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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 $\tau = \tau_{cf}$.

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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.

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Hint: In a $T_1$ space, finite sets are closed.

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