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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?

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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
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1 Answer

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 $x \in A$.
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