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Let $M$ be a complete non-compact manifold (possibly with boundary). Let $E$ be an open proper connected non-precompact subset of $M$ with smooth topological boundary, so that $\overline{E}$ is a non-compact complete manifold with boundary. Suppose that $E=M\setminus K$, where $K$ is a compact set that is the closure of a non-empty open set. If we now assume that $\overline{E}$ is a manifold with boundary of bounded geometry as described here, is it then also true that $M$ is of bounded geometry? I believe that it is true but does someone know a proof or a reference for that statement?

Added question: Is it essential that we assume that $\overline{E}$ is of bounded geometry and not $E$? What happends if we just assume that $E$ is of bounded geometry? Thanks in advance!

Let $M$ be a complete non-compact manifold (possibly with boundary). Let $E$ be an open proper connected non-precompact subset of $M$ with smooth topological boundary, so that $\overline{E}$ is a non-compact complete manifold with boundary. Suppose that $E=M\setminus K$, where $K$ is a compact set that is the closure of a non-empty open set. If we now assume that $\overline{E}$ is a manifold with boundary of bounded geometry as described here, is it then also true that $M$ is of bounded geometry? I believe that it is true but does someone know a proof or a reference for that statement?

Is it essential that $\overline{E}$ is of bounded geometry and not $E$? What happends if we just assume that $E$ is of bounded geometry? Thanks in advance!

Let $M$ be a complete non-compact manifold (possibly with boundary). Let $E$ be an open proper connected non-precompact subset of $M$ with smooth topological boundary, so that $\overline{E}$ is a non-compact complete manifold with boundary. Suppose that $E=M\setminus K$, where $K$ is a compact set that is the closure of a non-empty open set. If we now assume that $\overline{E}$ is a manifold with boundary of bounded geometry as described here, is it then also true that $M$ is of bounded geometry? I believe that it is true but does someone know a proof or a reference for that statement?

Added question: Is it essential that we assume that $\overline{E}$ is of bounded geometry and not $E$? What happends if we just assume that $E$ is of bounded geometry? Thanks in advance!

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Shaq155
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Let $M$ be a complete non-compact manifold (possibly with boundary). Let $E$ be an open proper connected non-precompact subset of $M$ with smooth topological boundary, so that $\overline{E}$ is a non-compact complete manifold with boundary. Suppose that $E=M\setminus K$, where $K$ is a compact set that is the closure of a non-empty open set. If we now assume that $\overline{E}$ is a manifold with boundary of bounded geometry as described here, is it then also true that $M$ is of bounded geometry? I believe that it is true but does someone know a proof or a reference for that statement?

ThanksIs it essential that $\overline{E}$ is of bounded geometry and not $E$? What happends if we just assume that $E$ is of bounded geometry? Thanks in advance!

Let $M$ be a complete non-compact manifold (possibly with boundary). Let $E$ be an open proper connected non-precompact subset of $M$ with smooth topological boundary, so that $\overline{E}$ is a non-compact complete manifold with boundary. Suppose that $E=M\setminus K$, where $K$ is a compact set that is the closure of a non-empty open set. If we now assume that $\overline{E}$ is a manifold with boundary of bounded geometry as described here, is it then also true that $M$ is of bounded geometry? I believe that it is true but does someone know a proof or a reference for that statement?

Thanks in advance!

Let $M$ be a complete non-compact manifold (possibly with boundary). Let $E$ be an open proper connected non-precompact subset of $M$ with smooth topological boundary, so that $\overline{E}$ is a non-compact complete manifold with boundary. Suppose that $E=M\setminus K$, where $K$ is a compact set that is the closure of a non-empty open set. If we now assume that $\overline{E}$ is a manifold with boundary of bounded geometry as described here, is it then also true that $M$ is of bounded geometry? I believe that it is true but does someone know a proof or a reference for that statement?

Is it essential that $\overline{E}$ is of bounded geometry and not $E$? What happends if we just assume that $E$ is of bounded geometry? Thanks in advance!

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Shaq155
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Let $M$ be a complete non-compact manifold (possibly with boundary). Let $E$ be an open proper connected non-precompact subset of $M$ with smooth topological boundary, so that $\overline{E}$ is a non-compact complete manifold with boundary. Suppose that $E=M\setminus K$, where $K$ is a compact set that is the closure of a non-empty open set. If we now assume that $\overline{E}$ is a manifold with boundary of bounded geometry as described here, is it then also true that $M$ is of bounded geometry? I believe that it is true but does someone know a proof or a reference for that statement?

Thanks in advance!

Let $M$ be a complete non-compact manifold (possibly with boundary). Let $E$ be an open proper connected non-precompact subset of $M$ with smooth topological boundary, so that $\overline{E}$ is a non-compact complete manifold with boundary. Suppose that $E=M\setminus K$, where $K$ is a compact set that is the closure of a non-empty open set. If we now assume that $\overline{E}$ is a manifold with boundary of bounded geometry as described here, is it then also true that $M$ is of bounded geometry? I believe that it is true but does someone know a proof or a reference?

Thanks in advance!

Let $M$ be a complete non-compact manifold (possibly with boundary). Let $E$ be an open proper connected non-precompact subset of $M$ with smooth topological boundary, so that $\overline{E}$ is a non-compact complete manifold with boundary. Suppose that $E=M\setminus K$, where $K$ is a compact set that is the closure of a non-empty open set. If we now assume that $\overline{E}$ is a manifold with boundary of bounded geometry as described here, is it then also true that $M$ is of bounded geometry? I believe that it is true but does someone know a proof or a reference for that statement?

Thanks in advance!

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