Every nonempty open subset of $S^{n-1}$ contains an (open) subset whose complement is a $(n-1)$-cell
This above result was used in the the proof of the Jordan-Brouwer Seperation Theorem in Introduction to Algebraic Topology by Rotman.
Note that $(n-1)$-cell in this case is any topological space homeomorphic to $D^{n-1}$, the $n-1$ disk.
I'm trying to prove this result, however I'm running into some difficulty. The first thing to note is if we choose a non-empty open subset $U$ of $S^{n-1}$, and we choose another open subset of $U$, which I'll call $V$, the complement $V^c$ must be relative to $S^{n-1}$ otherwise we could have $V^c$ being disconnected.
Now the one reason Rotman gives as to why this result is true is that
every open set contains a homeomorphic copy of $\mathbb{R}^{n-1}$
where I guess he's using the fact that $S^{n-1}$ is a $n-1$ dimensional topological manifold and so locally Euclidean, but even so, I don't see how this helps us show that $V^c$ is homeomorphic to $D^{n-1}$. The only thing I can think of is that since $V^c$ is closed and $S^{n-1}$ is compact we have $V^c$ to be compact and since $D^{n-1}$ is compact so there's some similarity already between
the two spaces.
Edit: Since $V^c$ is compact and $D^{n-1}$ is Hausdorff it suffices to find a bijective continuous map $f : V^c \to D^{n-1}$, but actually constructing such a $f$ is proving hard to do. Like in the case $n=1$ and $n=2$, I can sorta see what such a map should do, but still I don't know how to write down an explicit formula for it
How can I prove this result?
Best Answer
Let $U \subset S^n$ be open and $x \in U$. There exists a homeomorphism (in fact a rotation) $h : S^n \to S^n$ such that $h(x) = n = (0,\dots,0,1)$. Let $V = h(U)$. Define $S^n_+ = \{ x \in S^n \mid x_{n+1} > 0 \}$. This is an open hemisphere containing $n$. Let $B^n(r) = \{ y \in \mathbb{R}^n \mid \lVert y \rVert < r \}$.
The projection $p : S^n_+ \to B^n(1), p(x_1,\dots,x_{n+1}) = (x_1,\dots,x_n)$, is a homeomorphism. Choose $r \in (0,1)$ such that $V(r) = p^{-1}(B^n(r)) \subset V \cap S^n_+$. Then $S^n \setminus V(r)$ is an $n$-cell. To see this, use the stereographic projection $s : S^n \setminus \{ n \} \to \mathbb{R}^n$ which is a homeomorphism. Then obviously $s(S^n \setminus V(r))$ is a closed $n$-cell with center $0$ and radius $> 1$.
This shows that $U' = h^{-1}(V(r)) \subset U$ is the desired set.