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Let $X$ be a $T_1$ space. Let $\mathfrak {D}$ be a collection of subsets of $X$ that is maximal with respect to the finite intersection property. Show that there is at most one point belonging to $\bigcap_{D \in \mathfrak{D}} \bar D$

Here's my attempt. Suppose there're two points $x$, $y$ both belonging to $\bigcap_{D \in \mathfrak{D}} \bar D$. By $T_1$ axiom, there is one open set $A$, such that $x \in A$, while $y \notin A$. The maximality of $\mathfrak D$ implies every neighborhood of $x$ and $y$ belongs to $\mathfrak D$, and so is $A$. Thus every neighborhood of $y$ intersects with $A$, which shows that $y\in \bar A$.

I don't know how to proceed.

EDIT:This is an excercise from James Munkres' textbook Topology(2ed), page 235. However, I can't find a problem in K.Gosh's answer. Maybe something is wrong in this problem, or my paraphrasing of it.

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up vote 3 down vote accepted

This does not hold in $T_1$. Let $X$ be an infinite set endowed with the finite complement topology. Let $\mathfrak A$ be the collection of all nonempty open sets in $X$, i.e. $\mathfrak A$ is the collection of all sets whose complement is finite. Since $X$ is infinite, the intersection of any finite collection of elements of $\mathfrak A$ is nonempty; in fact, it is another element of $\mathfrak A$. Thus, $\mathfrak A$ has the finite intersection property. There exists a collection $\mathfrak D$ of subsets of $X$ such that $\mathfrak D$ has the finite intersection property, is maximal with respect to this property, and contains $\mathfrak A$. Since $\mathfrak D$ has the finite intersection property and contains $\mathfrak A$, every element $D$ of $\mathfrak D$ must intersect every element of $\mathfrak A$, i.e. every nonempty open set in $X$. Thus, $\bar D =X$ for all $D\in\mathfrak D$. It follows that $\bigcap_{D\in\mathfrak D}\bar D=X$.

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