open set in $\mathbb{S}^2$ whose boundary is a finite union of points

Question

I want to prove that an open subset of $\mathbb{S}^2$ whose boundary is a finite union of points must be $\mathbb{S}^2$ minus the points. Am I wrong to think that this is the case? Any ideas on how to prove it. Thanks.

Answer 1

Pick a point $p$ of $S^2$ which is not in your set and not one of the finitely many boundary points. The complement of the closure of your open set is an open set, so $p$ lies in a disk not containing any point of your open set. By Moebius transformation, we can assume that $-p$ lies in your open set. Then each great circle through $p$ strikes a different boundary point of your open set, so the boundary is not a finite set of points.

Answer 2

A finite subset is discrete, so you can find small disks around each boundary points that are disjoint. The complement of the union of the disks is then connected and does not meet the boundary of your open set $A$, thus it must be either contained in $A$ or in its complement. Taking possibly smaller disks, you can ensure that the complement of the disks meets $A$, so that it is contained in $A$. Your set must thus contain the union of the complements for arbitrary small radii, i.e. $A$ is the complement of its boundary.

Answer 3

Let $X$ be a topological space. Let $U$ be an open subset of $X$ and $\partial U=\overline U\setminus U$ be its boundary. Let $V=X\setminus \overline U$. Then $U$ and $V$ form a partition of $X\setminus \partial U$ in open subsets. If $X\setminus \partial U$ is connected, then either $U=\emptyset$ (in which case $\partial U=\emptyset$ as well), or $V=\emptyset$. You probably want to avoid the first case, and in the latter case, $\overline U=X$ and and $U=X\setminus \partial U$.

In your example one has $X=\mathbf S_2$ and the complement of any finite set of points is connected (for any reason you want, for example, because there is a family of circles passing through two points, and pairwise disjoint except for the fixed points, so that most of them avoid the given finite set).

Answer 4

(a) Let $X$ be an arbitrary topological space, and $U,V$ open subsets with the same boundary $B$, and $W=U\cap V$. Then the boundary of $W$ is contained in $B$ (immediate).

(b) Next, if $U,W$ are open subsets of $X$ with $W\subset U$ and the boundary of $W$ is contained in the boundary of $U$, then $W$ is open and closed in $U$ (immediate too).

(c) In the setting of (a), if $U$ is connected and $U\cap V\neq\emptyset$ (which is automatic if $U$ is dense and $V$ nonempty), we deduce from (b) that $V$ contains $U$. If $U$ is dense then we deduce that $V$ is connected, which implies by the same argument that $U$ contains $V$, so $U=V$.

So if $X$ is a topological space in which the complement of any finite subset is connected and dense, then any open subset with nonempty finite boundary is equal to the complement of its boundary.

Answer 5

Of course you have to assume that your open subset $\Omega$ is nonempty. In such case, notice that $S^2\setminus\partial\Omega$ is still connected (why?). Your original subset $\Omega$ is both open and closed in $S^2\setminus\partial\Omega$ (why?), so by connectedness $\Omega=S^2\setminus\partial\Omega$.