I am being asked to determine whether the function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$, defined by $f(x,y) = 2x$ is uniformly continuous or not. In the question $\mathbb{R}^2$ is equipped with the "Manhattan" metric $d$ which is defined as $$d((x_1,y_1),(x_2,y_2))=|x_1-x_2|+|y_1-y_2|$$ and $\mathbb{R}$ is equipped with the normal Euclidean metric $e$ so $$e(f(x_1,y_1),f(x_2,y_2))=|f(x_1,y_1)-f(x_2,y_2)|=|2x_1-2x_2|$$
I've produced this proof and I was looking for it to be verified:
Let $\epsilon>0$ and let $\alpha=\epsilon/3$. Then $$|x_1-x_2|+|y_1-y_2|<\alpha$$ Now $$|2x_1-2x_2|=2|x_1-x_2|<3(|x_1-x_2|+|y_1-y_2|)<3\alpha=\epsilon$$ so $$d((x_1,y_1),(x_2,y_2))<\alpha \Rightarrow e(f(x_1,y_1),f(x_2,y_2))<\epsilon\;\;\;\;\; \forall (x,y) \in \mathbb{R}^2$$ so $f$ is uniformly continuous on $\mathbb{R}^2$.
I understand delta is the conventional character to use where I've used alpha, however this appeared as epsilon when I typed this. I hope this formatting is okay and I've presented my answer with sufficient rigour.