I am thinking of forming a finer topology on a particular subset of the plane. Let $X\subseteq \mathbb R ^2$ be endowed with the Euclidean topology $\tau$. Let $A,B\subseteq X$. Let $\tau'$ be the topology generated by $\tau\cup \{A,B\}$. Then will $\tau'$ be metrizable? 


If not (very sad), then what assumptions about the sets $A$ and $B$ would guarantee metrizability of $\tau'$?



Is a plane set still metrizable if two new subsets are declared open?
Even adding one set can break metrizability, if that set is not $F\_\sigma$.


Let $\tau'$ be generated by $\tau$ and $A$, where $A$ is not $F\_\sigma$ with respect to $\tau$. (For instance, by the Baire category theorem, $A = (\mathbb{Q} \times \mathbb{Q})^c$ would do.) Now if $\tau'$ is metrizable, then the open set $A$ must be $F\_\sigma$ with respect to $\tau'$; indeed, we would have $A = \bigcup\_{n=1}^\infty \bigcap\_{x \in A^c} (B'(x, 1/n)^c)$. But I claim this is not so.


It's easy to verify that every open set $U'$ in $\tau'$ may be written as $U' = U \cup (A \cap V)$ where $U, V \in \tau$. (Check that the collection of all such sets is a topology which contains $\tau$ and $A$.) Now if $A$ is $F\_\sigma$ in $\tau'$, then $A^c$ is $G\_\delta$ in $\tau'$. So $A^c = \bigcap\_{n=1}^\infty U\_n'$ where $U\_n' = U\_n \cup (A \cap V\_n) \in \tau'$, with $U\_n, V\_n \in \tau$. But every $U\_n'$ must contain $A^c$, which means that $U\_n' = U\_n \cup V\_n \in \tau$. So we conclude that $A^c$ is $G\_\delta$ in $\tau$, a contradiction.


