Properties of the cofinite topology on an uncountable set 
Let $X$ be an uncountable set and let $\mathcal T = \{U \subseteq X : U = \varnothing\text{ or }U^c \text{ is finite} \}$.
Then is topological space $(X,\mathcal T)$

*

*separable?


*Hausdorff?


*second-countable (has a countable basis)?


*first-countable (has a countable basis at each point)?

I am confirmed about (2), $(X,\mathcal T)$ is not a Hausdorff space because we have a result $(X,\mathcal T)$ is Hausdorff iff $D = \{(x,x) : x \in X \}$ is closed , but $D$ is closed if $D^c$ is open and $D^c$ is open if $D^c$ is finite or $\varnothing$ which is not possible . so $(X,\mathcal T)$ is not a Hausdorff.
Please tell me about other three option. Thank you
 A: Hints:


*

*Show that every infinite set is dense; in particular, the countably infinite sets.  (Fix an infinite $A \subseteq X$, and show that $U \cap A \neq \varnothing$ for every nonempty open $U \subseteq X$.)

*Show that any two nonempty open sets have nonempty intersection.

*If $\mathcal{B} \subseteq \mathcal{T}$ is countable, then the set $A = \bigcup_{U \in \mathcal{B}} ( X \setminus U )$ is countable (since it is a countable union of finite sets). Pick $x \in X \setminus A$, and consider $V = X \setminus \{ x \}$.  (Is it a union of sets in $\mathcal{B}$?)

*Very similar to the above.



For your attempt at the non-Hausdorffness, you need to be a little bit lot more careful.  You need to show that $D$ is not closed in the square $X \times X$. In order to proceed as you have done, you would first have to show that the product $X \times X$ also has the co-fintie topology (since you make an appeal to this).  However this is not true.  If $A, B \subseteq X$ are finite nonempty, then $( X \setminus A ) \times ( X \setminus B )$ is open in $X \times X$, but it is not co-finite (since for any $a \in A$ the uncountable set $\{ \langle a , x \rangle : x \in X \}$ is disjoint from this set).
