What $n$ s.t. this topology satisfies the separation axiom $T_n$? Let $X$ be the real with the topology $T = \{ U \subset\mathbb{R}\:|\:0\notin U\text{ or }\mathbb{R}\setminus U\text{ finite}\} $. What is the largest value for $n \in \{0,1,2,3,3\frac{1}{2},4\}$ such that $X$ satisfies the separation axiom $T_n$?
First of all, the open sets in this topology are all cofinite sets and open intervals $I \subset \mathbb{R} \setminus \{0\}$. I would argue that the topology is hausdorff, by using only the open intervals (without using the cofinite sets) to find open neighborhoods $U$ and $V$  for some $a,b \in X$. But I'm not sure if I could go any higher into $T_3$.
 A: No, the open sets are all subsets of $\Bbb R\setminus\{0\}$ together with all cofinite sets. You’re right that $X$ is Hausdorff, but since you described the topology incorrectly, you need to think a little more about how to show this.
Note that $X$ is compact. If $\mathscr{U}$ is an open cover of $X$, there is some $U_0\in\mathscr{U}$ such that $0\in U_0$. Now $X\setminus U_0$ is finite, so only finitely other elements of $\mathscr{U}$ are needed to complete a subcover.
HINT: What do you know about the separation properties of compact Hausdorff spaces?
A: The possible values of $n$ are 0,1.
Proof of $X$ satisfies $T_1$ axiom: Let $x,y\in X$, $x\neq y$. Then
$$U=X\setminus \{y\}, \quad V=X\setminus \{x\}$$
satisfies $x\in U\not\ni y$, $y\notin V \ni y$. And $U$ is neighborhood of $x$ and $V$ is neighborhood of $y$.
Proof of $X$ is not $T_2$ : Let $x,y\in X$, $x\neq y$. If $U$ is neighborhood of $X$ such that does not contain element $y$. If $y$ has neighborhood $V$ disjoint form $U$, We get $V$ is uncountable and $V\subset X\setminus U$. But $V\subset X\setminus U$ is finite and we get contradiction.
