I want to compare two topologies on $\mathbf{R}$: $\tau_1$=finite complement topology and $\tau_2$=the topology having all sets $(-\infty,a)=\{x|x<a\}$ as basis. If I pick out the element $\mathbf{1}$, take $B_1=\mathbf{R}$-$\{0\}\in \tau_1$ and $B_2=(-\infty,2) \in \tau_2$ I see that $\mathbf{1}\in B_1$, $B_2$ but $B_2\not\subset B_1$ and vice-versa, so I say that the two topologies cannot be compared.

Now if I take another case where I change $B_1=\mathbf{R}$-$\{3\}\in \tau_1$ and keep the rest unchanged, again $\mathbf{1}\in B_1$, $B_2$ but now $B_2\subset B_1$. They not uncomparable anymore. So what do I conclude?