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This is a closing remark in a paper of Elon Lima, Separation Theorem for Smooth Hypersurfaces.

He proves a well known lemma: Suppose $M\subset\mathbb{R}^m$ is a compact, orientable, smooth hypersuface. Fix a smooth field of unit normal vectors $v:M\to\mathbb{R}^m$, and define a smooth map $h:M\times\mathbb{R}\to\mathbb{R}^m$ by $h(x,t)=x+t\cdot v(x)$. Then for some $\epsilon>0$, $h:M\times(-\epsilon,\epsilon)\to\mathbb{R}^m$ is a diffeomorphism onto an open subset of $\mathbb{R}^m$.

He later remarks that if $M$ is only known to be closed, the lemma can be modified to change $\epsilon$ from a constant to a continuous positive function $\epsilon: M\to\mathbb{R}$ with the property if $x|y$ are in $M$ then $x+s\cdot v(x)|y+t\cdot v(y)$ for all $s\in(-\epsilon(x),\epsilon(x))$ and $t\in(-\epsilon(y),\epsilon(y))$.

I'm just a layman, but is there a pedestrian proof on why such a function $\epsilon$ exists? Also, what does the notation $x|y$ mean? Thanks!

NB: I posted this on MSE a couple days ago, but didn't get much of a response.

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    $\begingroup$ I don't know what the notation $x|y$ means, but for the construction of tubular neighborhoods of non-compact submanifolds of $\mathbb{R}^n$ (and, more generally, submanifolds of general manifolds), see Section 4.5 of Hirsch's book "Differential Topology". By the way, observe that it is necessary to take a function $\epsilon(x)$ as in your statement; think about the real line $\mathbb{R}$ embedded in $\mathbb{R}^2$ with the two "ends" of $\mathbb{R}$ getting closer and closer to to each other as you go further and further out. $\endgroup$
    – Andy Putman
    Nov 4, 2013 at 19:23
  • $\begingroup$ From the context I'd say the vertical bar must mean that the two points are different. Additional to Andy Putmans remark, another reason for $\epsilon(x)$ dependence is that $\epsilon(x)$ cannot be larger than the curvature radius at each point of $M$; if $M$ is not compact then this quantity may not be globally bounded from below. $\endgroup$
    – Jaap Eldering
    Nov 4, 2013 at 20:08
  • $\begingroup$ Cover the manifold with countably many compact sets. On each compact set, you can find an epsilon. Globally, you have now constructed a function which is not yet continuous, but a step function. Think of some way to smoothen it out, and you are done. $\endgroup$
    – Matthias Ludewig
    Nov 5, 2013 at 16:27

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