Take any $K_{1}$ and $K_{2}$ closed sets. Define $A = \{||x-y||: x \in K_{1}, y \in K_{2}\}$. I'm asked to prove (or give a counterexample) that $A$ is closed, in general.
A classmate came up with the following counterexample:
Consider $K_{1},K_{2} \subseteq \mathbb{R}^2$ such that $K_{1} = \{(x,y) \in \mathbb{R}^2: x\geq 1, y=0\}$, i.e, the horizontal axis from $1$ on and $K_{2} = \{(x,y) \in \mathbb{R}^2: x\geq 1, y=1/x\}$. Clearly, both sets are closed. Notice that there are elements in $A$ that are arbitrarily close to $0$ but $0 \notin A$. Then, $A$ is not closed.
I could come up with some more examples, but all of them in the same spirit of the one above. I could also prove that if either $K_{1}$ or $K_{2}$ is compact, then $A$ is closed. So, a counterexample should rely on two unbounded sets, since I'm thinking about Euclidean spaces for the sake of simplicity.
In special, I was trying to find a counterexample in which $K_{1}$ and $K_{2}$ are subsets or the real numbers, but I had no success! Any ideas? Thanks a lot in advance!
