Comparability of topologies on $\Bbb R$ > Let $\cal C=\\{a,b):a<b; a,b\in \Bbb Q\\}$ be a basis for a topology on $\Bbb R$. [This says that topology generated by $\cal C$ is not comparable to the countable complement topology on $\Bbb R$. But I think that topology generated by $\cal C$, call it $\cal T(\cal C)$ is finer than the countable complement topology: the basis elements $B$ for the countable complement topology are at most countably many elements removed from $\Bbb R$, for example, $\Bbb R-\\{0\\}$ is a basis element for the countable complement topology. So, if $x\in B$, then there is an element of $\cal C$ such that $x\in [a,b)\subset B$ where $a$ and $b$ are rationals.

You only take _one_ basis element of the countable complement on $\mathbb R$ and prove that it is an element of $\tau(\mathcal C)$.

That is not enough.

You must at least prove that all elements of a a randomly chosen subbase of the cocountable topology are elements of $\tau(\mathcal C)$.

From the fact that $\mathbb R-\mathbb Q$ is evidently not an element of $\tau(\mathcal C)$ (in contrast to non-empty elements in $\tau(\mathcal C)$ it does not contain an interval) it can be concluded that $\tau(\mathcal C)$ is not finer than the cocountable topology.