In one of the properties proof for the Lebesgue measure, Stein and Shakarchi (2009) uses the fact that there is a strictly positive distance between compact and closed sets. I am following this mini-proof and have one question.
Claim:
Let $d(,)$ be the usual metric. If $F$ is closed, $K$ is compact, and they are disjoint, then $d(F,K)>0$Sketch of the Proof:
1. $F$ is closed, so take any point $x\in K$, then you can create a ball with some radius that doesn't touch $F$.
2. Arbitrarily, we can make the distance such that $d(x,F)>3\delta_x$.
3. Take the union of the balls with radius $2\delta_x$ that cover $K$.
4. By the Heine-Borle theorem, there is a finite subcover such that $\bigcup^N_{j=1} B_{2\delta_j}(x_j)$.
5. Let $\delta=\min(\delta_1,…,\delta_N)$, then we must have $d(K,F)\geq\delta>0.$
My Question:
- Why is #5 true?
Reference:
Real Analysis: Measure Theory, Integration, and Hilbert Spaces. Elias M. Stein, Rami Shakarchi. Princeton University Press, 2009.
Best Answer
From 4, we know that $K$ can be covered by a finite number of balls $B_{2\delta_j}(x_j)$ each disjoint from $F$. Thus if $a \in K$ and $b \in F$, then $a \in B_{2\delta_j}(x_j)$ for some $j$ and so by the triangle inequality $$d(x_j,b) \leq d(x_j,a) + d(a,b) < 2\delta_j + d(a,b)$$ $$\implies d(a,b) > d(x_j,b) - 2\delta_j > 3\delta_j - 2\delta_j = \delta_j.$$ Since $d(F,K) = \{\inf d(a,b) : a\in K, b \in F\}$, we have therefore shown that $d(F,K) \geq \delta$ where $\delta$ is the minimum of all the $\delta_j$'s.
Intuitively, you know that points in $K$ have to be close to the finite number of points $x_j$ by part 4., and each of them is at least some distance away from $F$ by part 1., so combining we can show that any element of $K$ is also some distance away from $F$.