Proving partial orders in set theory for $(x_1,y_1 )≤(x_2,y_2 )⇔〖(x〗_1 ≤_X x_2$ and $x_2 ≰_X x_1)$ or $(x_1 =_X x_2$ and $y_1 ≤_Y y_2)$

Question

I'm trying to prove partial order for a set, but I can't seem to be able to fully understand what I'm doing and weather I'm doing it right at all. I don't think I understand how to work with all the conditions between the two partial oders when proving antisymmetry and transitivity because of all the conditions I have to work with to get there.

The exercise is as follows:
Let X be a partial order with the sign $$≤_X$$ and Y be a partial order with the sign $$≤_Y$$ Prove that the following is a partial order for $$X\times Y$$ given that: $$(x_1,y_1 ) \le (x_2,y_2 ) \iff (x_1 \le_X x_2 \text{ and } x_2 \not \le_X x_1)\text{ or }(x_1 =_X x_2 \text { and } y_1 \le_Y y_2)$$

Answer 1

I preassume that $=_X$ denotes the equality relation on $X$.

Then relation $\leq$ on $X\times Y$ is defined by:$$(x_1,y_1)\leq(x_2,y_2)\iff [x_1<_X x_2\text{ or }(x_1=x_2\text{ and }y_1\leq_Y y_2)]$$where $x_1<_X x_2$ abbreviates $x_1\leq_X x_2\text{ and }x_1\neq x_2$.

Based on $x_1=x_1\text{ and }y_1=y_1$ we then find that $(x_1,y_1)\leq(x_1,y_1)$, proving reflexivity.

In order to prove transitivity let $(x_1,y_1)\leq(x_2,y_2)$ and $(x_2,y_2)\leq(x_3,y_3)$.

Now discern the following cases:

• $x_1<_X x_2$ and $x_2<_Xx_3$ leading to $x_1<_X x_3$
• $x_1<_X x_2$ and $[x_2=x_3\text{ and }y_2\leq_Y y_3]$ leading to $x_1<_X x_3$
• $[x_1=x_2\text{ and }y_1\leq_Y y_2]$ and $x_2<_X x_3$ leading to $x_1<_X x_3$
• $[x_1=x_2\text{ and }y_1\leq_Y y_2]$ and $[x_2=x_3\text{ and }y_2\leq_Y y_3]$ leading to $[x_1=x_3\text{ and }y_1\leq_Y y_3]$

In each case the conclusion $(x_1,y_1)\leq(x_3,y_3)$ is justified, so this proves transitivity.

Now it remains to prove antisymmetry, and for that let $(x_1,y_1)\leq(x_2,y_2)$ and $(x_2,y_2)\leq(x_1,y_1)$.

On base of this it must be shown that $(x_1,y_1)=(x_3,y_3)$.

As above discern $4$ cases and give it a try yourself.

Let me know if you run into troubles.