Task: Let $A = \{(n, k) ∈ N × N : k \leq n\}$ and $B = \{(n, k) ∈ N × N : n \leq k\}$. Consider the restriction of the lexicographic order $N ×_{lex} N $to these sets: a pair $(n_1, k_1)$ is less than a pair $(n_2, k_2)$ if $n_1 < n_2$ or $n_1 = n_2$ and $k_1 < k_2$. Are these orders isomorphic on sets A and B?
My thoughts: Let there be two pairs $(n_1, k_1) ~ and ~ (n_2, k_2)$ from the set A such that $(n_1, k_1) <_{lex} (n_2, k_2)$. Then this means that either $n_1 < n_2, ~ or ~ n_1 = n_2 ~ and ~ k_1 < k_2$.
The first case: $n_1 < n_2$. Then for pairs of pairs $(n_1, k_1) ~ and ~ (n_2, k_2)$, $n_1 \leq n_2$ is also performed (since $n_1 > k_1 ~ and ~ n_2 > k_2$). This means that the pair $(n_1, k_1)$ from the set A must also be a pair from the set B. Thus, the lexicographic order on set A is a subset of the order on set B.
The second case: $n_1 = n_2 ~ and ~ k_1 < k_2$. Then it follows from the condition for the set B that the pairs $(n_1, k_1) ~ and ~ (n_2, k_2)$ are not included in the set B. This means that the lexicographic order on set A contains pairs that are not in order on set B.
Thus, the lexicographic order on the set A is a superset of the order on the set B. These orders are not isomorphic.
The question is, is this a rigorous proof or should it have been done in some other way?