Let $(X,T_{\mathrm{cof}})$ be a cofinite topological space. I have to prove that $\forall x \in X$, the intersection of all the neighbourhoods of $x$ equals to the unitary set $\{x\}$:
$$\{x\}=\bigcap_{N \in \mathcal N_x} N.$$
It is clear that $\{x\}\subseteq\bigcap_{N \in \mathcal N_x} N$. In order to prove the other inclusion, I have tried to use reduction to absurdity supposing that there exists an $y \not = x$ such that $y\in \bigcap_{N \in \mathcal N_x} N$, but I didn't find any contradiction.
Besides, I don't know if the fact that every neighbourhood of a point is an open set in the cofinite topology could be useful for this problem.
Thanks in advance!
Best Answer
Let $y\in \bigcap_{N \in \mathcal N_x} N$ and $y\neq x $. Then $X\setminus \{y\} $ is a neighborhood of $x $. By assumption, we have $y\in X\setminus \{y\} $, a contradiction.