How to prove (x,y) is a minimal/maximal element of AxB if x∈A is a minimal/maximal element and y∈B is a minimal/maximal element

Question

Not exactly sure what to do here.

I'm assuming that since x ∈ A and y ∈ B then we can assume that (x,y) ∈ AxB.

Not sure what I'm stuck on. Can we not just say that since x and y are minimal, then (x,y) is minimal in AxB?

Answer 1

In the product order, we have

$$(x,y) ≤ (x′, y′) \qquad≡\qquad x ≤ x′ \;∧\; y ≤ y′$$

Recall $$ m \text{ minimal } ≡ (∀ m′ \,•\, m′ ≤ m \;⇒\; m′ = m) $$

So, for any $x′, y′$ we calculate: \begin{align} & (x′, y′) ≤ (x, y) \\ ≡ & \text{ \{ Product Order definition \} } \\ & x′ ≤ x \;∧\; y′ ≤ y \\ ≡ & \text{ \{ Both $x,y$ are minimal \} } \\ & x′ = x \;∧\; y′=y \\ ≡ & \text{ \{ Equality of pairs \} } \\ & (x′, y′) = (x, y) \end{align}

Hence $(x, y)$ is minimal provided each $x, y$ is minimal.