# Product Order of Total Orders

Given a product order $$\langle X_1 \times X_2 \times ... \times X_p, \preceq\rangle$$ of ordered sets $$\langle X_1, \preceq^\ast \rangle$$, $$\langle X_2, \preceq^" \rangle$$, ...

Is there a possibility to deduce whether these ordered sets ($$\langle X_1, \preceq^\ast \rangle$$, $$\langle X_2, \preceq^" \rangle$$, ...) are total orders from the properties of the product order?

Let $$A$$ and $$B$$ be ordered sets with $$b$$ in $$B$$, give $$A\times B$$ the product order and let $$p:A\times B \to A$$, $$(x,y) \mapsto x$$ be the first projection.
Show $$p(A\times\{b\})$$ is order isomorphic to $$A$$.
• Thanks for the answer. But I'm looking for some sort of property which tells me right away if the ordered sets are totally ordered sets. If I have the totally ordered sets $X_1, X_2, ..., X_p$ and give the product order $X_1 \times X_2 \times ... \times X_p$. Is it fair to say if the width of the product order is $p$ then the ordered sets $X_1, X_2, ..., X_p$ are all totally ordered? – andrestless Jan 26 '19 at 16:55
• @andrestless Suppose $X$ is the product of two two-element chains, and $Y$ is a one-element chain. In this situation, $X \times Y$ has width $2$, and you would conclude $X$ and $Y$ were chains, which is not the case. On the other hand, if $X$ and $Y$ are three-element chains, then $X\times Y$ has width $3$, and you would conclude that at least one of $X$ and $Y$ were not a chain, which they were. So that condition of yours is neither necessary nor sufficient. – amrsa Jan 27 '19 at 11:37