Talk:Complement of Bounded Set in Complex Plane has at most One Unbounded Component

This applies in any NVS for which the complement of closed balls around the origin are connected (this result is not true for $\R$ for instance, $\R \setminus \closedint {-R} R$ has two unbounded components for any $R > 0$) but I wasn't sure of a good way to characterise this. Caliburn (talk) 09:51, 14 August 2023 (UTC)

Maybe $\dim \ge 2$? --Usagiop (talk) 13:21, 14 August 2023 (UTC)

It should hold $\R^n$ and $\C^k$ with $n \ge 2$, $k \ge 1$ yes. But as to whether this is true for all normed vector spaces of dimension at least $2$, I have no idea. In $\R^n$ it should be possible to explicitly construct a path to show connectedness (this is quite easy in $\R^2$ or $\C$ - rotate the point so it makes the right angle with the origin then send it along a straight line), but I don't have any geometric intuition of the situation in infinite dimensions. Taking a wild guess, I would guess there are some geometrically crazy normed vector spaces where this might not be true, but visualising such a space would be visualising one that looks completely different to $\R^n$ which leaves me clueless. Caliburn (talk) 13:45, 14 August 2023 (UTC)

Got an almost instant answer from MSE. It's true with a very simple proof following directly from the $\R^2$ case. [1] Caliburn (talk) 14:25, 14 August 2023 (UTC)