Line with two origins

Question

I'm having trouble understanding why in the "line with two origins", every neighborhood of the zero in one copy of the real line intersects every neighborhood of the zero of the other copy. Can someone explain this point to me?

Thanks!

Answer 1

There are only two 'types' of base elements that contain the origins, and they are of the form $U_a=((-a,a)\setminus \{0\}) \cup \{p\}$ and $V_a=((-a,a)\setminus \{0\}) \cup \{q\}$ (with $a>0$).

If you have a neighbourhood $U$ of $p$ and a neighbourhood $V$ of $q$ then they must contain a base element $U_a \subset U, V_b \subset V$ as above. You can see that the point ${1 \over 2} \min(a,b) \in U_a \cap V_b$.