Counterexample of any infinite union of open balls is a complement of a compact set in $R^d$.

Question

I am curious about if there is a infinite union of open balls which is not a complement of a compact set in $R^d$.

Any help would be appreciated.

Answer 1

Yes. If all the balls are equal to the open unit ball, their union will be the open unit ball, which is not the complement of a compact set.

If you want the balls to be disjoint, take the balls with radius $\frac12$ centered at points such that all of its coordinates are integers.

Answer 2

Yes. Denote by $B(a,r)$ the ball around $a$ in radius $r$.

Take $B_n:=B(0,1-\frac{1}{n})$ then $\bigcup_{n\in\mathbb{N}} B_n = B(0,1)$. This is not a complement of a compact set.

Note moreover, that $B_n\backslash B_{n-1}$ contains an open ball. So it's even possible to construct a sequence of disjoint balls with that property.

Answer 3

All you need to do is to show that the union is bounded, since the complement of a bounded set is unbounded, which implies that the complement is not compact.

For example, consider the union of all sets $\cup_{n \in \mathbb{N}} \{0 < x < 1/n\}$. It’s complement is unbounded.