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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.

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  • $\begingroup$ Well, as a start if you choose anything bounded in R^n you get at best a manifold with boundary, not a manifold. $\endgroup$
    – Brevan Ellefsen
    Commented Feb 21, 2019 at 9:29
  • $\begingroup$ take a point in $\mathbb{R}^n$. $\endgroup$
    – Tsemo Aristide
    Commented Feb 21, 2019 at 9:30
  • $\begingroup$ @TsemoAristide yes, it look like it is not the case, intuitively, but I'm looking for a mathematical explanation. $\endgroup$
    – Our
    Commented Feb 21, 2019 at 9:31

1 Answer 1

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

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  • $\begingroup$ can I ask why ? because I'm not comfortable in stating such a thing. $\endgroup$
    – Our
    Commented Feb 21, 2019 at 9:30
  • $\begingroup$ @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?) $\endgroup$
    – Brevan Ellefsen
    Commented Feb 21, 2019 at 9:34
  • $\begingroup$ @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. $\endgroup$
    – Our
    Commented Feb 21, 2019 at 9:36
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    $\begingroup$ @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 $\mathbb{R}^2$. $\endgroup$
    – José Carlos Santos
    Commented Feb 21, 2019 at 9:41
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    $\begingroup$ @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. $\endgroup$
    – Brevan Ellefsen
    Commented Feb 21, 2019 at 19:22

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