# the geometric boundary of a manifold

Let $X$ be a topological manifold with geometric boundary $\partial_g X$ (here the subscript $g$ to indicate geometric boundary which is a different notion from topological boundary $\partial_t X$).

Can someone indicate an example of a manifold $X$ such that $\partial_g X\not = \emptyset$ and $\partial_g X\not =\partial_t X$?

Recall that the geometric boundary is the set of points $p\in X$ which have a neighborhood $N_p\subset X$ homeomorphic to the closed upper half plane, while in the definition of the topological boundary we consider $X$ as a subset of some bigger space $Y$, that is $X\subseteq Y$, and then define it as the set of points $p\in Y$ which has a neighborhood $N_p\subset Y$ that contains at least a point of $X$ and at least a point of $Y-X$.

I think that the topological boundary $\partial_t X$ is clear to me as it depends on the set $Y$ we chose and the topology we put on it. But I have confusion regarding the geometric boundary as it seems depending only on the manifold $X$. What I understand from the definition is that $\partial_g X$ is contained in $X$ unlike topological boundary points which may or may not belong to $X$. In this note , page 1, example 4, just before proposition 1.3 he says that $[0,1)$ is a $1$-manifold with geometric boundary $\{1\}$ while $1\not \in [0,1)$ thank you for your clarification!!

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Well.. it may be possibly a typo only.. In my understanding, $[0,1)$ is a 1-manifold (with boundary!) and its geometrical boundary is $0$.

Another example is the strip with 2 circles as (geom.)boundary or the Moebius strip with 1 circle as (geom.)boundary.

And, in general, an $n$-manifold with boundary is defined as a top.space such that all of its points have open neighborhoods homeomorphic to an open subset of the upper halfspace...

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Couldn't you replace "homeomorphic to $\mathbb{R}^n$ or to the upper halfspace" by "homeomorphic to an open subset of the upper halfspace"? – Michael Albanese Oct 12 '12 at 10:57
except in one-dimension where we also allow $(-\infty,0]$ as a half-space. Well, some books. And, I'd wager the upper halfspace is a convenience of notation. Any homeomorphic set ought to suffice. – James S. Cook Oct 12 '12 at 10:57
@JamesS.Cook: I am a little bit confused by your comment. Do you mean you can use either $(-\infty, 0]$ or $[0, \infty)$ in the definition, or that you have to use both? I don't think it is the latter because the two spaces are homeomorphic via the map $\phi : [0, \infty) \to (-\infty, 0]$, $\phi(x) = -x$. Having said that, I may be missing something. – Michael Albanese Oct 12 '12 at 13:39
@MichaelAlbanese I can't find the book at the moment, but, if memory serves me correctly, the convention of allowing both in the one-dimensional case was to reduce some clutter in certain arguments. I'll try to find the book and give you a more specific quote when I get back to my office. – James S. Cook Oct 13 '12 at 3:13
@MichaelAlbanese my student thinks it's in Conlon, but I cannot find it... sorry. – James S. Cook Oct 26 '12 at 22:46

Let $Y = [0, \infty)$ with the subspace topology inherited from $\mathbb{R}$. Then $X = [0, 1)$ is an open subset of $Y$ and has topological boundary $\{1\}$. However, the geometric boundary of $X$ is $\{0\} \neq \{1\}$.

Note, the topological boundary of a topological manifold $X$ does not make sense in general. First of all, $X$ may not be given as a subspace of some other topological space $Y$ (other than $Y = X$) and if you try to manufacture one, there may be different choices for $Y$, some of which may give different answers. For example, had I chosen $Y = \mathbb{R}$ with its usual topology, then $X$ would have topological boundary $\{0, 1\}$.

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The point Micheal is trying to make is that when you talk about topological manifolds, you think of them (most of the time) as not embedded in any space, that's why the definition of $\partial S = \overline{S}\setminus S^{\circ}$ for topological subspaces doesn't make sense: a manifold exists by itself, not obly if embedded. – Andy Oct 12 '12 at 11:14
@Michael Albanes : No, as a subspace of $[0,\infty)$, $\partial_t[0,1)=\{1\}$ and more generally, an open subset need not have an empty boundary.. all what we know is that it does not contain all its topological boundary points – palio Oct 12 '12 at 11:14
@palio: You are correct, any neighbourhood of $1$ contains points of both $X$ and $Y\setminus X$. I (mistakenly) was only considering points in the boundary of $X$ that were also in $X$. I will edit accordingly. – Michael Albanese Oct 12 '12 at 11:19