $$(x_1-a_1+x_2-a_2)^2\lt \epsilon-1$$ $$(x_1-a_1+x_2-a_2)^2+1\lt \epsilon$$ Then, using the fact that $n^2+1 \gt |x|$, $$|x_1+x_2-a_1-a_2|\lt \epsilon$$ $$|f(x_1,x_2)-f(a_1,a_2)|\lt \epsilon$$ And so the continuity of the function is proven.

My question is this: since I let $$\delta=\sqrt{\epsilon-1}$$ it must be that when $\epsilon \lt 1$, $\delta$ is imaginary. Is this a problem, or is the proof still valid?

Do you see any other flaws in it?

Thanks!

Answer 1

As it is mentioned in the comments $\delta$ might not be real.

If you want, you can do this:

Let $ \epsilon>0$

We have that $$|f(x,y)-f(x_0,y_0)|=|(x-x_0)+(y-y_0)| \leqslant$$ $$|x-x_0|+|y-y_0|= \sqrt{|x-x_0|^2}+ \sqrt{|y-y_0|^2} \leqslant$$ $$\sqrt{(x-x_0)^2+(y-y_0)^2}+ \sqrt{(x-x_0)^2+(y-y_0)^2}<2 \delta$$

Take $\delta= \epsilon /2 $ and you are done.

Answer 2

Set $\delta:=\frac{\epsilon}{2}$, which is always positive if $\epsilon$ is. Then we calculate $$|x_1+x_2-a_1-a_2|\leq|x_1-a_1|+|x_2-a_2|=\sqrt{(x_1-a_1)^2}+\sqrt{(x_2-a_2)^2},$$ which we can estimate by $$\sqrt{(x_1-a_1)^2+(x_2-a_2)^2}+\sqrt{(x_2-a_2)^2+(x_1-a_1)^2}=2d'(x,a)<2\delta=\epsilon.$$

This proves the continuity of $f$.

EDIT: Whoops, Ramiz Karaeski has submitted essentially the same answer while I was typing.