How to prove that $f(x,y)=y-x$ is continuous? Let $f:\mathbb{R}^2\rightarrow \mathbb{R}:(x,y)\mapsto y-x$ be a function.
How do I prove that $f$ is continuous?

[From a comment]
I'm trying to show the famous equality $\Gamma(z)\Gamma(w)=\Gamma(z+w)B(z,w)$ (Gamma and Beta functions). I found that a function which arises in that proof, is continuous, hence measurable if $\{(x,y):0<x<y\}$ is open in $\mathbb R^2$. So I want to know whether $f$ in my post is continuous. It's geometrically trivial, but I find it hard to prove it via $ϵ−δ$ criteria.
 A: Hint:
Take a point $(x_0,y_0)\in\mathbb{R}^2$. For any point $(x,y)\in\mathbb{R}^2$, we can write
$$
(x,y)=(x_0,y_0)+(\Delta x,\Delta y),\qquad \Delta x:=x-x_0,\qquad \Delta y:=y-y_0.
$$
Then
$$
f(x,y)-f(x_0,y_0)=(x-y)-(x_0-y_0)=(x-x_0)-(y-y_0)=\Delta x-\Delta y.
$$
Given $\epsilon>0$, you want to find $\delta>0$ such that whenever $d((x,y),(x_0,y_0))<\delta$, you have $\lvert f(x,y)-f(x_0,y_0)\rvert<\epsilon$.
Note, however, that 
$$
d((x,y),(x_0,y_0))=\sqrt{\Delta x^2+\Delta y^2},
$$
and by the above
$$
\lvert f(x,y)-f(x_0,y_0)\rvert=\lvert \Delta x-\Delta y\rvert\leq\lvert\Delta x\rvert+\lvert \Delta y\rvert.
$$
Can you see how to make $\lvert\Delta x\rvert$ and $\lvert \Delta y\rvert$ small by choosing $\delta$ small?
Of course, alternatively, you can prove that the functions $(x,y)\mapsto x$ and $(x,y)\mapsto y$ are continuous, and the (very direct) theorem that if $g$ and $h$ are continuous, then $g(x,y)-h(x,y)$ is also continuous.
A: Use the Euclidean norm on $\mathbb{R}^2$. That is $\|(x,y)-(\hat{x},\hat{y})\| \le \sqrt{(x-\hat{x})^2 + (y-\hat{y})^2}$.
In this case we have $|y-x-(\hat{y}-\hat{x})| \le |y-\hat{y}| + |x-\hat{x}| \le \|(x,y)-(\hat{x},\hat{y})\|+ \|(x,y)-(\hat{x},\hat{y})\|$,
so if we choose $\delta = {\epsilon \over 2}$, then we get
$|f(x,y)-f(\hat{x},\hat{y})| \le 2 \|(x,y)-(\hat{x},\hat{y})\| < \epsilon$.
