A problem in topology relating to the finite intersection property

Question

This is a problem from Munkres' Topology Exercise 37.1 (c)

Let $X$ be a space. Let $\mathscr{D}$ be a collection of subsets of $X$ that is maximal with respect to the finite intersection property.

(c) Show that if $X$ satistifes the T1 axiom, there is at most one point belonging to $\bigcap_{D\in \mathscr{D}}\bar D$.

The property that I'm attempting to use is the following Lemma.

Lemma 37.2

Let $X$ be a set; let $\mathscr{D}$ be a collection of subsets of $X$ that is maximal with respect to the finite intersection property. Then:

If $A$ is a subset of $X$ that intersects every element of $\mathscr{D}$ then $A$ is an element of $\mathscr{D}$.

To use this property, I've been trying to show that {$\bar D | D \in \mathscr{D}$}is equal to $\mathscr{D}$, however, I'm stuck here as you can see below. How can I work this out? Here is what I have:

To lead to a contradiction, assume there are distinct $x_1,x_2\in\bigcap_{D\in\mathscr{D}}\overline D$. Then we have $x_1,x_2\in\overline D$ for all $D\in\mathscr{D}$ and so $\{x_1\},\{x_2\}\subset\overline D$ for all $D\in\mathscr{D}$. By $T_1$, $\{x_1\}$ and $\{x_2\}$ are closed in $X$.

Let $\mathscr{D}'=\{D\mid D\in\mathscr{D}\}\cup\{\overline D\mid D\in\mathscr{D}\}$. Then $\mathscr{D}\subset\mathscr{D}'$. Claim: $\mathscr{D}'$ has f.i.p.

Take any finite elements $C_1,\ldots,C_n\in\mathscr{D}'$.

i. if all $C_1,\ldots,C_n\in\mathscr{D}$, then $C_1\cap\ldots\cap C_n\ne\varnothing$ by f.i.p. of $\mathscr{D}$.
ii. if all $C_1,\ldots,C_n\in\{\overline D\mid D\in\mathscr{D}\}$, then $C_1\cap\ldots\cap C_n\ne\varnothing$ since $x_1,x_2\in\bigcap_{D\in\mathscr{D}}\overline D$.
iii. if $D_1,\ldots,D_k\in\mathscr{D}$, $\overline{D_{k+1}},\ldots,\overline{D_n}\in\{\overline D\mid D\in\mathscr{D}\}$, $$D_1\cap\ldots\cap D_k\cap\overline{D_{k+1}}\cap\ldots\cap\overline{D_n}\supset D_1\cap\ldots\cap D_n\ne\varnothing$$ by f.i.p. of $\mathscr{D}$. Thus by maximality, $\mathscr{D}=\mathscr{D}'$, so $\{D\mid D\in\mathscr{D}\}\supset\{\overline D\mid D\in\mathscr{D}\}$.

Answer 1

The result is false; here’s a counterexample.

Let $X=\Bbb N\cup\{p,q\}$, where $p$ and $q$ are distinct points not in $\Bbb N$. Points of $\Bbb N$ are isolated. A set $U\subseteq X$ is a nbhd of $p$ if and only if $p\in U$ and $\Bbb N\setminus U$ is finite. Similarly, $U\subseteq X$ is a nbhd of $q$ if and only if $q\in U$ and $\Bbb N\setminus U$ is finite. This space is $T_1$.

Let $\mathscr{A}$ be a family of subsets of $\Bbb N$ containing the cofinite sets and maximal with respect to the f.i.p.; you can use Zorn’s lemma to show that such a family exists. For future reference note that every member of $\mathscr{A}$ is infinite. Let

$$\mathscr{D}=\big\{A\cup S:A\in\mathscr{A}\text{ and }S\subseteq\{p,q\}\big\}\;.$$

If $\{A_1\cup S_1,\ldots,A_n\cup S_n\}\subseteq\mathscr{D}$, where each $A_k\in\mathscr{A}$ and each $S_k\subseteq\{p,q\}$, then

$$\bigcap_{k=1}^n(A_k\cup S_k)\supseteq\bigcap_{k=1}^nA_k\ne\varnothing\;,$$

so $\mathscr{D}$ has the f.i.p. Suppose that $B\subseteq X$, and $\{B\}\cup\mathscr{D}$ has the f.i.p. Let $A_0=B\cap\Bbb N$; then

$$A_0\cap\bigcap\mathscr{F}=B\cap\bigcap\mathscr{F}\ne\varnothing$$

for each finite $\mathscr{F}\subseteq\mathscr{A}$, so $A_0\in\mathscr{A}$ by the maximality of $\mathscr{A}$. And clearly $B\setminus A_0\subseteq\{p,q\}$, so $B\in\mathscr{D}$. Thus, $\mathscr{D}$ is maximal with respect to the f.i.p.

Let $D\in\mathscr{D}$. Then $D\cap\Bbb N\in\mathscr{A}$, so $D\cap\Bbb N$ is infinite, and $p,q\in\operatorname{cl}(D\cap\Bbb N)\subseteq\operatorname{cl}D$. Thus,

$$p,q\in\bigcap_{D\in\mathscr{D}}\operatorname{cl}D\;.$$

(In fact it’s not hard to show that $\bigcap_{D\in\mathscr{D}}\operatorname{cl}D=\{p,q\}$.)

The result would be true if $X$ were $T_2$ rather than merely $T_1$; try proving that version instead.

Answer 2

Here I assume $X$ is T$_2$.

Suppose $x,y\in X$ and $x\neq y$. Let $U,V\subseteq X$ be disjoint open sets containing $x$ and $y$, respectively. Then either $U$ or $X\setminus U$ is in $\mathcal D$ (can you prove this?). If $X\setminus U \in \mathcal D$ then $x\notin\bigcap_{D\in\mathcal D} \overline D$. If $U\in\mathcal D$ then since $\overline U$ misses $y$ we have $y\notin \bigcap_{D\in\mathcal D} \overline D$.