Euclidean metric geometry of the tubular neighbourhood thm

Question

Let $M$ be a compact manifold of dimension $m$ and of class at least $C^2$ embedded in $R^n$. For $x\in M$ let $N_x(\varepsilon)$ be the intersection $T_xM^\bot\cap B(x,\varepsilon)$, where $T_xM^\bot$ is orthogonal complement of $T_xM$ in $T_x\mathbb R^n$ and $B(x,\varepsilon)$ is the open ball.

Under this assumptions the tubular neighbourhood theorem (the version I know) states that there exists $\varepsilon>0$ such that $$N(\varepsilon)=\bigcup_{x\in M} N_x(\varepsilon)$$ is an open subset of $\mathbb R^n$ and $N_x(\varepsilon)\cap N_y(\varepsilon)=\emptyset$.

Since we can identify $(-\varepsilon,\varepsilon)$ with $\mathbb R^n$ we can deduce that there exists a diffeomorphism between the normal bundle and an open neighbourhood of $M$. Let $BM(\delta) = \bigcup_{x\in M} B(x,\delta)$. Using compactness we can find $\delta>0$ such that $\delta$-neighbourhood $BM(\delta)$ of $M$ fits in $N(\varepsilon)$.

This gives us the standard formulation from the web: $\delta$-neighbourhood of $M$ is diffeomorphic with some open neighbourhood of the zero-section of the normal bundle.

My question is: Is it true that $BM(\varepsilon) \subseteq N(\varepsilon)$ ($\supseteq$ is obvious)?

I have never seen such a geometric formulation and it is obviously false for some noncompact manifolds even if the tubular neighbourhood thm is true for them [for example by extending them to compact manifolds] (consider an open interval in $\mathbb R^2$). But it seems so true...

Answer 1

Uhmm... It was extremely easy.

Consider any point $y$ from $BM(\varepsilon)$ - by compactness its distance from $M$ is realized as a distance from some $x\in M$. I claim that vector $y-x$ is orthogonal to $T_xM$.

Suppose conversely, that orthogonal projection $\pi$ of $y$ to $T_xM$ is not $x$. Then of course the vector $y-\pi(y)$ is shorter than $y-x$ (Pythagoras's) - the thing is that probably $\pi(y)\notin M$.

We may assume that locally (there are linearly equivalent orthogonal coordinates such that) $M$ is a graph of $T_xM$ (easy excercise) (actually it is enough to compose parametrization with inverse of its derivative at $0$). So if we move a little ($r>0$) from $0\in T_xM$ in direction of $\pi(y)-x$ in the domain of parametrization, we would move from $x\in M$ towards $\pi(y)$ by that little $r$ up to $o(r)$. By Pythagoras's thm we would end up with a point $x'\in M$ that is closer to $y$ than $x$ - contradiction.

Actually, the above proves the first part of the theorem without $C^2$ assumption (the only thing we use is existence of derivative).