Tychonoff's Theorem/General Case/Proof 2/Lemma 2

Lemma

Let $X$ be a set.

We say that a family $\CC$ of subsets of $X$ has the "finite intersection property" if:

$\ds \bigcap_{S \in \SS} S \ne \O$

for all finite subsets $\SS \subseteq \CC$.

Let:

$\ds S = \set {\CC \subseteq \AA : \AA \text { has the finite intersection property} }$

Let $\AA$ be a $\subseteq$-maximal element of $S$ given by Lemma 1.

Let $A, B \in \AA$.

Then $A \cap B \in \AA$.

Proof

Let $\set {C_1, C_2, \ldots, C_n}$ be a finite subset of $\AA$.

Then $\set {A, B, C_1, C_2, \ldots, C_n}$ is a finite subset of $\AA$.

Then since $\AA$ has the finite intersection property:

$\ds A \cap B \cap \bigcap_{i \mathop = 1}^n C_i = \paren {A \cap B} \cap \bigcap_{i \mathop = 1}^n C_i \in \AA$

for every finite subset $\set {C_1, C_2, \ldots, C_n}$ of $\AA$.

So $\AA \cup \set {A \cap B} \in S$ also has the finite intersection property.

Since $\AA$ is $\subseteq$-maximal in $S$, we have $\AA \cup \set {A \cap B} = \AA$.

That is:

$A \cap B \in \AA$

$\blacksquare$