Let $\mathscr{T}$ be the topology of $U \subset \mathbb{R}$ for which $U = \emptyset$ or $U^c$ is countable. Determine is this space is Hausdorff? [duplicate]

Let $\mathscr{T}$ be the topology of $U \subset \mathbb{R}$ for which $U = \emptyset$ or $U^c$ is countable. Determine is this space is Hausdorff?

So $\mathscr{T}$ is the cocountable topology on $\Bbb R$. In order for the space $(\Bbb R, \mathscr{T})$ to be Hausdorff I need that for $x,y \in \Bbb R$ and $O_x, O_y \in \mathscr{T}$ the intersection $O_x \cap O_y = \emptyset.$

Now this intersection is empty if $(O_x \cap O_y)^c = O_x^c \cup O_y^c$ is not countable. So I have $$(O_x \cap O_y)^c = O_x^c \cup O_y^c=(\Bbb R \setminus O_x) \cup (\Bbb R \setminus O_y) $$ but I don't have information about $O_x$ and $O_y$ other than that since they contain $x$ and $y$ respectively they're non-empty and thus the complement of both is countable? So why wouldn't $O_x^c \cup O_y^c$ be countable as the union of two countable sets?

Solution 1:

$x\neq y \in \Bbb{R}$ .

Suppose there exists $x\in O_x, y\in O_y \in \mathscr{T}$ and $O_x \cap O_y = \emptyset.$

Hence, $(O_x )^c, (O_y )^c$ are countable.

Then, $(O_x \cap O_y)^c = O_x^c \cup O_y^c =\Bbb{R}$

$(O_x )^c\cup ( O_y)^c =\Bbb{R}$

$\implies \Bbb{R}$ is countable, which is a contradiction.

Hence, the existence of two non empty disjoint set $O_x, O_y $ containing two disjoint points $x, y $ , is not possible.

Hence, $ (\Bbb R, \mathscr{T})$ is not Hausdorff space.