Tychonoff's Theorem/General Case/Proof 2/Lemma 1

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:


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

Then $S$ has a $\subseteq$-maximal element $\AA$.

Proof
We will use Zorn's Lemma for this.

Clearly $S \ne \O$, since $\CC \in S$.

Now let $\family {U_\alpha}_{\alpha \mathop \in A}$ be a non-empty chain in $S$.

Let:


 * $\ds U = \bigcup_{\alpha \mathop \in A} U_\alpha$

We have $U_\alpha \subseteq U$ for each $\alpha \in A$.

It remains to show that $U$ has the finite intersection property.

Let $F \subseteq U$ be a finite set, and write:


 * $F = \set {C_1, C_2, \ldots, C_n}$

For each $1 \le i \le n$, pick $\alpha_i \in A$ such that:


 * $C_i \in U_{\alpha_i}$

Since $\family {U_\alpha}_{\alpha \mathop \in A}$ is a chain, there exists $1 \le k \le n$ such that:


 * $U_{\alpha_i} \subseteq U_{\alpha_k}$ for each $1 \le i \le n$.

Then:


 * $C_i \in U_{\alpha_k}$ for each $1 \le i \le n$.

Since $U_{\alpha_k}$ has the finite intersection property, we have:


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

Since $F \subseteq U$ was an arbitrary finite set, $U$ has the finite intersection property.

So $U$ is an upper bound for $\family {U_\alpha}_{\alpha \in A}$.

So $\struct {S, \subseteq}$ is an ordered set such that every non-empty chain has an upper bound.

So by Zorn's Lemma $\struct {S, \subseteq}$ has a maximal element $\AA$.