Let $g:\mathbb{R^2}\rightarrow \mathbb{R},\quad \text{defined by}\quad g(\vec{x})=2x-y-1,\quad\text{where}\quad\vec{x}=(x,y)^t$
I want to show this function is continuous for every $p\in\mathbb{R^2}$
I start by fixing $p=(a,b)^t$. Fix $\epsilon>0.$ Then take $\delta =...$ (not sure what $\delta$ to take)
Then for $\vec{x} =(x,y)^t\in\mathbb{R^2}$ with $||\vec{x}-\vec{p}||<\delta$, we have $||g(\vec{x})-g(\vec{p})|| =|2x-y-1-(2a-b-1)|=|2(x-a)-(y-b)|$
I'm quite stuck after fiddling around with the triangle inequality. I believe I can set $\delta$ as a function of my point $\vec{p}$ and $\epsilon$, but I can't find the right choice of function in order to satisfy $||g(\vec{x})-g(\vec{p})|| <\epsilon$
Could someone provide an idiot proof answer (saying why you're doing even basic things) - so that I can improve my conceptual understanding here.