# why does a closed subset of a top. n-manifoldis not again a top. n-manifold?

In the book of Int. Smooth Manifolds by Lee, at page 4, it is stated that

it follows easily from these two exercises that any open subset of a topological n-manifold is itself a topological n-manifold (with the subspace topology, of course).

However, I cannot understand exactly what and how things go wrong when that subset is closed ? i.e why does a closed subset of a top. n-manifoldis not again a top. n-manifold ?

To be clear, the closed subset is again Hausdorff and second countable, but why are they not homeomorphic to $$\mathbb{R}^n$$ locally ?

Edit:

I'm looking mainly for a mathematical explanation, rather than only an intuitive one.

• Well, as a start if you choose anything bounded in R^n you get at best a manifold with boundary, not a manifold. – Brevan Ellefsen Feb 21 at 9:29
• take a point in $\mathbb{R}^n$. – Tsemo Aristide Feb 21 at 9:30
• @TsemoAristide yes, it look like it is not the case, intuitively, but I'm looking for a mathematical explanation. – onurcanbektas Feb 21 at 9:31

## 1 Answer

Take a closed disk in $$\mathbb{R}^2$$, for instance. No neighborhood of a point $$p$$ of its boundary is homeomorphic to $$\mathbb{R}^2$$.

• can I ask why ? because I'm not comfortable in stating such a thing. – onurcanbektas Feb 21 at 9:30
• @onurcanbektas it's just topology. Every open set in R^n about a point p contains a disk about p, and this disk will intersect the exterior. Can you see why this is a problem? (Hint: what must happen to such a neighborhood when we consider the subspace topology by intersecting it with our disk?) – Brevan Ellefsen Feb 21 at 9:34
• @BrevanEllefsen I know topology; but no I can't see why this is a problem since we are looking that set as in the subspace topology. – onurcanbektas Feb 21 at 9:36
• @onurcanbektas Right now, I don't see a simple proof. But note that if we have a small neighborhood of $p$, then if we remove from it its intersection with the boundary of the closed disk, what remains is homeomorphic to $\mathbb{R}^2$. But, in $\mathbb{R}^2$, if we remove something homeomorphic to $\mathbb R$, then what's left is never both connected and simply connected, and therefore it is not homeomorphic to $\mathbb{R}^2$. – José Carlos Santos Feb 21 at 9:41
• @onurcanbektas let it be noted, the one point example is even easier: any neighborhood is just the point, which is compact, so not homeomorphic to R^n. – Brevan Ellefsen Feb 21 at 19:22