This is exercise 1 on page 177 in Chapter 27 of James R. Munkers’ Topology 2nd Edition textbook.

Question

Prove that if \(X\) is an ordered set in which every closed interval is compact, then \(X\) has the least upper bound property.

Proof

Suppose \(X\) is an ordered set in which every closed interval is compact. Suppose \(C\) is a closed set of \(X\). Because \(C\) is compact by assumption, we know the union of some finite collection of open sets equals \(C\). Thus, \(C\) must also be open.

Now, consider the closed ray \((-\infty, a]\) where \(a \in X\). This must also be open. Therefore, the only open interval we can form with an inclusive upper boundary consists of those of the form (a, b] where \(a < b \in X\) and \(b\) is the largest element of \(X\).

Because \(X\) has a largest element, the set \(C\) must be bounded above. To find the least upper bound of \(C\), we may take the smallest element of all elements that are bounded above \(C\). Thus, showing the set \(C\) has the least upper bound property, completing the proof.

\[\tag*{$\blacksquare$}\]

Extras

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