Question

Show that any open disc in xy-plane is a surface

Answer 1

Definition:

A subset $S$ of $\R ^3$ is a surface if , for every points $p \in S,$ there is an open set $U$ in $\R ^2$ and an open set $W$ in $\R ^3$ containing $p$ such that $S \bigcap W$ is homomorphic to $U$.

Let $S$ be an open disk in $xy-$ plane. This implies that $S\subset \R^3$ .

We basically need a map $f$ from $\R^2$ to $\R^3$ , that has $S$ as its range.

We can pick the (so called "canonical") map that is basically the identity, just mapping from $\R^2$ to $\R^3$. That is $f:(x,y)↦(x,y,z|z=0)$.

Now, suppose we pick a point $p\in S$. We need an open set $W$ that contains $p$. Take W$=S$. We have that $S\bigcap W=S=\{(x,y,z)|z=0\}$. We can pick $U=f^{-1}(W)=f^{-1}(S),$ which is the same open set in $\R^2$ and $\R^3.$

Since the function $f:(x,y)↦(x,y,0)$ is continuous and bijective. Then, $S\bigcap W$ is homomorphic to $U$. Hence, $S$ is a surface.