Are spheres and balls perfect sets? What is their boundary?

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.

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.

• And similar "relativity" exists for terms like open and closed and interior and closure. For example, if you take $S^n$ as a subset of the topological space $S^n$ itself, then it is an open subset (in fact clopen). Every space is open (clopen) as a subset of itself (that is a part of the definition of a space). But $S^n$ as a subset of $\mathbb{R}^{n+1}$ is not open. Similarly, the interval $\left( \frac12 , 1 \right]$ is open in space $[0,1]$, but $\left( \frac12 , 1 \right]$ is not open in $\mathbb{R}$. May 7, 2020 at 14:56
• If $X=S^n$ then the improper subset $S^n$ would be the space itself and hence clopen by definition. A set is clopen iff its boundary is the empty set. I don't like this argument though. Do you know a different way to see why the improper subset $S^n$ of $X=S^n$ admits no boundary?
– Rem
May 7, 2020 at 16:07

There are (at least) two independent meanings of the word "boundary" in the literature of topology.

First Meaning: Given a topological space $$X$$ and a subspace $$A \subset X$$, the boundary of $$A$$ in $$X$$ is defined to be $$\overline A - \hat A$$. One can equivalently define the boudnary to be $$\overline A \cap \overline{X-A}$$, and I prefer that definition because it emphasizes an important feature of the boundary, namely that it is a relative property of $$A$$ meaning a property of $$A$$ relative to the space $$X$$.

However, that first meaning is not the intended meaning in the sentence "a sphere has no boundary". Instead:

Second Meaning: This notion of "boundary" is defined for the theory of manifolds. An $$m$$-dimensional manifold-with-boundary is a topological space that is locally modelled on $$\overline H^m = \{(x_1,...,x_m) \in \mathbb R^m \mid x_m \ge 0\}$$ In other words, every point of $$M$$ has an open neighborhood homeomorphic to some open subset of $$\overline H^m$$ (if one is studying manifolds in the context of calculus there are additional requirements regarding smoothness of overlap maps, which I am ignoring; so, I'm talking solely about the theory of topological manifolds-with-boundary). Inside $$\overline H^m$$ we have its "interior" which is the subset $$H^m = \{(x_1,...,x_m) \in \mathbb R^m \mid x_m > 0\}$$ If $$M$$ is a manifold with boundary then the manifold interior of $$M$$, denoted $$\text{int}(M)$$, is defined the set of points in $$M$$ that have an open neighborhood homeomorphic to some open subset of $$H^m$$. Finally, the manifold boundary of $$M$$ is defined to be $$\partial M = M - \text{int}(M)$$. Often, when all one is talking about are "manifolds", then one often drops the word itself and speaks just of the interior of $$M$$ and the boundary of $$M$$. This is the sense of boundary that is intended in the sentence "a sphere has no boundary".

Regarding this terminological confusion, For the "First Meaning" I prefer to use the term frontier instead of boundary, as I learned from Munkres book "Topology", and as is explained here.