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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!

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    $\begingroup$ Find a neighbourhood that does not contain $y$. $\endgroup$
    – Qi Zhu
    Jan 2 '20 at 18:18
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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.

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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\} $.

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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 $X \setminus \left( U \setminus\{y\} \right)$?)

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