How ae these three order types on $Z_+ \times Z_+$ different?

Question

Let $Z_+$ denote the set of positive integers. Consider the following relations on $Z_+ \times Z_+$:

1. The dictionary order; that is, $(x_0,y_0) < (x_1,y_1)$ if either $x_0 < x_1$, or $x_0 = x_1$ and $y_0 < y_1$.

2. $(x_0, y_0) < (x_1,y_1)$ if either $x_0 - y_0 < x_1 - y_1$, or $x_0 - y_0 = x_1 - y_1$ and $y_0 < y_1$.

3. $(x_0, y_0) < (x_1,y_1)$ if either $x_0 + y_0 < x_1 + y_1$, or $x_0 + y_0 = x_1 + y_1$ and $y_0 < y_1$.

In these order relations, which elements have immediate predecessors?

In which order relations, does the set $Z_+ \times Z_+$ have a smallest element?

How to show that all the above three order types are different?

Let $A$ and $B$ be two non-empty sets with the order relations $<_A$ and $<_B$, respectively. Then the sets $A$ and $B$ are said to have the same order types if there exists a bijective function $f \colon A \to B$ such that $a_1 <_A a_2$ implies $f(a_1) <_B f(a_2)$ for any pair of elements $a_1$, $a_2$ in $A$. Otherwise, $A$ and $B$ are said to have different order types.

Answer 1

I think I've arrived at the answers to my own questions.

In the dictionary order relation, each element of $Z_+ \times Z_+$ has an immediate predecessor, except for elements of the form $(m, 1)$ for any $m \in Z_+$. Thus, the elements $(1,1)$, $(2,1)$, $(3,1)$, $...$ have no immediate predecessors; all other elements have immediate predecessors. The element $(1,1)$ is the smallest element in the dictionary ordering of $Z_+ \times Z_+$.

In the order defined as $(x_0,y_0) < (x_1,y_1)$ if $x_0-y_0 < x_1-y_1$, or $x_0-y_0 = x_1-y_1$ and $y_0 < y_1$, each element $(m,n)$ has as its immediate predecessor the element $(m-1, n-1)$, provided that $m>1$ and $n>1$; the elements of the form $(m,1)$ and $(1,n)$ have no immediate predecessors. For every element $(m,n)$, the element $(m,n+1)$ precedes it in this ordering; so there's no smallest element under this ordering of $Z_+ \times Z_+$.

In the order defined as $(x_0,y_0) < (x_1,y_1)$ if $x_0+y_0 < x_1+y_1$, or $x_0+y_0 = x_1+y_1$ and $y_0 < y_1$, every element $(m,n)$ has an immediate predecessor, namely the element $(m+1,n-1)$, except the element $(1,1)$, which has no immediate predecessor. The element $(1,1)$ is the smallest element.

From the above observations, it is now easy to show these three orderings to be different.