I'd like to undersand better Brouwer separation theorem given in Massey (Proposition 6.5 p.215) since has some smoky parts to me. To lighten the notation we set $D^n := \mathbb{D}^n, S^n := \mathbb{S}^n$
Lemma $6.2$ would be cited so I'm going to cite the statement:
Lemma: Let $Y$ be a subset of $S^n$ which is homeomorphic to $I^k$ where $0 \leq k \leq n$. Then $\tilde{H}_i(S^n-Y) = 0$ for all $i$.
Furthermore let's denote $C_0,C_1$ the two connected components of the follwing:
Proposition: Let $A \subset S^{n}$ homeomorphic to $S^{n-1}$, then $A$ is the boundary of each components of $S^n-A$.
Proof: In order to prove the Proposition we must show that :
- The boundary of each component is a subset of $A$.
- Any point $a \in A$ is a boundary point of each component of $S^n-A$.
To prove the first, Massey asserts that it's sufficient to know that $S^n-A$ is open (but I don't understand why, any help would be appreciated). I'd like to give my explanation and ask if it's correct:
Let's suppose for the sake of contradiction that $\exists p \in \overline{C_0}^{S^n} \cap C_1$.
Since $p$ belongs to the first set, for every neighborhood $U$ of $p$ exists a sequence $(x_n)_n \in C_0$ such that $x_n \to p$ and $x_n \in U$ definitely. Since $C_1$ is open in $S^n$, exists $V$ neighborhood, with $p \in V \subset C_1$. This leads to contradiction since we then would have $C_0 \cap C_1 \ne \varnothing$.
Is this reasoning correct?
The proof prooceds as follows to prove the second point: Let N be any open neighborhood of $a$ in $S^n$; we must show that $N \cap C_i \ne \varnothing$ for $i=0,1$.
Note that $N \cap A$ is an open neighborhood of $a \in A$. Since $A$ is homeomorphic to $S^{n-1}$, we can find a decomposition $$A = A_1 \cup A_2$$ where $A_i$ is homeomorphic to $D^{n-1}$ and $A_1 \cap A_2 \simeq S^{n-2}$, with $A_2 \subset N \cap A$.
We know from Lemma $6.2$ that $S^n - A_1$ is arcwise connected. Let $p_0 \in C_0,p_1 \in C_1$ and choose an arc $ f: I \longmapsto S^n - A_1$ that joins them. Since $f(I) \cap A \ne \varnothing$ we must have $f(I) \cap A_2 \ne \varnothing.$
Consider the subset $f^{-1}(A_2) \subset I$; this is a compact subset of $I$ and hence it munst have a least a point $t_0$ and a greatest point $t_1$. Obviously $t_0,t_1$ are boundary points of $f^{-1}(A_2)$, and $f^{-1}(N)$ is an open subset of $I$ which cointains both. From this follows by an easy arguments that $f^{-1}(N) \cap f^{-1}(C_i) \ne \varnothing$ for $i=0,1$ which completes the proof.
Question : Here I don't understand what's the easy argument and why we have the necessity to use $t_0,t_1$ which arises from compactness of $f^{-1}(A_2)$. My argument was the following:
Since $I = f^{-1}(N) \cup f^{-1}(C_0) \cup f^{-1}(C_1)$ those are three open subset of $I$ which is connected, so we must have intersection, but $f^{-1}(N)$ necessarly $f^{-1}(N)$ has to lie in the middle of $I$ since $f^{-1}(C_1)$ and $f^{-1}(C_0)$ can't intersect; so we have $f^{-1}(N) \cap f^{-1}(C_i) \ne \varnothing$ for $i=0,1$ as requested.
Where this fails?
This argument doesn't use that $f^{-1}(A_2)$ is compact since it only uses that $A_2 \subset N$, hence $f^{-1}(A_2) \subset f^{-1}(N)$. I don't really see where this argument fails when $f^{-1}(A_2)$ is not compact.
Any help to understand the necessity of the assumptions and correctness of my arguments would be appreciated.