An example of an algebraically open set in $\mathbb{R}^2$ which is not open?

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.

Solution 1:

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.

Solution 2:

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.