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Definition. A subset $U$ of a real vector space $V$ is algebraically open if the sets $\{t\in\mathbb{R}:x+tv\in U\}$ are open for all $x,v\in V$.

In the real vector space $\mathbb{R}^2$ equipped with the usual topology, it is clear that every open set is algebraically open, but how to find a algebraically open set which is not open? The hint says that a line intersects the unit circle in at most two points.

mathreadler
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Xiang Yu
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2 Answers2

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To use the hint... take the plane, and subtract a countable dense subset of the unit circle. It is not open, but it is still algebraically open by the hint.

GEdgar
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  • I seem to recall some truly weird examples (perhaps due to Sierpinski?) like: a set in the plane that meets every line in exactly two points... – GEdgar Dec 07 '15 at 13:40
  • The set does not have to be countable, right? Just non-exhaustive. You can literally take the plane minus the unit circle but with one point added back in. I also don't think dense is necessary, as long as it is not closed. – SmileyCraft Jan 03 '19 at 21:09
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The sets $\{x+tv\mid t \in \mathbb R\}$ for $x, v \in V$ fixed represent lines in $V$, and the sets $\{t \in \mathbb R \mid x+tv \in U\}$ represent the intersections of those lines with $U$, pulled back to $\mathbb R$. So a subset is defined to be "algebraically open" if it is open along every line.

The hint says that the unit circle intersects with every line at at most two points; this suggests that you take the complement of the unit circle, because in a line, the complement of finitely many points (e.g. at most two points) is automatically open. Of course, the complement of the unit circle is actually open, so you need to make some kind of adjustment. Hint: if you take the complement of any $S \subset S^1$, the same reasoning will still show that the resulting set is algebraically open.

Dustan Levenstein
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