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$\newcommand{\angles}[1]{\langle { #1 } \rangle}$ I did the following exercise, can you tell me if I got it right? Thanks:

EXERCISE 31(G): Consider $\angles{ [0,1) , \leq } \otimes^s \angles{ [0,1] , \leq }$, $\angles{ [0,1) , \leq } \otimes^l \angles{ [0,1] , \leq }$, and $\angles{ [0,1) , \leq } \otimes^a \angles{ [0,1] , \leq }$. Find all minimal, minimum, maximal and maximum elements of these p.o.'s.

where we have the simple product

One can meaningfully define three kinds of products of p.o.'s. The simple product $\angles{ A , \preceq_A } \otimes^s \angles{ B , \preceq_B }$ (or shorthand $A \otimes^s B$) is the p.o. $\angles{ A \times B , \preceq_s }$, where the relation $\preceq_s$ is given by $\angles{ a , b } \preceq_s \angles{ c , d } \quad\text{iff}\quad a \preceq_A b \text{ and } b \preceq_B d.$

the lexicographic order

The lexicographic product or lexicographic order $\angles{ A , \preceq_A } \otimes^l \angles{ B , \preceq_B }$ (or shorthand $A \otimes^l A$) is the p.o. $\angles{ A \times B , \preceq_l }$, where the relation $\preceq_l$ is given by $ \angles{ a , b } \preceq_l \angles{ c , d } \;\;\text{iff}\;\; \begin{cases} a \prec_A c \quad\text{or}\\ a = c \text{ and } b \preceq_B d. \end{cases}$

and the antilexicographic order

The antilexicographic product or antilexicographic order $\angles{ A , \preceq_A } \otimes^a \angles{ B , \preceq_B }$ (or shorthand $A \otimes^a A$) is the p.o. $\angles{ A \times B , \preceq_a }$, where the relation $\preceq_a$ is given by $ \angles{ a , b } \preceq_a \angles{ c , d } \;\;\text{iff}\;\; \begin{cases} b \prec_B d \quad\text{or}\\ b = d \text{ and } a \preceq_A c. \end{cases}$

(Quoted text -- and linked images -- are from W. Just and M. Weese, Discovering Modern Set Theory, vol.1, p.25.)


For the simple product I got:

minimum = $(0,0)$ and hence only one minimal element. No maximum but maximal elements $\{ (a,1) \mid a \in [0,1) \}$.

For the lexicographic order I got:

minimum = $(0,0)$ and hence only one minimal element. No maximal elements and no maximum.

For the antilexicographic order I got:

minimum = $(0,0)$ and hence only one minimal element. No maximal elements and no maximum.

Thanks for your help!

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    I've converted the images to TeX(t); please check to ensure that it has been a faithful transcription.2012-10-25

1 Answers 1

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$[0,1)\otimes^s[0,1]$ has no maximal elements: for any $\langle a,b\rangle\in[0,1)\times[0,1]$ you have

$\langle a,b\rangle\prec^s\left\langle\frac12(a+1),b\right\rangle\;,$

since $a<\frac12(a+1)<1$. The rest is correct.

By the way, it may be easier to think about $[0,1)\otimes^a[0,1]$ if you realize that it’s isomorphic to $[0,1]\otimes^l[0,1)$, since the lexicographic order comes up a bit more often and is likely to be a bit more familiar.

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    @user14111: Yes, that probably explains the choice. Whether the cure is worse than the disease is a matter of individual taste.2013-05-29