Diffeomorphism between open surfaces

Question

Is it true that for every pair of surfaces $S_1, S_2$ in $\mathbb{R}^3$, there exist two non-empty open sets $W_1 \subseteq S_1$ and $W_2 \subseteq S_2$ that are diffeomorphic?

Here is my attempt:

Let $S$ be a surface in $\mathbb{R}^3$. For every $p$ in $S$, there exists an open neighborhood $V$ of $p$ in $\mathbb{R}^3$ and a parametrization $\varphi: U \subseteq \mathbb{R}^2 \rightarrow S \cap V$ with $U$ being an open set in $\mathbb{R}^2$. Take $q$ in $U$ such that $\varphi(q) = p$. Since $U$ is open, there exists an $r > 0$ such that $B_r^2(q)$ is an open ball in $\mathbb{R}^2$ of radius $r$ centered at $q$ and contained in $U$. As $\varphi$ is a diffeomorphism onto its image, the restriction $\varphi_{|{B_r^2(q)}}: B_r^2(q) \rightarrow \text{Im}(\varphi_{|{B_r^2(q)}}) \subseteq S$ (the restriction of a diffeomorphism to a subspace of the domain is still a diffeomorphism) is a diffeomorphism, and $\text{Im}(\varphi_{|{B_r^2(q)}})$ is an open subset of $S$. Therefore, given two surfaces $S_1$ and $S_2$ in $\mathbb{R}^3$, let's consider two open balls in $\mathbb{R}^2$: $B_{r_1}^2(q_1)$ and $B_{r_2}^2(q_2)$. Define $W_1 = \text{Im}(\varphi_{|{B_{r_1}^2(q_1)}})$ and $W_2 = \text{Im}(\varphi_{|{B_{r_2}^2(q_2)}})$. As mentioned earlier, $W_1$ is diffeomorphic to $B_{r_1}^2(q_1)$, and $W_2$ is diffeomorphic to $B_{r_2}^2(q_2)$. Since open balls in $\mathbb{R}^2$ are diffeomorphic, $B_{r_1}^2(q_1)$ is diffeomorphic to $B_{r_2}^2(q_2)$, implying that $W_1$ and $W_2$ are diffeomorphic (as compositions of diffeomorphisms yield a diffeomorphism).

Is it ok?

Answer 1

Your proof is certainly fine.

One thing to keep in mind is that the definition of a surface $S \subset \mathbb R^3$ has several equivalent formulations; your proof hovers very close to the proof of equivalence of these formulations.

1. (The definition you wrote:) For every $p \in S$ there exists an open neighborhood $V \subset \mathbb R^3$ of $p$ and a parameterization $\phi : U \subseteq \mathbb R^2 \to S \cap V$ with $U \subset \mathbb R^2$ an open subset.
2. (A very commonly used definition:) For every $p \in S$ there exists an open neighborhood $V \subset \mathbb R^3$ of $p$ and a parameterization $\phi : \mathbb R^2 \to S \cap V$.
3. (A minor variation:) For every $p \in S$ there exists an open neighborhood $V \subset \mathbb R^3$ of $p$ and a parameterization $\phi : B^2_1(\mathcal O) \to S \cap V$, where $\mathcal O=(0,0) \in \mathbb R^2$.
4. (Another minor variation:) For every $p \in S$ there exists an open neighborhood $V \subset \mathbb R^3$ of $p$ and a parameterization $\phi : B^2_r(q) \to S \cap V$ for some $r>0$ and $q \in \mathbb R^2$.

Your proof could be regarded as a proof of the implication $1 \implies 4$. The equivalence of 2, 3, 4 are each very short, assuming one has already proved that $\mathbb R^2$, $B^2_1(\mathcal O)$, and $B^2_r(q)$ are all diffeomorphic to each other. Also, any one of 2,3,4 immediately implies 1.