Recently I started working on a problem in Differential Geometry (where I'm not a specialist, so I apologize if this question turns out to have a trivial answer) and I had to consider the following situation.

Let $M \subset \mathbb{R}^n$ be a smooth submanifold of dimension $n-k$. By a *normal tube* of radius $\varepsilon$ centered at $M$, I mean a tubular neighborhood of $M$ given by a *disjoint* union $$\mathscr{B}(M, \, \varepsilon):=\bigsqcup_{p \in M} B(p, \, \varepsilon),$$ where $B(p, \, \varepsilon)$ is a $k$-dimensional ball of radius $\varepsilon$ centred at $p \in M$ and contained in the (embedded) normal subspace $N_pM \subset \mathbb{R}^n$.

Q.Under which conditions on $M$ does a normal tube $\mathscr{B}(M, \, \varepsilon)$ exist (for $\varepsilon$ sufficiently small)?

It seems to me that this is always the case when $M$ is compact, because then we can identify $\mathscr{B}(M, \, \varepsilon)$ with the open neighborhood $B_M(\varepsilon)$ of $M$ in $\mathbb{R}^n$ given by $$B_M(\varepsilon) := \{x \in \mathbb{R}^n \, | \, d(x, \, M) < \varepsilon\}.$$ On the other hand, there are also examples where $\mathscr{B}(M, \, \varepsilon)$ exists even if $M$ is not compact, for instance in the case where $M$ is a linear subspace (in this case, in fact, the normal spaces to $M$ are pairwise disjoint).

My feeling is that $\mathscr{B}(M, \, \varepsilon)$ should always exist when the curvature of $M$ is bounded in some suitable sense, but I'm not able to specify it better.

Any answer, example/counterexample and reference to the relevant literature will be greatly appreciated.