Let $U \subseteq \mathbb{R}^2$ be an open set and let $\sigma : U \rightarrow \mathbb{R}^3$ be a parametrization of an oriented surface $S$ embedded in $\mathbb{R}^3$ whose unit normal in $\sigma (u,v)$ is $N(u,v)$. $\forall \delta \in \mathbb{R}$ we define the map $\tau_\delta : U \rightarrow \mathbb{R}^3$ as
$\tau_\delta (u,v) := \sigma (u,v)+ \delta N(u,v)$
Prove that, if $V \subseteq U$ is an open set with compact closure in $U$, there exist $\epsilon > 0$ such that $\tau_\delta|_V$ is the parametrization of a regular embedded surface $\forall \delta \in (-\epsilon , \epsilon)$.
How can I approach this kind of problem?
Let's see if it works...
Let $N_S(\eta)$ be a tubular neighborhood of $S$ (where $\eta:S\rightarrow(0,+\infty)$ is a continuous function) and let $N_{\sigma(\overline V)}(\eta)$ be the restriction of this tubular neighborhood to the set $\sigma(\overline V)$ (observe that $\overline {N_{\sigma(\overline V)}(\eta)}$ is a compact set).
Suppose for contradiction that $\forall\epsilon>0$ exists $\delta\in(-\epsilon,\epsilon)$ such that $\tau_\delta(V)$ has a singular point. We choose $n\in\mathbb{Z}_+$ and set $\epsilon=\frac{1}{n}$, then there is $\delta_n$ with $|\delta_n|<\frac{1}{n}$ such that $\tau_{\delta_n}(V)$ has a singular point, let's call it $p_n$.
Now the sequence $\{p_n\}_{n\in\mathbb{Z}_+}$ is definitively contained in the compact set $\overline {N_{\sigma(\overline V)}(\eta)}$, so there is a subsequence $\{p_{n_k}\}$ which converges in $\overline {N_{\sigma(\overline V)}(\eta)}$ to a certain point $\overline p$; also $\overline p\in\sigma(\overline V)\subseteq\sigma(U)$.
Let $A\subseteq\sigma(U)$ be any open neighborhood of $\overline p$, let $a$ be a point of $A$ and let $I_S(a,\eta(a))$ be the segment $a+(-\eta(a),\eta(a))N(a)$ of lenght $2\eta(a)$ centered in $a$ and normal to $T_PS$. Then in $A$ there are at least two points $x,y$ such that $I_S(x,\eta(x))\cap I_S(y,\eta(y))\neq\emptyset$, contradicting the fact that $N_S(\eta)$ is a tubular neighborhood of $S$.
Do you think it is correct?