# Closure of a Manifold is a Manifold with Corners?

Is there a general theorem that shows that if you have a manifold $S$ then its closure $\overline{S}$ is a manifold with corners? I am dealing with a specific set $S$ (I would rather not say which set it is) which is an open subset of a vector space, but really I'm much more interested in $\overline{S}$. The only methods I know that I can use to show that $\overline{S}$ is a manifold with corners is either to construct coordinate charts (which seems rather daunting to me), or to find a clever group action $G \times X \to X$ of a group $G$ on some manifold with corners $X$ that will yield the quotient manifold $X/G \approx \overline{S}$. This question might be rather vague, but I'm wondering if there is a general theorem for showing that the closure of a manifold is a manifold with corners (obviously with or without corner points).

For reference I'm reading some material out of Differential Topology by Margalef-Roig/Outerelo-Dominguez for the necessary background.

Edit: Thank you for the cool counterexamples that you're all contributing. I'd like to augment my post to ask of potentially good methods, suggestions, and references to read up on manifolds with corners, specifically dealing with group actions. Constructing coordinate charts seems like my best option, but if there are theorems involving group actions in addition to those presented in Roig/Dominguez I'd really like to know of them.

• How about $S=\text{ two tangent circles minus the common point}$ ? $\overline S$ should not be a topological manifold with boundary, or am I missing something? – user228113 Jun 18 '15 at 19:58
• Or is connectedness an implicit hypothesis? – user228113 Jun 18 '15 at 20:03
• @G.Sassatelli That's a great counterexample to the general claim, thank you! It so happens that both my sets $S$ and $\overline{S}$ are simply connected, but since this is a bit of a research problem on my part I can't go into what they are. – Mnifldz Jun 18 '15 at 20:09
• The open region inside the Koch snowflake is a smooth $2$-manifold, but its closure is certainly not a smooth manifold with corners. – Jim Belk Jun 18 '15 at 20:31