Prove that if \(X\) is an ordered set in which every closed interval is compact, then \(X\) has the least upper bound property.
Solution
Let \(A\subset X\) be nonempty and bounded above. Choose \(a_0\in A\) and an upper bound \(b\) of \(A\). For each \(a\in A\), consider the closed subset \[ F_a=[a,b]\subset[a_0,b]. \] The family \(\{F_a:a\in A\}\) has the finite intersection property: for finitely many \(a_1,\dots,a_n\in A\), one of them is largest, say \(a_j\), and then \[ F_{a_1}\cap\cdots\cap F_{a_n}=[a_j,b]\ne\varnothing. \] Since the closed interval \([a_0,b]\) is compact by hypothesis, \[ F=\bigcap_{a\in A}F_a\ne\varnothing. \] The set \(F\) is exactly the set of upper bounds of \(A\) that lie in \([a_0,b]\).
It remains to show that the nonempty compact ordered subspace \(F\) has a least element. Suppose it had none. For each \(x\in F\), choose \(y_x\in F\) with \(y_x<x\). Then the sets \[ (y_x,\infty)\cap F \] form an open cover of \(F\): given \(z\in F\), choose \(x\in F\) with \(x<z\), and then \(z\in(y_x,\infty)\cap F\) since \(y_x<x<z\). By compactness, finitely many suffice. Let \(y\) be the least of the finitely many left endpoints \(y_{x_1},\dots,y_{x_n}\). Then \(y\in F\), but \(y\) belongs to none of the sets \((y_{x_i},\infty)\cap F\), contradiction. Thus \(F\) has a least element.
That least element is the least upper bound of \(A\). Therefore \(X\) has the least upper bound property.