How to understand "boundary" and "open set" at the point of the view "manifold"? Let's consider the 3-dimension for a surface $M \subset  \mathbb{R^3}$
I haven't studied "Manifold-theory" yet, But When I studied Gauss-bonnet theorem in Differential geometry, People regard the boundary of a surface at the point of the manifold.
Then, What is the open set of the 3-dimensional at the point of the manifold? 
There are two cases. Could you explain the reason why does the below fact like that?
(Please give me intuitive way to understand those, because I haven't studied the manifold theory.)
$(1)$ Sphere(hollow ball) : People said there are no boundary for this surface.
$(2)$ Ball : People said this has a boundary that a sphere like  $\{ (x,y) \vert x^2 + y^2+z^2 = r^2\}$
p.s.) In my guess, the open set regarding the manifold point is $M \cap U $ for the surface, $M$ (Here the $U$ is a open set in Usual topology in $(\mathbb{R}^3, U)$). But this thought can't applied for the case $(2)$ in my trial.
Please help me.
 A: Imagine the situation one dimension lower: with the "1-dimensional" circle and the "solid" disc.

On the left, there is a circle.
It is shown that the neighborhood of a points on the cicrle is homeomorphic to $\Bbb R$.
This is true for every point, so the circle is a 1-dimensional manifold without boundary.
The disc on the right has two kinds of points.
For some, a small neighborhood is homeomorphic to $\Bbb R^2$, and for some homeomorphic to $\Bbb R\times\Bbb R_+$ (the positive half-plane).
The latter points are the ones on the boundary of the disc, which is now seen to be a 2-dimensional manifold with boundary.
The situation on the sphere is equivalent to the one on the circle:
every point has a neighborhood homeomorphic to $\Bbb R^2$, making the sphere a 2-dimensional manifold without boundary.

You probably know now how it works for the solid ball, but this is trickier to draw.
A: If $X\subseteq Y$, where $Y$ is a topological/metric space then typically $X$ is considered with the subspace topology. Meaning $U\subseteq X$ is open if and only if there is an open subset $V\subset Y$ such that $U=X\cap V$.
A boundary of a manifold is not the same thing as topological boundary. A topological boundary of the sphere is the sphere itself, while it has no manifold boundary.
Normally a manifold (without boundary) is a topological space $M$ such that each point $p\in M$ has an open neighbourhood $U\subseteq M$ such that $U$ is homeomorphic to $\mathbb{R}^n$. On the other hand a manifold with boundary is a topological space $M$ such that each point $p\in M$ has an open neighbourhood homeomorphic to $\mathbb{R}^n_+=[0,\infty)^n$. In that situation those points that are mapped to $(x_1,\ldots, x_n)\in\mathbb{R}^n_+$ with some $x_i=0$ are called boundary points (this definition does not depend on the choice of mappings).
The simpliest example is: $(0,1)\subseteq\mathbb{R}$ is a $1$-dimensional manifold without boundary while $[0,1]\subseteq\mathbb{R}$ is a $1$-dimensional manifold with boundary $\{0,1\}$. Also $[0,1)\subseteq\mathbb{R}$ is a $1$-dimensional manifold with boundary $\{0\}$. Analogously the sphere $S^1\subseteq\mathbb{R}^2$ is a $1$-dimensional manifold without boundary while the (closed) ball $D\subseteq\mathbb{R}^2$ is a $2$-dimensional manifold with boundary. The boundary of $D$ is the sphere $S^1$.
The intuition here is that if we take the ball $D=\{v\in\mathbb{R}^n\ |\ \lVert v\rVert\leq 1\}$ then points from $\partial D=\{v\in D\ |\ \lVert v\rVert =1\}$ don't have an open neighbourhood homeomorphic to $\mathbb{R}^n$. Unlike points from $int(D)=\{v\in D\ |\ \lVert v\rVert < 1\}$. The boundary prevents that. However if we keep magnifying the disk near $\partial D$ then we can see that it looks more and more like $\mathbb{R}^n_+$.
Side note: if $M$ is an $n$-dimensional manifold with boundary, then the boundary $\partial M$ is an $n-1$-dimensional manifold without boundary.
A: Simply put, a 3-dimensional manifold looks locally like a neighborhood in $\mathbb{R}^3$ where a $3$-dimensional manifold with boundary $M$ looks locally like a neighborhood in the half plane 
$$ \mathbb{H}^3 = (-\infty, 0]\times \mathbb{R}^{2}. $$
In particular, each neighborhood of a boundary point on $M$ corresponds to a neighborhood of a boundary point of the half plane.
For more detail and formalism, compare the formal definitions of manifolds and manifolds with boundary.
