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 $A$.

Thus knowing the product order, one knows the order of the components.

  • $\begingroup$ 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? $\endgroup$ – andrestless Jan 26 '19 at 16:55
  • $\begingroup$ @andrestless. Does the sum of the widths equal the width of a not empty product? $\endgroup$ – William Elliot Jan 26 '19 at 23:20
  • $\begingroup$ @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. $\endgroup$ – amrsa Jan 27 '19 at 11:37
  • $\begingroup$ @amrsa Why is the width of the first X one? $\endgroup$ – William Elliot Jan 28 '19 at 0:31
  • $\begingroup$ @amrsa. Prove the width of {0,1,2}×{0,1,2} is three. $\endgroup$ – William Elliot Jan 28 '19 at 0:34

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