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!!
 A: 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...
A: 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 $\{0, 1\}$.
