Closure of a solid triangle in $\mathbb{R}_{(-)} \times \mathbb{R}_{\textit{finite complement}}$

Question

Consider the product topology $\mathbb{R}_{(-)} \times \mathbb{R}_{\textit{finite complement}}$. What is the closure of a solid triangle in this space?

I know that if we consider the topological basis, we have $\mathcal{B_1} = \{(a,b):a<b\}$ and $\mathcal{B}_2 = \{\mathbb{R}-F:F \text{ is finite}\}$ so I think I have an idea of what an open set in the space looks like, but how do I proceed with the problem?

Answer 1

HINT: If $a\in\Bbb R$, and the vertical line $x=a$ intersects the solid triangle in infinitely many points, the whole vertical line is in the closure of the triangle. (Why? What does a basic open nbhd of a point on that line look like?)