Let $B_1$ and $B_2$ be the basis in the topological space $(X,\tau)$. Will $B_1\cup B_2$ and $B_1\cap B_2$ be the base?
I think intuitively it would be $B_1\cup B_2$. But $B_1\cap B_2$ may not be either. Would you help me with this topic?
I know the base definition. However, since I did not understand the subject well in the course, I have difficulties in the proofs or examples here.
I would be very grateful if anyone could help explain.
Many different collections of sets may be bases for a given topology.
Notice that $B_{0} \cup B_{1} \subseteq \tau$. This means every element of $B_{i}$ is the union of finite intersections of elements in $B_{1 - i}$. You can't get any new open sets that were not in $\tau$.
To see that $B_{0} \cap B_{1}$ need not be a basis for $\tau$. Consider two basis for the usual topology of $\mathbb{R}^{2}$. The elements of $B_{0}$ are open disks. The elements of $B_{1}$ are open squares. Since no disk is a square and no square is a disk, the intersection is empty. The topology generated by $\varnothing$ is $\{ \varnothing, \mathbb{R}^{2}\}$, not the usual topology.