[Sophie wrote](https://forum.azimuthproject.org/discussion/comment/18107/#Comment_18107):
> > in other words, we have \\((x, y) \le (x', y')\\) iff \\(x \le x' \land y \le y'\\)
> I assume you used \\(\land\\) to mean "and" and not "join". Is that correct?

I can definitely see how that's ambiguous! I did mean "and". (Though technically, "and" is a join on some lattice of propositions, so it's not an _incorrect_ reading, as long as the right parentheses are inferred.)

(EDIT: Also, lest I fall back into bad habits, I think \\(\wedge\\) is the symbol for “meet”, the greatest lower bound. I do find it more natural sometimes to think of “true” as the informationless bottom element though.)

> Second I am still unclear about what ordering you are getting on \\(\mathbb R ^2 \\) from product. You wrote:
> > this lifts to the product straightforwardly
> Any chance you could make that explicit?

Sure thing. Generally when working with products, things "lift" componentwise. So we say \\((x, y) \le (x', y')\\) iff both \\(x \le x'\\) and \\(y \le y'\\). The lifting I'm referring to in the quote is that the fact that \\(+\\) respects the order on \\(R\\) lifts in the “usual” sense to how the component-wise \\(+\\) respects the order on \\(R^2\\). Explicitly, if \\((x, y) \le (x', y')\\) and we have some \\((r, s) \in \mathbb{R}^2\\), we know that \\((x, y) + (r, s) \le (x', y') + (r, s)\\) because \\(x + r \le x' + r\\) and \\(y + s \le y' + s\\) individually.