Tychonoff's Theorem/General Case/Proof 2/Lemma 3

Lemma

Let $X$ be a set.

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

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

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 $A \subseteq X$ be such that:

$A \cap B \ne \O$

for all $B \in \AA$.

Then $A \in \AA$.

Proof

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

By Lemma 2, we have:

$\ds \bigcap_{i \mathop = 1}^n C_i \ne \O$

Hence, by hypothesis, we have:

$\ds A \cap \bigcap_{i \mathop = 1}^n C_i \ne \O$

So $\AA \cup \set A \in S$ has the finite intersection property.

Since $\AA$ is $\subseteq$-maximal in $S$, we have:

$\AA \cup \set A = \AA$

so that $A \in \AA$.

$\blacksquare$