Cartesian relation $\preceq$ on $N\times N$ where $(a,b)\preceq (c,d)$ if $a\leq c$ and $b\leq d$.

Question

Studding for midterm. Lets define the Cartesian relation $\preceq$ on $N\times N$ where $(a,b)\preceq (c,d)$ if $a\leq c$ and $b\leq d$.

I'm trying to prove that the partially arranged set $\left(N \times N, \preceq\right)$ maintains the minimum condition.

As I understand $\left(A,\leq\right)$ is partially arranged set that maintains the minimum condition if for each $B\subseteq A$, $B\neq \varnothing$ contains a minimal number (at least one).

I understand the definition but how do I use it to prove the statement? Do I need to use induction for that?

Answer 1

HINT: Suppose that $A\subseteq\Bbb N\times\Bbb N$ has no $\preceq$-minimal element. Then there must be a sequence $\big\langle\langle a_n,b_n\rangle:n\in\Bbb N\big\rangle$ such that $\langle a_{n+1},b_{n+1}\rangle\prec\langle a_n,b_n\rangle$ for each $n\in\Bbb N$. Now consider the sequence $\langle a_n:n\in\Bbb N\rangle$ and $\langle b_n:n\in\Bbb N\rangle$ to get a contradiction.

Answer 2

Take $T \subseteq N \times N$. Define $T_a= \{a \in N : \exists b \ s.t. (a,b) \in T\}$. Since $N$ is well ordered exists $a_{min}=\min T_a$. Set $T_b=\{b \in B: (a_{min},b)\in T\}$. Since $N$ is well ordered exists $b_{min}=\min T_b$. Then it is easy to see that $(a_{min},b_{min})$ is a minimal element of $T$