This is exercise 5 on page 100 in Chapter 17 of James R. Munkers’ Topology 2nd Edition textbook.

Question

Let X be an ordered set in the order topology. Show that \(\overline{(a, b)} \subset [a, b]\). Under what conditions does equality hold.

Proof

Show \(\overline{(a, b)} \subset [a, b]\)

Let \(x \in \overline{(a, b)}\) in an order topology with no smallest or largest element. Furthermore, \((-\infty, a] \cup [b, \infty)\) are open in \(X\). Thus, \([a, b]\) itself must be closed in \(X\). Since, by definition of the intersection, the only elements available in the closure are those within \([a, b]\), \(x \in [a, b]\).

For the case with a smallest element:

Let \(x \in \overline{(a, b)}\) in an order topology with a smallest element \(a_0\). Thus, the set \((a_0, a) \cup (b, \infty)\) also open in \(X\). Thus, \([a, b]\) is a closed set. With same reasoning as above, \(x \in [a, b]\).

For the case with a largest element:

Similar reasoning applies to the largest element.

Under what conditions does equality hold

Equality holds in continuous domains, like \(\mathbb{R}\), but not discrete domains, like \(\mathbb{Z}\), where single point sets are considered open.

In the case of an order topology of a discrete domain, the single point set \(\{a\}\) is open. With the set \((a, b)\) in mind, the set \((-\infty, a) \cup (a, \infty)\) is closed. This is a set containing \((a, b)\) but not the point \(a\). Thus, \(x \in [a, b]\) such that \(x \notin \overline{(a, b)}\).

Continuous domains lack the single point set. Equality holds then.

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

Extras

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