Lesson 7

To establish the isomorphism of two orders, it is sufficient to provide the corresponding bijection. But how can one prove that two orders are not isomorphic? Let us give a few examples.

The orders \(([0,1], \le)\) and \(((0, 1), \le)\) (here standard number comparison is used) are not isomorphic because the first order has \(0\) as the minimal element: there is no other element smaller than zero. A bijection would have to map such an element to a minimal one as well.

Under isomorphism, a segment of the form \([x,y]=\{z \colon x \preceq z \preceq y\}\) must map to a segment. For example, for this reason, \((\mathbb{Z}, \le)\) is not isomorphic to \((\mathbb{Q}, \le)\): in \(\mathbb{Z}\) all segments are finite, while in \(\mathbb{Q}\) they are infinite.

Similarly, \((\mathbb{Z}, \le)\) is not isomorphic to \(\mathbb{Z}+\mathbb{Z}\): what segment could map to the infinite segment \([0,0']\)?