Consider the following sets in the dictionary order. Which are linear continua?
\(\mathbb{Z}_+ \times [0, 1)\)
\([0, 1) \times \mathbb{Z}_+\)
\([0, 1) \times [0, 1]\)
\([0, 1] \times [0, 1)\)
Solution
The linear continua are exactly (a) and (c).
\(\mathbb Z_+\) is well ordered, so by E-U00IE, \(\mathbb Z_+\times[0,1)\) is a linear continuum.
The subset \[ \{0\}\times\mathbb Z_+ \] of \([0,1)\times\mathbb Z_+\) is bounded above, for example by \((1/2,1)\). It has no least upper bound: any upper bound must have positive first coordinate, and there is no least positive element of \([0,1)\). Hence the least-upper-bound property fails.
The order is dense. For the least-upper-bound property, let \(A\) be nonempty and bounded above. Let \(s\) be the supremum of the first coordinates of points of \(A\). If \(s\) occurs as a first coordinate in \(A\), take the supremum of the corresponding second coordinates in the complete interval \([0,1]\); this gives the least upper bound. If \(s\) does not occur, then \((s,0)\) is the least upper bound; it lies in \([0,1)\times[0,1]\) because \(A\) has an upper bound with first coordinate \(b<1\), so \(s\le b<1\). Thus \([0,1)\times[0,1]\) is a linear continuum.
The subset \[ \{0\}\times[0,1) \] of \([0,1]\times[0,1)\) is bounded above, for instance by \((1/2,0)\), but has no least upper bound: an upper bound must have first coordinate \(>0\), and there is no least positive real number. Thus (d) is not a linear continuum.