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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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$(a,1)$ isn't maximal in the simple product, as $(a + \epsilon,1)$ is larger $\epsilon > 0$ so small, that $a+ \epsilon < 1$. –  martini Oct 25 '12 at 11:56
    
@martini Oops, you are right of course. Thank you! –  Matt N. Oct 25 '12 at 13:06
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I've converted the images to TeX(t); please check to ensure that it has been a faithful transcription. –  Arthur Fischer Oct 25 '12 at 13:34
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1 Answer

up vote 1 down vote accepted

$[0,1)\otimes^s[0,1]$ 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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Thank you for the hint. I was going to post a related question (I still might): to define ordinal multiplication one apparently uses the anti-lexicographic order. One could use the lexicographic order instead if they are order isomorphic, right? –  Matt N. Oct 25 '12 at 13:09
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@MattN.: Yes, for thinking about it. For reading and writing about it you have to remember that for example $\omega\cdot2$ is $2$ copies of $\omega$ end to end, not $\omega$ copies of $2$ end to end. In other words, the order-type of $\alpha\cdot\beta$ is the lexicographic order on $\beta\times\alpha$. –  Brian M. Scott Oct 25 '12 at 13:34
    
Wait, I'm very puzzled by which is a consequence of which: of course if we define multiplication as the product with the anti-lexicographic order then $\omega \cdot 2$ looks like $\omega + \omega$. If we define $\omega \cdot 2$ to be $\omega \times 2$ with the lexicographic order then it looks like $\omega$ copies of $2$. So it makes no difference in the sense that they do the same thing. But the lexicographic order seems much more intuitive. So why go through the pain of defining something using the non-intuitive order when we get the same operation? (except with operands swapped) –  Matt N. Oct 25 '12 at 17:09
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@MattN.: We define $\alpha\cdot\beta$ to be the ordinal with the order-type of $\alpha\times\beta$ ordered antilexicographically, which is the same as the order-type of $\beta\times\alpha$ ordered lexicographically. I don’t know how that convention came to be established; I’ve always thought it a bit backwards, but that’s the way it is. –  Brian M. Scott Oct 25 '12 at 18:41
    
: ) Thank you! I wasn't sure whether there was a particular property we want this multiplication to have which we only get if we define it anti-lexicographically. Disturbing. I'd like it much better lexicographically. –  Matt N. Oct 25 '12 at 18:44
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