If a topology $\mathcal{T}$ on $\mathbb{R}$ is strictly finer than both the lower limit topology and the cocountable topology on $\mathbb{R}$ , then $\mathcal{T}$ must be the discrete topology on $\mathbb{R}$.
My attempt :
Lower limit topology $\mathcal{T_{l}}$,
has basis of the form [a,b) and cocountable topology $\mathcal{T_{c}}$ has basis of the form $U_{i}\backslash (x_{1},x_{2},…)$. Singletons form a basis in discrete topology $\mathcal{T_{d}}$
$ \forall x \in X, \forall \mathcal{B} \in \mathcal{T_{l}},
\exists \{ x \} \in \mathcal{T_{d}},
\ni \{ x \} \subset \mathcal{B}$
Hence discrete topology is finer than lower limit topology. But it can't be strictly finer as no basis element of lower limit topology can sit inside singleton set of discrete topology (lower limit topology is not finer than discrete topology).
Similarly discrete topology is finer than cocountable topology since singleton set of discrete topology is a subset of all sets in basis of cocountable topology but cocountable topology is not finer than discrete topology.
So it is very clear to me that discrete topology is strictly finer than both cocountable and lower limit topology. But my doubt is how can I guarantee that there exist no other topology other than discrete topology which is strictly finer than both the given topologies. Or rather what other counter example can I give to disprove the given statement.
Best Answer
This statement appears to be false. The topology $\mathcal T$ generated by the sub-basis consisting of half open intervals and co-countable sets is certainly strictly finer than both the given topologies (since it includes all of the open sets in each one, and there are open sets in either that are not open in the other). However, every nonempty intersection of finitely many half-open intervals is uncountable, hence a further intersection with finitely many co-countable sets remains uncountable.
Since every open set is a union of finite intersections of sub-basis elements, it follows that every non-empty open set in $\mathcal T$ is uncountable, hence singletons are not open.
Remark
In general, given two topologies $\mathcal T_1$ and $\mathcal T_2$ on a set $X$, the topology $\mathcal T$ generated by $\mathcal T_1\cup \mathcal T_2$ (which has a basis consisting of sets $B_1\cap B_2$ for $B_i\in \mathcal B_i$, where $\mathcal B_i$ is a basis for $\mathcal T_i$) is the coarsest topology that is finer than the original two topologies. If neither $\mathcal T_1$ nor $\mathcal T_2$ is coarser or finer than the other, then $\mathcal T$ will therefore be the coarsest topology that is strictly finer than the original two, meaning every other topology strictly finer than the original two must be finer than $\mathcal T$. In particular, if $\mathcal T$ is discrete, then it is the only topology strictly finer than $\mathcal T_1$ and $\mathcal T_2$.
Therefore, $\mathcal T$ was the natural topology to check in this situation, as the original statement was guaranteed to be true if and only if $\mathcal T$ was discrete.