# Neighbourhoods in the cofinite topology

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.

• Find a neighbourhood that does not contain $y$. Jan 2 '20 at 18:18
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.
Given any point in the space $$y\neq x$$, there exists a neighborhood of $$x$$ that does not contain $$y$$. This is enough to conclude that the intersection is $$\{x\}$$.
You just need to apply the definition: $$U \subset X$$ is a neighborhood of $$x \in X$$ if, and only if, $$x \in U$$ and $$X \setminus U = \{x_1, \dots, x_n\}$$ is a finite set. Now, if $$y \in U$$ and $$y \ne x$$, why is $$U \setminus \{y\}$$ also a neighborhood of $$x$$ in $$X$$? (What is $$X \setminus \left( U \setminus\{y\} \right)$$?)