I read the following definition from this paper.
Definition: Let $N$ be a Riemannian manifold with boundary $\partial N$. We say $N$ is a manifold with boundary and cylindrical ends if there exists an open subset $V$ of $N$ which is isometric to $(-\infty, 0) \times X$, where $X$ is a compact manifold with boundary, such that $N \backslash V$ is compact.
My questions is whether the manifold $A=(-\infty, 0) \times S$, where $S$ is a compact manifold without boundary, is a manifold with boundary and cylindrical ends in the above definition.
In the definition, it's stated that $X$ has a boundary and I am not sure that we are allowed it to take it to be empty. I was thinking of taking $N=A$, so $\partial N = \{0\} \times S$, and $X$ to be the same as $S$ except with some portion removed so that it has a boundary...
If you took $A = [-\infty, 0] \times S$, it'd be fine; then $V = (-\infty, 0) \times S$ works nicely.
In your version (with the open interval), presumably you'd take $V = A$, and then $N\setminus V$ would be empty, hence compact. I can't believe that the authors want this, but who knows?