Two proper subsets of the real numbers, $A$ and $B$, that have the following conditions: $A$ and $B$ are closed, $A \cap B$ is empty, +1

Solution 1:

This was tricky! The idea, is you can't think of the sets as intervals. As there are not two intervals for which this is true.

Consider $A = \{2,3,4,5, \ldots\}$. This is a closed set. Consider $B =\{2+1/2, 3+1/3, 4+1/4, 5+1/5, \ldots\}$. This is also a closed set.

Now, $A$ and $B$ are clearly disjoint and for $n \ge 2$, we have that $n \in A$ and $n + 1/n \in B$. Thus the distance between these points is $1/n$. This is true for all integers $n \ge 2$ and hence they have distance $0$ between the sets.

Solution 2:

$(X, d) $ be a metric space.

$A$ and $B$ be two subsets of $X$.

1. $A\cap B =\emptyset$

2. $A$ is compact and $B$ is closed.

(1) and ( 2) implies $d(A, B) >0 $. (The proof is not difficult at all)

So, to find two closed disjoint sets $A$ and $B$ such that $d(A, B) =0$ , you have to consider both sets to be non-compact . Otherwise it is not possible. Think about it.