Exercise 1. Let \(X\) be a space. Let \(\mathcal{D}\) be a collection of subsets of \(X\) that is maximal with respect to the finite intersection property (FIP).

(a) Show that \(x \in \bar{D}\) for every \(D \in \mathcal{D}\) if and only if every neighborhood of \(x\) belongs to \(\mathcal{D}\). Which implication uses maximality of \(\mathcal{D}\)?

Proof:

1. Prove in the forward direction

   Let \(x \in \bar{D}\) for every \(D \in \mathcal{D}\). Let \(U\) be any neighborhood of \(x\) in \(X\). According to Theorem 17.5 (a) in Section 17, if \(x \in \bar{D}\), we have \(U \cap D \neq \varPhi\). This means any neighborhood \(U\) of \(X\) intersects every element in the maximal collection \(\mathcal{D}\). According to Lemma 37.2 (b), \(U \in \mathcal{D}\). The maximality of \(\mathcal{D}\) is used when applying this lemma.

2. Prove in the backward direction

   If there exists a \(D_0 \in \mathcal{D}\) such that \(x \notin \bar{D}_0\), \(x\) belongs to the complement of \(\bar{D}_0\), which is open in \(X\). According to the given condition \(U \in D\) for all \(D \in \mathcal{D}\), \(\bar{D}_0^c\) also belongs to \(\mathcal{D}\). Then \(\bar{D}_0^c \cap D_0 = \varPhi\) contradicts the fact that \(\mathcal{D}\) has the FIP.

(b) Let \(D \in \mathcal{D}\). Show that if \(A \supset D\), then \(A \in \mathcal{D}\).

Proof: Because \(\mathcal{D}\) has the FIP, for all \(D' \in \mathcal{D}\), \(D \cap D' \neq \varPhi\). Because \(D\) is contained in \(A\), \(A \cap D' \neq \varPhi\). According to Lemma 37.2 (b), \(A \in \mathcal{D}\).

(c) Show that if \(X\) satisfies the \(T_1\) axiom, there is at most one point belonging to the intersection of all elements in \(\mathcal{D}\), i.e., \(\bigcap_{D \in \mathcal{D}} \bar{D}\).

Proof: Assume that there are at least two points \(x_1\) and \(x_2\) in \(\bigcap_{D \in \mathcal{D}} \bar{D}\). If \(X\) is a Hausdorff space, there are disjoint open sets \(U_1\) and \(U_2\) in \(X\) containing \(x_1\) and \(x_2\) respectively. According to part (a) of this exercise, we have \(U_1 \in \mathcal{D}\) and \(U_2 \in \mathcal{D}\). Then, \(U_1\) and \(U_2\) being disjoint contradicts the fact that \(\mathcal{D}\) has the FIP.

Unfortunately, the given condition in this exercise, i.e. \(X\) satisfies the \(T_1\) axiom, is weaker than the above assumption that \(X\) is Hausdorff. Hence the above proof does not work. However, there seems no obvious or direct proof for the claim in the exercise. This may imply that the original statement is erroneous.

According to the discussion here, a counter example involving the cofinite topology \(\mathcal{T}_c\) on the set of natural numbers \(\mathbb{N}\) is given. It further shows that the intersection of all the elements in the maximal collection \(\mathcal{D}\) is actually \(\mathbb{N}\) itself. This contradicts the claim in the exercise. In the following, the construction of the counter example will be given.

Definition of the cofinite topology

Definition (Cofinite topology) Let \(\mathcal{T}_c\) be the cofinite topology of the space \(X\). Then for all \(U \in \mathcal{T}_c\), either \(U\) is empty or its complement \(U^c\) is finite.

Next, we'll show \(\mathcal{T}_c\) satisfying the conditions in the above definition really defines a topology on \(X\).

1. It is obvious that \(\varPhi\) belongs to \(\mathcal{T}_c\).

2. When \(U = X\), \(U^c = \varPhi\), which is finite. Hence \(X\) belongs to \(\mathcal{T}_c\).

3. Check the closeness of the union operation.

   Let \(\{U_i\}_{i \in I}\) be a collection of open sets in \(\mathcal{T}_c\). If some \(U_i\) in the collection is empty, it has no contribution to the union. Hence we assume all the \(U_i\) in the collection are non-empty.

   Then we have
   \[
   \left( \bigcup_{i \in I} U_i \right)^c = \bigcap_{i \in I} U_i^c,
   \]
   where each \(U_i^c\) is finite. The above intersection of \(\{U_i^c\}_{i \in I}\) is a subset of finite set, which is also finite. Therefore \(\bigcup_{i \in I} U_i \in \mathcal{T}_c\).

4. Check the closeness of the finite intersection operation.

   For a finite collection of open sets in \(\mathcal{T}_c\), we have
   \[
   \left( \bigcap_{k = 1}^n U_k \right)^c = \bigcup_{k = 1}^n U_k^c.
   \]
   Because each \(U_i^c\) is a finite set, the union of a finite number of finite sets is still finite. Hence \(\bigcap_{k = 1}^n U_k \in \mathcal{T}_c\).

Due to the above analysis, \(\mathcal{T}_c\) is really a topology for \(X\). We also know that because every finite set in \(X\) assigned with the topology \(\mathcal{T}_c\) is closed, \(X\) satisfies the \(T_1\) axiom.

Counter example derived from the cofinite topology on \(\mathbb{N}\)

Let the set of natural numbers \(\mathbb{N}\) be assigned with the cofinite topology \(\mathcal{T}_c\). \(\mathbb{N}\) satisfies the \(T_1\) axiom. Let \(\mathcal{C}\) be a collection of all those subsets in \(\mathbb{N}\), each of which has a finite complement. This means all the open sets in \(\mathcal{T}_c\) except \(\varPhi\) are included in \(\mathcal{C}\). Accordingly, the following can be obtained.

1. For all \(U \in \mathcal{C}\), because \(\mathbb{N} - U\) is finite while \(\mathbb{N}\) is infinite, \(U\) is an infinite subset of \(\mathbb{N}\).

2. Let \(\{U_k\}_{k = 1}^n\) be a finite collection arbitrarily selected from \(\mathcal{C}\). Then we have
   \[
   \left( \bigcap_{k = 1}^n U_k \right)^c = \bigcup_{k = 1}^n U_k^c.
   \]
   Because \(U_k^c\) for each \(k\) from \(1\) to \(n\) is a non-empty finite set, their finite union is still finite. Because \(\mathbb{N}\) is infinite, \(\bigcap_{k = 1}^n U_k\) must be infinite, which is a non-empty open set. Therefore \(\mathcal{C}\) has the FIP.

Next, by applying the Zorn's Lemma, a maximal collection \(\mathcal{D}\) exists, which contains \(\mathcal{C}\) as its sub-collection and also has the FIP. For all \(D \in \mathcal{D}\), \(D\) must have infinite number of elements. Otherwise, if \(D = \{d_i\}_{i = 1}^m\), we can select a sub-collection \(\{C_i\}_{i = 1}^m\) from \(\mathcal{C}\), such that \(d_i \notin C_i\). Then \(D \cap C_1 \cap \cdots \cap C_m = \varPhi\), which contradicts the fact that \(\mathcal{D}\) has the FIP.

Select an arbitrary \(x\) in \(D^c\), for any open set \(U\) in \(\mathcal{T}_c\) containing \(x\), it has non-empty intersection with \(D\) because \(D\) is an infinite set. This means any point \(x\) in \(D^c\) is a limiting point of \(D\), so \(\bar{D} = D \cup D^c = \mathbb{N}\). Hence \(\bigcap_{D \in \mathcal{D}} \bar{D} = \mathbb{N}\), which obviously has more than one point.