Consider the set of points $U$ in $\Bbb{R}^2$ that lie above the line $y = x$, i.e. points $(a,b)$ such that $b>a$. Prove that $U$ is open and connected.
The method that is recommended is showing that there is a continuous function $\Bbb{R}^2\to\Bbb{R}$ that is the inverse of $U$.
This is pretty difficult for me and I don't know how to go about proving this. Can someone explain to be the most basic and intuitive proof?
On
You can argue in this way: your set $U$ is homeomorphic to the open upper plane $\{(a,b), b>0\}$ via a rotation of angle $\pi/4$. Now, this is open, since it can be written as $\bigcup_{n=1}^{\infty}\{(x,y), x\in\mathbb{R}, 0<y<n\}$, and every set in this union is obviously open. Now, for connectedness, it's more! It's path connected, you can link any two points in the upper plane by a segment entirely contained in the upper plane.
Let $$f:\mathbb{R}^2\to \mathbb{R},\quad f(x,y)=y-x.$$ Then, $f$ is continuous and $(0,\infty)$ is open, so $$f^{-1}((0,\infty))=\{(x,y)\in\mathbb{R}^2:y>x\}=U$$ is open. Now, let $$g:\mathbb{R}^2\to\mathbb{R}^2,\quad g(x,y)=(x,x+e^{y}).$$ Then, $g$ is continuous and $\mathbb{R}^2$ is connected, so $$g(\mathbb{R}^2)=\{(x,y)\in\mathbb{R}:y>x\}=U$$ is connected.