I am trying to solve the following problem.
Give an example of a topological space $X$ and an infinite sequence of bases, $\mathcal{B}_1, \ldots, \mathcal{B}_2, \ldots$ such that $$ \mathcal{B}_1 \not \supset \mathcal{B}_2 \not \supset \mathcal{B}_3 \not \supset \ldots $$ all generating the same topology $\mathcal{T}$.
The setup of the problem is a bit confusing, because though I'm choosing a sequence and the order matters, it seems that I can have, for example, $\mathcal{B}_1 \supset \mathcal{B}_3$ and this restriction only applies to consecutive elements. I'm trying to understand exactly why that is significant, because it isn't that the sequence is anti-nested in the sense that $\mathcal{B}_i \subset \mathcal{B}_{i+1}$ for every $i$.
The most natural topological space to me seems to be $X = \mathbb{R}$ with its usual topology. The most natural basis is the set of all open intervals. I'm wondering if I can just "enlarge the intervals" and generate another countable basis that doesn't include all open such open intervals. For example, if I have $(a,b)$ and $(b,c)$, I can take their union and include $b$. This isn't rigorous at this point, though.
I'd appreciate if someone could point me in the right direction.