> A bunch of these are fairly easy to find. You just need to take Matthew's idea and generalize it a little. So, I bet you folks can find them!

I mentioned yesterday that I had in mind a proper poset for an answer, rather than a preorder. Jonathan gave a poset - the [product order](https://en.wikipedia.org/wiki/Product_order) of \\(\langle R, \leq \rangle\\) with itself.

Another monoidal poset that works is the [lexicographic order](https://en.wikipedia.org/wiki/Lexicographical_order) of \\(\langle R, \leq \rangle\\) with itself.

$$

x + i y \leq_{lex} p + i q \iff x < p \text{ or } (x = p \text{ and } y \leq_{\mathbb{R}} q)

$$

We can again get \\(2^{\aleph_0}\\) answers from this by just observing a different trick - this answer works even if we *translate* the complex plane around:

$$

a \leq_{lex, k} b \iff a + k \leq_{lex} b + k

$$

I have a little hunch this might be part of what John has in mind...

I mentioned yesterday that I had in mind a proper poset for an answer, rather than a preorder. Jonathan gave a poset - the [product order](https://en.wikipedia.org/wiki/Product_order) of \\(\langle R, \leq \rangle\\) with itself.

Another monoidal poset that works is the [lexicographic order](https://en.wikipedia.org/wiki/Lexicographical_order) of \\(\langle R, \leq \rangle\\) with itself.

$$

x + i y \leq_{lex} p + i q \iff x < p \text{ or } (x = p \text{ and } y \leq_{\mathbb{R}} q)

$$

We can again get \\(2^{\aleph_0}\\) answers from this by just observing a different trick - this answer works even if we *translate* the complex plane around:

$$

a \leq_{lex, k} b \iff a + k \leq_{lex} b + k

$$

I have a little hunch this might be part of what John has in mind...