Is $(\mathbb{R}^2,\mathcal{Z})$ first countable?

Question

Let $\mathcal{Z}$ be the topology on $\mathbb{R}^2$ such that every nonempty open set of $\mathcal{Z}$ is of the form $$\mathbb{R}^2 \setminus \{\text{at most finitely many points}\}.$$

I have already proved that $(\mathbb{R}^2,\mathcal{Z})$ is not Hausdorff. Now I am trying to show that $(\mathbb{R}^2,\mathcal{Z})$ is not first countable.

What I have already:

Suppose it is first countable. Let $x_0$ be an arbitrary element of $\mathbb{R}^2$. Then there exists a neighborhood basis for $X$ at $x_0$. In other words, there exists a collection $\mathcal{B}_{x_0}$ such that every neighborhood of $x_0$ contains some $B_i\in\mathcal{B}_{x_0}$, where $I$ is countable. By the definition of $\mathcal{Z}$, we have $$B_i= \mathbb{R}^2\setminus\{x_{i_1},\dots,x_{i_m}\}$$ for all $i\in I$ and some $m\in\mathbb{N}$. Consider $$\bigcup_{i=1}^{\infty} B_i= \mathbb{R}^2\setminus \{x_{11},\dots,x_{1_m}\}\cup\{x_{21},\dots,x_{2_m}\}\cup\cdots$$ Since $(\mathbb{R}^2,\mathcal{Z})$ is not Hausdorff, there exists $y_0\neq x_0\in \bigcup_{i=1}^{\infty} B_i$. Moreover $\mathbb{R}^2\setminus \{y_0\}$ is a neighborhood of $x_0$.

I am not sure if my claim at the end is correct and how else I can reach a contradiction.

Answer 1

There are several problems with your attempted proof. For instance, you claim that $I$ is countable without telling us what $I$ is. And from the fact that $(\mathbb{R}^2,\mathcal{Z})$ is not Hausdorff you cannot deduce that there's a $y_0\neq x_0\in\bigcup_{i=1}^\infty B_i$.

Suppose that a point $x_0\in\mathbb{R}^2$ has a countable fundamental system of neighborhoods $\{U_n\,|\,n\in\mathbb{N}\}$. For each $n\in\mathbb N$, since $x_0\in U_n$, $U_n\neq\emptyset$. So, $U_n\supset\mathbb{R}^2\setminus F_n$, for some finite set $F_n$. The union $\bigcup_{n=1}^\infty F_n$ is finite or countable and therefore distinct from $\mathbb{R}^2$. Take $x\in\mathbb{R}^2\setminus\left(\left(\bigcup_{n=1}^\infty F_n\right)\cup\{x_0\}\right)$. Then $\mathbb{R}^2\setminus\{x\}$ is a neighborood of $x_0$ which contains no $F_n$. Thereby, a contradiction is reached.

Note that the only property of $\mathbb{R}^2$ that I used was that it is an infinite uncountable set.

Answer 2

$\mathcal{Z}$ isn't the Zariski topology, but the cofinite topology, of $\mathbb{R}^2$ (the Zariski topology contains more open sets).

The cofinite topology is $T_0$, meaning for every $x\neq y$ there is an open set $U$ containing exactly one of $x,y$. Use this fact and $\mathbb{R}$ is not countable to derive a contradiction. (Hint: recall the intersection of a countable local base gives you s common subset of all open sets containing that point)