MHB Problem of the Week #120 - September 15th, 2014

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The problem discusses the relationship between the distance of two subsets, A and B, in a metric space and their intersection. It establishes that if A is compact and B is closed, then A and B are disjoint if and only if the distance between them is greater than zero. The proof demonstrates that if A intersects B, the distance is zero, while if they do not intersect, the distance must be positive. The solutions provided by participants Ackbach and Euge confirm the correctness of this relationship. This highlights the importance of compactness and closedness in determining the properties of subsets in metric spaces.
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Here's this week's problem!

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Problem
: Let $A$ and $B$ be subsets of a metric space $(X,\rho)$. Define
\[\mathrm{dist}(A,B) = \inf\{\rho(u,v)\mid u\in A,\,v\in B\}.\]
If $A$ is compact and $B$ is closed, show that $A\cap B=\emptyset$ if and only if $\mathrm{dist}(A,B)>0$.

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This week's problem was correctly answered by Ackbach and Euge. You can find Euge's solution below.

[sp]Suppose $A\cap B \neq \emptyset$. Then there is an $x \in X$ such that $x\in A$ and $x\in B$; hence, $0 = \rho(x,x) \le \text{dist}(A,B)$. So, $\text{dist}(A,B) = 0$. Conversely, suppose $A\cap B = \emptyset$. The mapping from $A$ to $\Bbb R$ given by $a \mapsto \rho(a,B)$ is continuous on the compact set $A$, so it has a minimum value at some $a_0 \in A$. If $\rho(a_0, B) = 0$, then since $B$ is closed, $a_0 \in B$, contradicting the assumption $A\cap B = \emptyset$. Hence, $\rho(a_0, B) > 0$. For all $a\in A$ and $b\in B$, $\rho(a,b) \ge \rho(a, B) \ge \rho(a_0, B)$. Therefore, $\text{dist}(A,B) \ge \rho(a_0, B) > 0$.[/sp]
 

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