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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 $\tau'$?

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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.

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  • $\begingroup$ 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. $\endgroup$ – bof Dec 15 '18 at 7:04
  • $\begingroup$ Thanks Nate. But if the 2 sets are $F_\sigma$'s in $\tau$, then will the extension be metriable? $\endgroup$ – aposyndetic Dec 15 '18 at 21:35
  • $\begingroup$ @aposyndetic: I don't know. $\endgroup$ – Nate Eldredge Dec 16 '18 at 23:03
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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.

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