Description:
A proof that [14 AA] implies [14 Z]
Content:
A proof that Alexander's subbase lemma [14 AA] ( A topological space is compact if and only
if there is a subbase \(S\)for the topology such that every cover by elements of \(S\) has a finite subcover.) implies
[14 Z] (Tychonoff's Compactness Theorem for families of finite spaces.)
[14 AA] implies [14 Z]
Let \(\{X_i: i\in I\}\) be a family of non-empty finite sets and let \(X = \prod_{i\in I}X_{i}\). If \(X\) is empty we are done; otherwise
let \(g\in X\) and for each \(j\in I\) and \(a\in X_j\) let \(B_{j,a}= \{f\in X: f(j)\in X_{j} \backslash\{a\}\}\). \({\cal B} = \{B_{a,j}: j\in I\)
and \(a\in X_j\}\) is a subbase for the topology on X. We show every cover of \(X\) by elements of \({\cal B}\) has a finite subcover. Let
\(\cal C\) be such a cover. If for some \(j\in I\), \(B_{a,j}\) and \(B_{a',j}\) are in \(\cal C\) with \(a\neq a'\), then these two sets
are the required finite subcover. If not then we can define an element of \(X\) that is not covered by \({\cal C}\) as follows:
\(f(j)= \begin{cases} a, & \hbox{ if } B_{j,a} \in {\cal C} \\
g(j), & \hbox{ otherwise }
\end{cases}.
\)
Howard-Rubin number:
34
Type:
Proof
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