Exercise 3.3.8

Question

Let $K$ and $L$ be nonempty compact sets, and define
  $$
  d=\inf \{|x-y|: x \in K 	ext { and } y \in L\}
  $$
  This turns out to be a reasonable definition for the distance between $K$ and $L$.
  \begin{enumerate}[label=(\alph*)] 
  \item If $K$ and $L$ are disjoint, show $d>0$ and that $d=\left|x_{0}-y_{0}ight|$ for some $x_{0} \in K$ and $y_{0} \in L$.
  \item Show that it's possible to have $d=0$ if we assume only that the disjoint sets $K$ and $L$ are closed.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item The set $|K - L| = \{|x-y| : x \in K, y \in L\}$ is compact since $K-L$ is compact by 3.3.6 (b) and $|\cdot|$ preserves compactness. Thus $d = \inf |K-L|$ has $d = |x_0 - y_0|$ for some $x_0 \in K$ and $y_0 \in L$.
  \item $K = \{n : n \in \mathbf N\}$ and $L = \{n + 1/n : n \in \mathbf N\}$ have $d = 0$, and both are closed since every limit diverges.
   \end{enumerate}

Exercise 3.3.8

Answers

Comments

Exercise 3.3.8

Answers

Comments

Add answer