# Partial order involving cartesian product: P x Q

So I'm a bit confused with the concept of a partial order.

P = {1, 4}

Q = {1, 2}

How do you define a partial order Z on (P x Q)? and how would the hasse diagram of Z look like? is it even possible with these two sets?

I know a partial order has to be reflexive, antisymmetric, and transitive but I'm still confused with the idea of defining a partial order.

Thank you

• You can define a partial order on any set. If the set has more than one element, you can define more than one partial order on it. Are there any additional conditions, or do you simply have to define any old partial order on this set? – Brian M. Scott Oct 7 '15 at 21:09
• No conditions. Just a partial order on this set. How do you define one? – Mathy Oct 7 '15 at 21:14
• The relation of equality is a partial order. Or you can work from the obvious orders on $P$ and $Q$ and write $\langle p_0,q_0\rangle\preceq\langle p_1,q_1\rangle$ if and only if $p_0\le p_1$ and $q_0\le q_1$. (Of course you need to verify that both of these really are partial orders.) – Brian M. Scott Oct 7 '15 at 21:16
• How about if the condition was the subset ordered by inclusion? (I think that's one of the properties right?) – Mathy Oct 7 '15 at 21:21
• Would Z = (1, 1) be a partial order of (P x Q)? It passes all the restrictions; reflexive, antisymmetric, and transitive – Mathy Oct 7 '15 at 21:27

The elements of $P\times Q$ are the ordered pairs $\langle 1,1\rangle,\langle 1,2\rangle,\langle 4,1\rangle$, and $\langle 4,2\rangle$. Make any Hasse diagram with those four entries. For instance, you could have this Hasse diagram:

          <1,1>
|
<1,2>
|
<4,1>
|
<4,2>


In that case your partial order would actually be a total (or linear) order. Or you could have this one:

          <4,2>
/   \
/     \
<4,1>   <1,1>
\     /
\   /
<1,2>


And there are many, many other possibilities. The relation corresponding to this second Hasse diagram is the set containing the following ordered pairs of elements of $P\times Q$:

\begin{align*}&\big\langle\langle 1,2\rangle,\langle 1,2\rangle\big\rangle,\big\langle\langle 1,2\rangle,\langle 1,1\rangle\big\rangle,\big\langle\langle 1,2\rangle,\langle 4,1\rangle\big\rangle,\big\langle\langle 1,2\rangle,\langle 4,2\rangle\big\rangle,\\ &\big\langle\langle 4,1\rangle,\langle 4,1\rangle\big\rangle,\big\langle\langle 4,1\rangle,\langle 4,2\rangle\big\rangle,\big\langle\langle 1,1\rangle,\langle 1,1\rangle\big\rangle,\big\langle\langle 1,1\rangle,\langle 4,2\rangle\big\rangle,\\ &\big\langle\langle 4,2\rangle,\langle 4,2\rangle\big\rangle \end{align*}

Try to see why, then try to write out the relation corresponding to the first Hasse diagram and to the one below:

    <1,1>        <1,2>        <4,1>        <4,2>