Consider the n-sphere $S^n$ in $\mathbb{R}^{n+1}$ or a ball (sometimes referred to as closed disk) $D^n$ in $\mathbb{R}^{n}$ with usual topology. The first inquiry is regarding whether these sets are considered perfect sets, i.e. they are closed (containing all their limit points) and consist only of limit points.
The fact that $S^n$ and $D^n$ are closed (and compact) is clear to me. However, I think they are also perfect since they don't have any isolated points. Is this true?
The second point has to do with their boundaries. Often I read that a sphere has no boundary. However, if we define the boundary of a set to be the set of points in the complement of the interior of the set with respect to its closure, we get that the boundary of the sphere is the sphere itself.
That is, for a set $A$, let $
\bar{A}$ denote the closure of $A$ given by $\bar{A}=A\cup A'$ where $A'$ is the set of all limit points of $A$. Alternatively, $\bar{A}$ is the smallest closed set that has $A$ as a subset.
Let the interior of $A$ be denoted by $\hat{A}$, which is the set of all interior points of $A$, i.e. $p$ is an interior point of $A$ iff $\exists$ an open subset $U$ s.t. $p \in U\subset A$. Alternatively, $
\hat{A}$ is the largest open set contained in $A$.
The boundary of $A$ is just $\partial A=
\bar{A}-\hat{A}$, i.e. points in the closure of $A$ that are not in the interior of $A$.
The same definition yields the boundary of a ball is a sphere as well. The only way I can think of to reconcile the no boundary statement for a sphere with this definition is to restrict neighborhoods to the sphere itself. That is, redefine what we mean by open set and no longer have the usual topology on $
\mathbb{R}^n$. In this way the sphere would be clopen and hence admit no boundary.
Best Answer
You are right $S^n$ and $D^n$ are perfect, and have no isolated points.
I think some of your confusion regarding boundary can be resolved if you clearly state which topological space $X$ you consider, and which subset $A\subseteq X$ you want to know the boundary of.
For example, if $X=\mathbb{R}^{n+1}$ and $A=S^n$, then the boundary of $S^n$ in $\mathbb{R}^{n+1}$ is all of $S^n$ itself.
However, if you, as one often does, take $X=S^n$ to be the topological space, and you consider the nonproper subset $S^n$ in $S^n$, then the boundary is empty! Can you see why?
Similarly, the boundary of the disk $D^n$ as a subset of the space $\mathbb{R}^n$ is the sphere $S^{n-1}$; while the boundary of $D^n$ in the space $D^n$ is empty.