$E_{a,b,t_0} = \{(x,y) \in \mathbb{R}^2 \vert x =at, y = bt, t>t_0\}$

Question

Let $E_{a,b,t_0} = \{(x,y) \in \mathbb{R}^2 \vert x =at, y = bt, t>t_0\}$ with a,b $\in \mathbb{R}$ and $t_0 \in \mathbb{R}^{+}$. Then let $\mathcal{B} = \{ E_{a,b,t_0} \vert a,b t_0 \in \mathbb{R} \} \bigcup \{(0,0)\}$. I want to know whether or not $\mathcal{B}$ is a base for a topology on $\mathbb{R}^2$. I have some doubts on the element of the base, are they quadrants? If they are I think it's trivial because then the two conditions to prove that it is a base for the topology are satisfied, but I'm having some doubts.

Answer 1

You see that $E_{a,b,t_0}$ are all halflines that are a "section" of a line that pass through the origin, then they are a base for a topology of $\mathbb R^2$, but it's clear that the topology that induce, is not the classical. (See for example that 0 is an open set). That base can be also reduced to elements of this type: $$E_{a,t_0}=\{(t,at)\vert t>t_0\}$$.

$$U_{q_0}^+=\{(0,q)\vert q>q_0\}, U_{q_0}^-=\{(0,-q)\vert q>q_0\}$$

So $\mathcal B=\bigcup_{a\in\mathbb R, t_0\in \mathbb R^+}E_{a,t_0}\cup \{(0,0)\}\cup \bigcup_{q_0\in \mathbb R^+}(U_{q_0}^+\cup U_{q_0}^-)$ In fact, they cover all points of the $\mathbb R^2$ and the intersection of any set of this tree distinct tipes of sets in the base is the $\emptyset$, that clearly $\emptyset\in \mathcal B$. But $E_{a,t_0}\cap E_{a,t_1}=E_{a, \max(t_0, t_1)}$ and $U_{q_0}^+\cap U_{q_1}^+=U_{\max(q_0,q_1)}^+$.