# Is a plane set still metrizable if two new subsets are declared open?

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 $$\tau'$$?

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.

• On the other hand, I suspect the topology generated by $\tau$ and $A$ and $B=\mathbb R^2\setminus A$ will be metrizable, even if $A$ is a Borel set. – bof Dec 15 '18 at 7:04
• Thanks Nate. But if the 2 sets are $F_\sigma$'s in $\tau$, then will the extension be metriable? – aposyndetic Dec 15 '18 at 21:35
• @aposyndetic: I don't know. – Nate Eldredge Dec 16 '18 at 23:03

If you add the set $$\mathbb{Q}^2$$ to the topology of the plane then the resulting topology is not regular: a new basic neighbourhood of $$(0,0)$$ is of the form $$B((0,0),r)\cap\mathbb{Q}$$ and its closure in the new topology is $$\{(x,y):\|(x,y)\|\le r\}$$. If you also add $$\mathbb{Q}^2+(\pi,\pi)$$ then the resulting space is still not regular.

As to adding both a set $$A$$ and its complement $$B=\mathbb{R}^2\setminus A$$: that will result in the topological sum of the subspaces $$A$$ and $$B$$ of the plane, which is metrizable.