Proving the boundary of a disc is a circle As the title suggests, I already know that the boundary of a disk is the circle with the same radius, but what I'm interested in is actually proving that fact.
I recently got around to the section of my topology book that deals with closure, interior, and boundaries of a subset and I've been having a great deal of trouble actually finding any of these for a given subset. I know what they are, but actually finding these things for a given subset is fairly difficult for me. 
I want to emphasize that when I say it's difficult, I don't mean that I have no idea what they are in reference to that subset, but that I don't know how to rigorously find them or prove one set is the closure/boundary/interior of the other.
I used the disk with circular boundary mostly as a springboard example, if you know of another example which may elucidate this a little more, feel free to use that one.
I'm aware there is no one process that will definitely find it every time, but if there is some way of thinking about the question that can help make it a little easier or some useful theorems to aid in finding these things then I'd be immensely grateful.
 A: I find the following characterizations extremely helpful for both intuition and also computations: 


*

*The interior of a set $A$ is the largest open set that is contained in $A$. 

*The closure of a set $A$ is the smallest closed set containing $A$.
So for instance, when you try to argue that the disk $$D=\{x \in \mathbb{R}^n \mid  \Vert x \Vert \leq1 \}$$ is the closure of the set $$U=\{x \in \mathbb{R}^n \mid \Vert x \Vert < 1 \},$$ you can ask yourself the following questions: "Is $D$ closed? And if yes, are there any other intermediate sets $U \subseteq B \subsetneq D$ that are closed?" If you can check that the answer of the latter is no, then you are done by 2.
In order to conclude that the boundary is $$S=\{x \in \mathbb{R}^n \mid \Vert x \Vert=1\}$$ you just need to observe that the interior of $U$ is $U$, because $U$ is already open (see 1.).
A: Hope the following proof helps.
So with $B=B_{r}(a)=\{x\in{\bf{R}}^{n}:|x-a|<r\}$, we are to prove that $\partial B=S_{r}(a)=\{x\in{\bf{R}}^{n}:|x-a|=r\}$.
Given $\epsilon>0$ and $x\in S_{r}(a)$, we need to find $x_{\epsilon}\in B$, $y_{\epsilon}\in B^{c}$ such that $x_{\epsilon},y_{\epsilon}\in B_{\epsilon}(x)$, this will prove the inclusion that $S_{r}(a)\subseteq \partial B$.
Let $\eta\in(0,\min\{r,\epsilon\})$, we define that $x_{\epsilon}=a+\eta\cdot\dfrac{x-a}{|x-a|}=a+\eta\cdot\dfrac{x-a}{r}$, $y_{\epsilon}=x+\eta\cdot\dfrac{x-a}{|x-a|}=x+\eta\cdot\dfrac{x-a}{r}$, then we have $|x_{\epsilon}-x|=\eta<\min\{r,\epsilon\}$, so $x_{\epsilon}\in B_{\epsilon}(x)\cap B$, $|y_{\epsilon}-a|=|x-a|\left(1+\dfrac{\eta}{r}\right)=r\left(1+\dfrac{\eta}{r}\right)=r+\eta>r$, so $y_{\epsilon}\in B^{c}$, also that $|y_{\epsilon}-x|=\eta<\epsilon$, so $y_{\epsilon}\in B_{\epsilon}(x)$, and hence $y_{\epsilon}\in B_{\epsilon}(x)\cap B^{c}$, as promised.
Similar reasoning we can show that $\partial B\subseteq S_{r}(a)$.
A: A point $ b $ is in the  boundary of a set $A$ if every open ball B(b;r) contains 
 a point inside $A$ and a point outside $A$.
Note that in metric topology drawing a graph is often helpful. 
For a disk $B(b;r )$ the boundary is the circle $d(x,b)=r $.
Because every open ball centered at $x$ has points for which $d(x,b)>r $ 
and points for which  $d(x,b)<r$.
Thus $x$ is a boundary point.
