Dense sets in complement-countable topology I`m supposed to verify wether this statement is true or not:
Statement: Let $(X, \tau)$ be a topological space where $\tau$ is complement countable topology. If $D\subset X$ is dense then $D$ is open.}
My intuition says it`s not true. For a counter-example I was thinking about figuring out a uncountable set which could contain at least two uncountable sets whose complements are also uncountable.
My attempt: Consider $(\mathbb{R}^2,\tau )$ where $\tau$ is as defined above.
Now, consider the sets $$A_{Q,I}\dot{=}\{(q,i)\in \mathbb{R}^2:q\in\mathbb{Q}\wedge i\in \mathbb{R}\backslash\mathbb{Q} \}$$
$$A_{I,Q}\dot{=}\{(i,q)\in \mathbb{R}^2:q\in\mathbb{Q}\wedge i\in \mathbb{R}\backslash\mathbb{Q} \}$$
Note that, as in both cases we have irrational coordinate so we have uncountable many coordinates. They are not open as both complements are uncountable (neither closed). 
Questions: Is the statement really false? Does this example do the job? Would you give me more counter-examples? Is there a set which could be divided in exactly two uncountable parts?(Be cause the sets I defined are not complement of each other, they are contained in the complement of each other).
 A: Your example works, but it's more complicated than is necessary. I don't understand why you say you need "at least two uncountable sets whose complements are also uncountable". 
All you need is a single uncountable subset $X$ of your space whose complement is also uncountable. For example, take $X = (0,\infty)\subseteq \mathbb{R}$. It's dense in the complement-countable topology, since any open set $U$ can only miss countably many points of $X$, so $U\cap X \neq \emptyset$. But it's not open, since it has uncountable complement.

Actually every uncountable set $A$ has an uncountable subset with uncountable complement. You can well-order $A$ and put just the elements indexed by even ordinals in $X$. So whenever a space with the complement-countable topology is not discrete, it has a subset which is dense but not open. Of course, this argument relies on the axiom of choice. Without choice, there can be uncountable sets where every subset is countable or co-countable (called $\aleph_1$-amorphous sets). See this question.
