I'm trying to understand the proof of Lemma 7 from Terence Tao's blog, i.e.,
Let $X$ be an open subset of $\mathbb R^n$ and $f:X \to \mathbb R^n$ be differentiable such that $\partial f (x)$ is invertible for all $x\in X$. Assume that
- $f(0)=0$ and $\partial f (0)= I$ where $I$ is the identity map on $\mathbb R^n$.
- $|x| = 1 \implies |f(x)| > \frac{1}{2}$.
- $|x| = \frac{1}{20} \implies |f(x)| < \frac{1}{10}$.
- $x \in B(0, \frac{1}{20}) \implies f(x) \in B(0, \frac{1}{10})$. Equivalently, $f [B(0, \frac{1}{20})] \subset B(0, \frac{1}{10})$.
- $x_1, x_2 \in B(0, \frac{1}{20})$ such that $y :=f(x_1) = f(x_2) \in B(0, \frac{1}{10})$.
Let $$ K_r := \{x \in \Omega : |f(x)-y| \le r\} \quad \forall r \ge 0. $$
Then $K_r$ is closed and $x_1, x_2 \in K_r$. Let $K_r^1$ be the connected component of $K_r$ that contains $x_1$. Clearly, $r \le s$ implies $K^1_r \subset K_s^1$.
Lemma 5: $K_0^1 = \{x_1\}$.
Lemma 6: $B(0, \frac{1}{20}) \subset K_r^1 \subset B(0,1)$ for all $\frac{2}{10} \leq r \leq \frac{4}{10}$. In particular, $K_r^1$ is compact for all $0 \leq r \leq \frac{4}{10}$, and contains $x_2$ for $\frac{2}{10} \leq r \leq \frac{4}{10}$.
Lemma 7 If $0 \leq r<\frac{1}{20}$ and $\epsilon>0$, then for $r<r^{\prime}<\frac{1}{20}$ sufficiently close to $r, K_{r^{\prime}}^1$ is contained in an $\epsilon$-neighbourhood of $K_r^1$.
Proof of Lemma 7:
By the finite intersection property, it suffices to show that $\bigcap_{r^{\prime}>r} K_{r^{\prime}}^1=K_r^1$. Suppose for contradiction that there is a point $x$ outside of $K_r^1$ that lies in $K_{r^{\prime}}^1$ for all $r^{\prime}>r$. Then $x$ lies in $K_{r^{\prime}}$ for all $r^{\prime}>r$, and hence lies in $K_r \cap B(0,1)$.
As $x$ and $x_1$ lie in different connected components of the compact set $K_r \cap \overline{B(0,1)}$ [recall that $K_r$ is disjoint from $\partial B(0,1)$], there must be a partition of $K_r \cap \overline{B(0,1)}$ into two disjoint closed sets $F$, $G$ that separate $x$ from $x_1$ (for otherwise the only clopen sets in $K_r \cap \overline{B(0,1)}$ that contain $x_1$ would also contain $x$, and their intersection would then be a connected subset of $K_r \cap \overline{B(0,1)}$ that contains both $x_1$ and $x$, contradicting the fact that $x$ lies outside $K_r^1$ ).
By normality, we may find open neighbourhoods $U, V$ of $F, G$ that are disjoint. For all $x$ on the boundary $\partial U$, one has $\|f(x)-y\|>r$ for all $x \in \partial U$. As $\partial U$ is compact and $f$ is continuous, we thus have $\|f(x)-y\|>r^{\prime}$ for all $x \in \partial U$ if $r^{\prime}$ is sufficiently close to $r$. This makes $U \cap K_{r^{\prime}}$ clopen in $K_{r^{\prime}}$ and so $x$ cannot lie in $K_{r^{\prime}}^1$ giving the desired contradiction.
My questions:
Could you explain the reasoning behind "By the finite intersection property, it suffices to show that $\bigcap_{r^{\prime}>r} K_{r^{\prime}}^1=K_r^1$"?
Why does the author consider $K_r \cap \overline{B(0,1)}$, but not $K_r \cap B(0,1)$? It seems to me with $K_r \cap B(0,1)$, we don't know if $F, G$ are closed in $\mathbb R^n$, and thus we can not apply the normality of $\mathbb R^n$ to get $U, V$.
Why does [$K_r$ is disjoint from $\partial B(0,1)$] imply [$x$ and $x_1$ lie in different connected components of the compact set $K_r \cap \overline{B(0,1)}$]?
Why does [$U \cap K_{r^{\prime}}$ clopen in $K_{r^{\prime}}$] imply [$x$ cannot lie in $K_{r^{\prime}}^1$]?
Thank you so much for your help!
Question 1:
Let $W$ be an $\epsilon$-neighbourhood of $K_r^1$. Assume we know that $\bigcap_{r^{\prime}>r} K_{r^{\prime}}^1=K_r^1$. Since $K^1_r \subset K_s^1$ for $r \le s$, we see that $\bigcap_{4/10 \ge r^{\prime}>r} K_{r^{\prime}}^1=K_r^1$. We have $\bigcap_{4/10 \ge r^{\prime}>r} (K_{r^{\prime}}^1 \setminus W) = (\bigcap_{4/10 \ge r^{\prime}>r} K_{r^{\prime}}^1) \setminus W = K_r^1 \setminus W = \emptyset$. By Lemma 6 $K_{r'}^1$ is compact for $r' \le 4/10$, hence all $K_{r^{\prime}}^1 \setminus W$ are closed in the compact space $K_{4/10}^1$. Their intersection is empty, thus by the finite intersection property there must be finitely many $r'_1,\ldots,r'_n > r$ such that $\bigcap_{i=1}^n (K_{r_i^{\prime}}^1 \setminus W) = \emptyset$. Denote by $r^*$ the smallest of these $r_i'$. Then $\bigcap_{i=1}^n (K_{r_i^{\prime}}^1 \setminus U) = K_{r^*}^1 \setminus W$. Therefore $K_{r^*}^1 \subset W$ so that $K_{r'}^1 \subset W$ if $r < r' \le r^*$.
Question 2:
Since $K_r$ is disjoint from $\partial B(0,1)$, we have $K_r \cap \overline{B(0,1)} = K_r \cap B(0,1)$. The LHS of this equation shows that this set is compact; using the RHS would not guarantee this. This enables us to use There is a partition of $X$ into disjoint closed sets $F, G$ such that $a\in F$ and $b\in G$.
Question 3:
Concerning the use of "$K_r$ is disjoint from $\partial B(0,1)$" see 2. Since $K^1_r$ is the connected component of $K_r$ that contains $x_1$ and $x \notin K^1_r$, the point $x$ must lie in a different connected component of $K_r$. This makes it impossible that $x, x_1$ lie in the same connected component of $K_r \cap \overline{B(0,1)} \subset K_r$.
Question 4:
Tao is imprecise here.
Tao does not say in which set $X \supset K_r \cap \overline{B(0,1)}$ the disjoint sets $U \supset F, V \supset G$ are supposed to be open. So let us take $X = \overline{B(0,1)}$ which is compact; then $\partial U \subset \overline{B(0,1)}$ is in fact compact and does not intersect $K_r$. Tao's argument now shows that $U \cap K_{r^{\prime}} \cap \overline{B(0,1)}$ is clopen in $K_{r^{\prime}} \cap \overline{B(0,1)}$ for $r' \le 4/10$ sufficiently close to $r$. We have $x \in F \subset U \cap K_{r^{\prime}} \cap \overline{B(0,1)}$ and $x_1 \in G \subset V \cap K_{r^{\prime}} \cap \overline{B(0,1)}$. Thus $x_1 \notin U \cap K_{r^{\prime}} \cap \overline{B(0,1)}$ and we conclude that $x, x_1$ do not lie in the same connected component of $K_{r^{\prime}} \cap \overline{B(0,1)}$. But we know that $K_{r^{\prime}}$ is disjoint from $\partial B(0,1)$, thus all connected components of $K_{r^{\prime}} \cap \overline{B(0,1)}$ are also connected components of $K_{r^{\prime}}$. Therefore $x, x_1$ do not lie in the same connected component of $K_{r^{\prime}}$. This contradicts the assumption that $x \in K_{r^{\prime}}^1$ for all $r' > r$.