I am looking for examples where topologies generated by a subbasis and a basis yield to the same topologies, preferably in a finite topological space.
For instance let $X=\{1,2,3\}$. The collection $\mathcal{S} = \{\{1\}, \{2\}, \{1,3\} \}$ is a subbasis as the union of elements equals $X$. The topology generated by this subbasis is all unions of finite intersections of elements of $\mathcal{S}$ which gives us $\mathcal{T} = \{\phi, \{1\}, \{2\}, \{1,2\}, \{1,3\}, \{1,2,3\} \}$.
Moreover $\mathcal{S}$ is itself a basis as it satisfies the conditions of being a basis. So we can generate a topology with finding the collection of all unions of elements of $\mathcal{S}$, which is worded differently compared to the case where we generate a topology on $\mathcal{S}$ as a subbasis. In this case the generated topology on $\mathcal{S}$ as a basis is $\mathcal{T'} = \{\phi, \{1\}, \{2\}, \{1,2\}, \{1,3\}, \{1,2,3\} \}$. We note $\mathcal{T} = \mathcal{T'}$.
Is it always the case that a topology generated by a subbasis yield to a same topology when the topology is generated by the same collection $\mathcal{S}$ but as a basis and not as a subbasis? What are other differences/similarities between the topologies in these case? I understand similar questions have been asked before but I am specifically asking about the differences/similarities that the same collection of sets $\mathcal{S}$ make when it is viewed as a subbasis and when viewed as a basis.
Best Answer
The topology generated by the subbase $\mathcal{S}$ is not $\mathcal{T}$, but rather $\mathcal{T}'$: in the second paragraph you failed to take the union of $\{1\}$ and $\{2\}$.
If a family $\mathcal{S}$ of subsets of $X$ is a base for a topology, that is also the topology generated by $\mathcal{S}$ as a subbase: it is already closed under finite intersections, so whichever way you view it, you’re just taking the closure under arbitrary unions. That is, every base for a topology is a subbase for that same topology.