Show that if \(X\) is a well-ordered set, then \(X \times [0, 1)\) in the dictionary order is a linear continuum.
Solution
Let \[ L=X\times[0,1) \] with dictionary order, where \(X\) is well ordered.
First we prove the least-upper-bound property. Let \(A\subseteq L\) be nonempty and bounded above. Let \[ C=\{x\in X:\text{some }(x,t)\text{ is an upper bound of }A\}. \] The set \(C\) is nonempty, so it has a least element \(x_0\).
If \(A\) contains points with first coordinate \(x_0\), let \[ T=\{t:(x_0,t)\in A\}. \] Because some \((x_0,s)\) is an upper bound, \(T\) is bounded above by \(s<1\). Let \(t_0=\sup T\). Then \((x_0,t_0)\) is the least upper bound of \(A\).
If \(A\) contains no point with first coordinate \(x_0\), then every point of \(A\) has first coordinate \(<x_0\), and \((x_0,0)\) is the least upper bound. Thus \(L\) has the least-upper-bound property.
Now let \((x,s)<(y,t)\). If \(x=y\), choose \(u\) with \(s<u<t\). If \(x<y\), choose \(u\) with \(s<u<1\); then \[ (x,s)<(x,u)<(y,t). \] Hence the order is dense. Therefore \(X\times[0,1)\) is a linear continuum.