I wanted to ask if these orders are isomorphic – namely if my solutions are correct. Here $\leq$ means the standard order on integer or rational numbers and $\leq_\text{lex}$ means lexicographical defined with the help of standard order on axes.
- $(\mathbb{Q},\leq)$ and $(\mathbb{Z},\leq)$ – I think these orders aren't isomorphic, because the set of rational numbers is dense and of integers not,
- $(\mathbb{Z}\times \mathbb{Z},\leq_\text{lex})$ and $(\mathbb{Z},\leq)$ – I think these orders are isomorphic (I do not know how to argue it formally),
- $(\mathbb{Q},\leq)$ and $((0,\infty)\cap \mathbb{Q},\leq)$ – I think these orders aren't isomorphic, because the first set hasn't got the least element in the opposite to the second.
I would really appreciate checking the correctness of these solutions.
Best Answer
The first one is correct.
In the second one the orders are not isomorphic. Note that in the lexicographic order on $\mathbb{Z}\times\mathbb{Z}$ there are infinitely many elements between $(1,1)$ and $(2,1)$. Indeed, for each $n\geq 2$ we have $(1,1)<(1,n)<(2,1)$. On the other hand, in $\mathbb{Z}$ there are only finitely many elements between each two elements. Use this to show that there can't be an isomorphism here.
The third one: the orders are isomorphic by Cantor's theorem, which says that any two orders on countable sets which are linear, dense and have no first and last elements are isomorphic to each other.