Let \(X\) be an ordered set in the order topology. Show that if \(X\) is connected, then \(X\) is a linear continuum.
Solution
Let \(X\) be connected in its order topology.
First, the order must be dense. If \(x<y\) and no point lies strictly between them, then \[ (-\infty,y)\qquad\text{and}\qquad(x,\infty) \] are disjoint nonempty open sets whose union is \(X\), a separation. Thus for every \(x<y\) there is \(z\) with \(x<z<y\).
Now let \(A\subseteq X\) be nonempty and bounded above, and let \(B\) be the set of upper bounds of \(A\). If \(B\) had no least element, then both \(B\) and \(X-B\) would be nonempty open sets. Indeed, if \(b\in B\), choose a smaller upper bound \(b'<b\); then \((b',\infty)\) is a neighborhood of \(b\) contained in \(B\). If \(x\notin B\), choose \(a\in A\) with \(x<a\); then \((-\infty,a)\) is a neighborhood of \(x\) contained in \(X-B\). These two sets would separate \(X\), contradiction.
Therefore \(B\) has a least element, which is exactly \(\sup A\). Hence \(X\) has the least-upper-bound property and a dense order, so it is a linear continuum.