0
$\begingroup$

Let $(X, d)$ be a metric space and let $A$ and $B$ be subsets of $X$. Define $d(A,B) = \inf\{d(a, b) : a \in A, b\in B\}$. Pick out the true statements.

a. If $A$ and $B$ are disjoint, then $d(A,B) > 0$.

b. If $A$ and $B$ are closed and disjoint, then $d(A,B) > 0$.

c. If $A$ and $B$ are compact and disjoint, then $d(A,B) > 0$.

My answer is- a is not true if $A$ & $B$ are open and they have a common limit point . b & c are true. Am I correct?

  • 0
    If you take A = \{(x,\frac{1}{x}) \:|\: x > 0\} \subseteq \mathbb{R}^2 and $B$ the $x$-axis then $A$ and $B$ are closed and disjoint with $d(A,B) = 0$. This shows that b is also false.2012-09-11

1 Answers 1

5
  1. $[0,1)$ and $[1, 2]$ are disjoint
  2. As Matthias pointed out, $\{(x, 1/x):x>0\}$ and $\{(x, 0):x>0\}$ are disjoint and closed.
  3. Proof. Let $A, B$ be compacts. Say $(x_n) \in A$ is a sequence such that $\lim d(x_n, B) = 0$. Take the limit of any convergent subseqence $x = \lim x_{n_i}$ then by continuity of distance function we have $d(x, B) = 0$. The rest is intuitive for closed sets. Construct a sequence in $B$ converging to $x$ to show that $x \in B$ as well as $x \in A$.