Topology generated by open balls minus countable sets I'm given the topology $\mathcal{T}$ on $\mathbb{R^2}$ such that the open sets are generated by sets $B\setminus A$ where $B$ is an open ball and $A$ is countable. It seems a bit difficult to me work with the topology using its base, in part because I'm asked to prove connectedness and then open sets cannot be substituted by basic open sets. Do you think I can characterize this topology im other way to work better with it? 
I thought before that $\mathcal{T}$ was the family  $\mathcal{T_0}$ of usual open sets minus a countable set, but looking carefully it doesn't seem to be. Or at least I don't know how to prove it. Is $\mathcal{T}\neq \mathcal{T_0}$? Please could you give an example of open set in $\mathcal{T}$ that is not a usual open set minus a countable set ? And could you show me a comfortable way to work with this topology? I had thought a proof for conectedness and not path-connectedness, but that was when I thought $\mathcal{T}= \mathcal{T_0}$ . Now, I've got nothing
 A: Your intuition is correct, though it takes a bit of work to prove it. Let $$\mathscr{T}'=\{U\setminus C:U\in\mathscr{T}_0\text{ and }C\text{ is countable}\}\;;$$ I’ll show that $\mathscr{T}'=\mathscr{T}$.
Recall that $\mathscr{T}_0$ has a base $\mathscr{B}_0$ consisting of the open balls $B(\langle x,y\rangle,r)$ such that $x,y,r\in\Bbb Q$ and $r>0$. Let $B$ be any open ball and $C$ a countable subset of $B$. Let 
$$\mathscr{U}=\{V\in\mathscr{B}_0:V\subseteq B\}\;;$$
then $B=\bigcup\mathscr{U}$, so $B\setminus C=\bigcup\{V\setminus C:V\in\mathscr{U}\}$, and it follows that 
$$\mathscr{B}=\{V\setminus C:B\in\mathscr{B}_0\text{ and }C\text{ is countable}\}$$
is a base for $\mathscr{T}$.
Now let $U\in\mathscr{T}$; there is a family $\mathscr{B}_U\subseteq\mathscr{B}$ such that $U=\bigcup\mathscr{B}_U$. Let 
$$\mathscr{B}_U=\{B_\alpha\setminus C_\alpha:\alpha\in A\}$$ 
for some index set $A$, where $B_\alpha\in\mathscr{B}_0$ and $C_\alpha$ is countable for each $\alpha\in A$. For each $B\in\mathscr{B}_0$ let $A_B=\{\alpha\in A:B_\alpha=B\}$; then 
$$\bigcup_{\alpha\in A_B}(B_\alpha\setminus C_\alpha)=\bigcup_{\alpha\in A_B}(B\setminus C_\alpha)=B\setminus\bigcap_{\alpha\in A_B}C_\alpha\;,$$
which is in $\mathscr{B}$ since $\bigcap_{\alpha\in A_V}C_\alpha$ is countable. Thus, for each $\alpha\in A$ we may as well replace $\{V_\alpha\setminus C_\alpha:\alpha\in A_V\}$ by the single set $V\setminus\bigcap_{\alpha\in A_V}$, and we can then assume that $V_\alpha\ne V_\beta$ whenever $\alpha,\beta\in A$ and $\alpha\ne\beta$ and hence that $A$ is countable. Let $V=\bigcup_{\alpha\in A}V_\alpha\in\mathscr{T}_0$, and let $C=\bigcup_{\alpha\in A}C_\alpha$; $C$ is countable, and $V\setminus U\subseteq C$, so $V\setminus U$ is countable. But then $U=V\setminus(V\setminus U)\in\mathscr{T}'$, and we’ve shown that $\mathscr{T}\subseteq\mathscr{T}'$.
Showing that $\mathscr{T}'\subseteq\mathscr{T}$ and hence that $\mathscr{T}'=\mathscr{T}$ is easier. Let $U\in\mathscr{T}'$, so that $U=V\setminus C$ for some $V\in\mathscr{T}_0$ and countable $C$. Then there is a family $\mathscr{B}_V$ of open balls such that $V=\bigcup\mathscr{B}_V$, and clearly 
$$U=V\setminus C=\bigcup\{B\setminus C:B\in\mathscr{B}_V\}\in\mathscr{T}\;.$$
