# Is any Morse trajectory contained in a contractible open set?

Suppose $f$ is a Morse function on a Riemannian Hilbert manifold $M$. Let $p_{\pm}\in \text{Crit}(f)$ be given and fix some $u:R\rightarrow M$ which is an integral curve of $-\nabla f$ connecting $p_-$ and $p_+.$ Is it true that there is an open contractible set $U\subseteq M$ containing the image of $u$?

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you may want to email Lizhen Qin about this question ( qinl@math.purdue.edu ). He has though a lot about Morse theory on Hilbert manifolds. – John Klein May 3 '12 at 1:43

Yes, this follows from what's usually called the "$\epsilon$-neighbourhood theorem" in textbooks like Guillemin and Pollack's Differential Topology.
Specifically, given a submanifold $N$ of a manifold $M$ there is an open neighbourhood $V$ of $N$ in $M$ which is diffeomorphic to a vector bundle over $N$. So $V$ is open in $M$.
The only restriction on this theorem is that $N$ can not be a manifold with boundary. It can be a non-compact manifold with empty boundary. But it's perfectly fine for $N$ not to be closed in $M$. The proof of this theorem is basically the same as the tubular neighbourhood theorem, except you give up on the idea of having a uniform injectivity radius for the normal bundle's exponential map, and you let the injectivity radius vary smoothly along $N$.