A Topological Space $(X,\tau)$ is $T_{1}$ if, and only if $\tau$ is finer than the cofinite topology A topological space $(X,\tau)$ is $T_{1}$ if and only if for every pair of distinc points $x,y$ $\in$ $X$, there exist $U$ $\in$ $\tau$ such that $x$ $\in$ $U$, and $y$ $\notin$ $U$
and $V$ $\in$ $\tau$ such that $y$ $\in$ $V$, and $x$ $\notin$ V. 
if $\tau_{1}$ and $\tau_{2}$ are topologies on X. We say that
 $\tau_{1}$  is finer than $\tau_{2}$ if 
 $\tau_{2} \subset \tau_{1}$.
But in this problem, I do not know how to start.
 A: The following are all equivalent:


*

*$(X,\tau)$ is $T_1$ 

*For every $x \in X$, $\{x\}$ is closed.

*For every finite subset $F$ of $X$, $F$ is closed.

*$\tau_{\text{cof}} \subseteq \tau$, i.e. $\tau$ is finer than the cofinite topology on the set $X$.


1 to 2: Let $X$ be $T_1$ and consider the set $\{x\}$ for some $x \in X$. Then for every $y \neq x$ we have (by $T_1$-ness) an open set $U_y$ such that $y \in U_y$ and $x \notin U_y$. Then $X\setminus \{x\} = \bigcup \{U_y : y \neq x\}$ which makes $X\setminus \{x\}$ open (as a union of open sets) and thus $\{x\}$ closed. 
2 to 3: closed sets are closed under finite unions, and finite sets are finite unions of singleton sets.
3 to 4: obvious, as all non-empty open sets of the cofinal topology are of the form $X\setminus F$ where $F$ is finite (thus closed in $\tau$ by assumption) and so also open in $\tau$ (as the complement of a closed set of $\tau$).
4 to 1. Let $x \neq y$ be two points of $X$. Then $U = X\setminus\{y\}$ is open in the cofinal topology, so open in $\tau$ by the assumption, and $x \in U$, $y \notin U$. We can take $V= X\setminus\{x\}$ to get $y \in V, x \notin V$ as well.
A: $"\Rightarrow" $Say $\tau_1$ be any $T_1$ topology on $X$ and $\tau_2$ cofinite topology. Take any $U\in \tau_2$. Then $U^C$ must be finite it means closed in any $T_1$ space, so $U\in \tau_1$.
$"\Leftarrow"$ $\tau_2\subset \tau_1$ while $\tau_2$ cofinite topology. Take any $x\in X$. Then $\{x\}^C\in \tau_2\subset \tau_1$. It means $\{x\}^C$ must be open in $\tau_1$. So $\{x\}$ is closed in $\tau_1$.
