Munkres, Lemma 16.2: Existence of $C_x$ s.t it is disjoint from $D_{i-2}$

Question

In the book of Analysis on Manifolds by Munkres, at page 137, it is given that

I have doubts about the existence of such a $C_x$, so just to make sure, I wanted the prove the existence of such a set $C_x$.

Proof:

Let assume that for some $x_0 \in B_i$, every neighbourhood $C_{x_0}$ of $x_0$ has a non-empty intersection with $D_{i-2}$. Then since $B_i \subseteq Ext(D_{i-2})$, and $C_{x_0} \cap B_i \not = \emptyset$, then it must be true that $$x_0 \in \partial D_{i-2} \subseteq D_{i-2},$$ but $B_i$ and $D_{i-2}$ are disjoint, a contradiction.

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Is there any flaw in the proof ? I'm particularly asking this because it took me 1 day to figure out this contradiction out clearly.

Answer 1

I don't feel like reading your proof, but here's the proof that comes to my mind. $B_i$ and $D_{i-2}$ are disjoint compact sets, so they are some positive distance apart from each other. So given any $x \in B_i$, we can take a small enough cube centered at $x$ that is disjoint from $D_{i-2}$ (we can insist the cube is contained in $A$, since $A$ is open).

Answer 2

I think your proof is correct.

Proof:

Let assume that for some $x_0 \in B_i$, every neighbourhood $C_{x_0}$ of $x_0$ has a non-empty intersection with $D_{i-2}$. Then since $B_i \subseteq Ext(D_{i-2})$, and $C_{x_0} \cap B_i \not = \emptyset$, then it must be true that $$x_0 \in \partial D_{i-2} \subseteq D_{i-2},$$ but $B_i$ and $D_{i-2}$ are disjoint, a contradiction.

I understood your proof as follows:

Assume that there exists $x_0\in B_i$ such that every closed cube $C_{x_0}$ centerd at $x_0$ is not disjoint from $D_{i-2}$.
Since $D_{i-2}\subset\operatorname{Int}D_{i-1}$, $\mathbb{R}^n-\operatorname{Int}D_{i-1}\subset\mathbb{R}^n-D_{i-2}.$
And $B_i\subset\mathbb{R}^n-\operatorname{Int}D_{i-1}.$
So, $B_i\subset\mathbb{R}^n-D_{i-2}.$
$\mathbb{R}^n-D_{i-2}$ is an open set in $\mathbb{R}^n$.
So, $\mathbb{R}^n-D_{i-2}=\operatorname{Ext}D_{i-2}.$
Therefore, $B_i\subset\operatorname{Ext}D_{i-2}.$

Let $C_{x_0}$ be an aribitrary closed cube centerd at $x_0$.
Since $x_0\in C_{x_0}\cap B_i$, $C_{x_0}\cap B_i\neq\emptyset.$
And $C_{x_0}\cap B_i\subset C_{x_0}\cap\operatorname{Ext}D_{i-2}.$
So, $C_{x_0}\cap\operatorname{Ext}D_{i-2}\neq\emptyset.$
So, $C_{x_0}\cap (\mathbb{R}^n-D_{i-2})\neq\emptyset$ since $\mathbb{R}^n-D_{i-2}=\operatorname{Ext}D_{i-2}.$

By our assumption, $C_{x_0}\cap D_{i-2}\neq\emptyset.$

So, $x_0\in\operatorname{Bd}D_{i-2}$ by the definition of the boundary of a set.
Since $D_{i-2}$ is an closed set in $\mathbb{R}^n$, $\operatorname{Bd}D_{i-2}\subset D_{i-2}$.
So, $x_0\in D_{i-2}.$

So, $x_0\in B_i\cap D_{i-2}.$
Since $B_i\subset\operatorname{Ext}D_{i-2}$, this is a contradiction.