Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
Great...proofs? I mean, but for (a), which follows almost immediately from the very definitions, I think you must write down some arguments justifying your choices. –  DonAntonio Sep 11 '12 at 11:49
    
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. –  Matthias Klupsch Sep 11 '12 at 11:50

1 Answer 1

up vote 5 down vote accepted
  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 $x \in A$.
share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.