# Proof that the weakest $T_{1}$ topology on infinite set is cofinite topology

Topological space $$(X,\tau)$$ is $$T_{1}$$ space if for any two points $$a,b \in X$$ there exist an open neibourhood $$U_{a}$$ of $$a$$ with no $$b$$ in it.

Cofinite topology on set $$X$$ is $$\tau_{cf} = \{ A \in 2^{X} | X \setminus A$$ is finite$$\} \cup \{ \emptyset \}$$

I need to proof that there is no weaker topology $$\tau$$ then cofinite topology on an infinite set $$X$$ that turns $$(X,\tau)$$ into $$T_{1}$$ space.

My attempt was to go from the contradiction, that is suppose there is a topology $$\tau \subset \tau_{cf}$$ that turns $$X$$ into $$T_{1}$$ space. $$\tau$$ consists only of sets from $$\tau_{cf}$$ and for any two points there exists an open neibourhood of one with no another. But then I stucked. I think I should show that $$\tau = \tau_{cf}$$.

In a $$T_1$$-space all single-point sets are closed, thus any $$T_1$$-topology $$\tau$$ on $$X$$ must contain the set $$\tau_0 = \{X \setminus \{x\} \mid x \in X \}$$. $$\tau$$ must also contain the set $$\tau_1$$ of all finite intersections of members of $$\tau_0$$. Thus $$\tau_{cf} \subset \tau$$. Since $$\tau_{cf}$$ is a topology such that $$(X,\tau_{cf})$$ is $$T_1$$, we see that it is the coarsest toplopgy with this property.
Hint: In a $$T_1$$ space, finite sets are closed.