Topological boundary vs geometric boundary Let $M_1=B((0,0),1)=\{(x,y) \mid x^2+y^2<1\}$
$M_2=\{(x,y) \mid x^2+y^2\le1\}$
What are the interior of $M_1$ and $M_2$ ?
And  what are the boundary of $M_1$ and $M_2$ ? 
How do I find them? Please show me so I can understand. 
I know the answers but not the solutions.  
Moreover - considering these examples - how can we say that topological boundaries are different from the geometric boundaries? 
Please explain this in a clear and instructive way.
These are examples from my notebook which I need to understand well.
Thank you for help. 
 A: Interior (topological) $\operatorname{int} A$ of a subset $A$ of a topological space $X$ is the largest open set contained in $A$.
Topological boundary $\partial A$ of a subset $A$ of a topological space $X$ can be defined in various, equivalent ways, for instance as the difference $\overline A\setminus \operatorname{int} A$ between the closure of $A$ and its interior, or as the intersection $\overline A\cap \overline{A^c}$of closure of $A$ and closure of its complement, or more directly as the set of points whose every neighbourhood intersects $A$ and its complement.
Geometric boundary is something more inherent: when you have a manifold $M$ with boundary, it comes with an associated atlas of homeomorphisms (maps) $\varphi_\alpha:U_\alpha \to M$, where $U_\alpha$ is an open subset of $R^n_+=\{\overline x=(x_1,\ldots,x_n)\in {\bf R}^n\mid x_1\geq 0\}$ which agree in some way. Boundary $\partial M$ is defined then to be $\bigcup_\alpha \varphi_\alpha[U_\alpha\cap \{\overline x\mid x_1=0\}]$. $M\setminus \partial M$ can be referred to as the (geometric) interior of $M$.
For topological boundary and interior, $M_1$ is open and is in fact the interior of $M_2$ (because any ball containing points from the circle contains points from outside it), and you can use any of the two definitions to see that the boundary of both is the unit circle.
For geometric boundary and interior, with the natural manifold structure $M_1$ has no boundary (it is an open subset of ${\bf R}^2$), while $M_2$ has as its boundary, again, the unit circle; consequentially, geometric interior of both is $M_1$.
A: Let $X$ be a topological space.  The interior of a set $S\subset X$, denoted $S^{\circ},$ is the collection of all points $s\in S$ such that there is an open neighborhood of $s$, say $U$, with $s\in U\subset S$.
Notice that for any point in $(x_0,y_0)\in M_1$, we have $x_0^2+y_0^2<1$, hence letting $\epsilon=1-x_0^2+y_0^2$, and $U=B((x_0,y_0),\epsilon)$, we have $(x_0,y_0)\in U\subset M_1$.  This shows that $M_1=M_1^{\circ}$ as is true for any open set.
See if you can show that $M_2^{\circ}=M_1$.
As for boundaries, the topological boundary, $\partial S,$ of a set $S\subset X$ can be characterized in a few ways.  Here are the two that are most helpful for me.


*

*$\partial S=\overline{S}\setminus S^{\circ}$, that is, the boundary is the closure minus the interior.

*The boundary of a set $S$ consists of all points in $x\in X$ such that all neighborhoods of $x$ intersect both $S$ and $S^c$.
The first definition is more technical, while the second is a little more intuitive.  Suppose $(x_0,y_0)$ satisfies $x_0^2+y_0^2=1$.  Notice that any open ball around $(x_0,y_0)$ will contain points whose distance from the origin is less than $1$, and points whose distance from the origin is greater than $1$.  In other words, any open ball around $(x_0,y_0)$ intersects both $M_1$ and $M_1^c$.  It follows that $\partial M_1=\{(x,y)\mid x^2+y^2=1\}$.
See if you can show that the boundary of $M_2$ is the same.
To your question about topological boundary vs. geometric boundary, see the other answers.
A: The geometric boundary is the set of points where you have to use (a relatively open subset of) half-space $\mathbb H^2$ for a chart at that point. For $M_1$ this is clearly the empty set since every point has an open ball around it inside $M_1$. For $M_2$ it's the unit circle since charts for the points "on the edge" cannot use coordinates coming from an open $U \subseteq \mathbb R^2$.
