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 $\AA$.
$\blacksquare$