Theorem 27.1 Let \(X\) be a linearly ordered set with the least upper bound property in the order topology. Every closed interval in \(X\) is compact. Proof Let \(a,b \in X\) with \(a Let \(\mathcal{A}\) be a covering of \([a,b]\) by sets open in \(X\). \( \langle 1 \rangle 1 \) For any \(x \in [a,b)\), there exists \(y \in (x,b]\) such that \([x,y]\) can be covered by at most two elements of \(\mathcal{A}\). Let \(x \in [a,b)\). Pick \(U \in \mathcal{A}\) such that \(x \in U\). Pick \(y > x\) such that \([x,y) \subseteq U\). Assume w.l.o.g. \(y \leq b\). Pick \(V \in \mathcal{A}\) such that \(y \in V\). \([x,y]\) is covered by \(U\) and \(V\). Let \(C = \{y \in (a,b] \mid [a,y] \text{ can be covered by finitely many elements of } \mathcal{A} \}\). Let \(c = \sup C\). \(C\) is nonempty. Pick \(y \in (a,b]\) such that \([a,y]\) can be covered by at most two elements of \(\mathcal{A}\). \(\...
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