TITLE: Can one disjoin any submanifold in $\mathbb R^n$ from itself by a $C^{\infty}$-small isotopy?
QUESTION [8 upvotes]: Let $M$ be a manifold and $V$ be an oriented vector bundle. It's well known that if the Euler class of $V$ is non zero, then $V$ can't have a non-vanishing section. The converse is not true, see Vanishing of Euler class , or the answer by John Klein below, given to the first version of this question. So:
Question. Recall that the normal bundle of a smooth orieantable submanifold in $\mathbb R^n$ always has zero Euler class.  Does such a normal bundle always has a non-vanishing section? If yes, how to prove this? If no, what is a counterexample? 
PS. It turns out that this question is a subquestion of the following much more informed one: 
Embeddings without nonvanishing normal vector fields

REPLY [9 votes]: Here's a counter example. Take any embedding of $\mathbb{C} P^2$ in $\mathbb{R}^7$ (such embeddings exist by 
Steer, B., On the embedding of projective spaces in Euclidean space, Proc. Lond. Math. Soc., III. Ser. 21, 489-501 (1970). ZBL0206.25501;
the construction is summarised in this answer on MSE).
If such an embedding admits a normal vector field, then by the Compression Theorem of Rourke and Sanderson it is isotopic to an embedding with normal field parallel to the last coordinate of $\mathbb{R}^7=\mathbb{R}^6\times\mathbb{R}$, and then the projection is an immersion $\mathbb{C} P^2\looparrowright \mathbb{R}^6$. Such immersions cannot exist, by András Szűcs' answer here.