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!!