Let poset A = ( {a, b}, { (b, a), (a, a), (b, b) } ) and B = ( {0, 1}, { (0, 1), (0, 0), (1, 1) } )
In abuse of notation, I'll also write A = {a, b}, B = {0, 1}, and \$$\leq_A \$$ = { (b, a), (a, a), (b, b) }, and \$$\\leq_B \$$ = { (0, 1), (0, 0), (1, 1) }.

You can form a new poset by taking a pairwise union of sets:

C = (A \$$\cup\$$ B, \$$\leq_A \cup \leq_B\$$) = ( {a, b, 0, 1}, { (0, 1), (0, 0), (1, 1), (b, a), (a, a), (b, b) } )

Like you said, this isn't a total order. The order can be flattened by saying "all letters are less than numbers". In notation, this is done by setting \$$\leq_C \$$ to \$$\leq_C \cup \$$ { (a, 0), (a, 1), (b, 0), (b, 1), (c, 0), (c, 1) }

For what it's worth, I don't really like thinking of relations as certain special types of sets. But it's consistent with this book. Your question seemed like a nice way to practice manipulating posets as sets.

**Edit** And it seems that although in this case we do get a poset by performing a union, this isn't always true. [David Tanzer](https://forum.azimuthproject.org/discussion/comment/16303/#Comment_16303) gave an example of where a normal union fails, and some more typical examples of products between posets.

[John Baez](https://forum.azimuthproject.org/discussion/comment/16321/#Comment_16321) Gave insights for how subtle these operations can be. Taking the union as I did is apparently complicated in general. So the procedure I described doesn't generalize well. A "disjoint union" is apparently a much safer operation. So I think your basic relations combined nicely, because the underlying sets are different.