- #1
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If [tex]M_1[/tex] and [tex]M_2[/tex] are ordered sets, the ordered sum [tex]M_1+M_2[/tex] is the set [tex]M_1\cupM_2[/tex] with the ordering defined as:
If [tex]a,b \epsilon M_1[/tex] or [tex]a,b \epsilon M_2[/tex] then order them as they would be in the original orderings. If [tex]a \epsilon M_1[/tex] and [tex]b \epsilon M_2[/tex] then [tex]a<b[/tex]
The question then is if [tex]a \epsilon M_1[/tex] and [tex]a \epsilon M_2[/tex], then we get [tex]a< a[/tex] which is impossible. In general, it seems you'll get a is less than and greater than some elements, which means [tex]M_1+M_2[/tex] isn't really ordered at all
(I use epsilon as the 'element of' symbol as I couldn't find a more appropriate one in the latex pdfs)
If [tex]a,b \epsilon M_1[/tex] or [tex]a,b \epsilon M_2[/tex] then order them as they would be in the original orderings. If [tex]a \epsilon M_1[/tex] and [tex]b \epsilon M_2[/tex] then [tex]a<b[/tex]
The question then is if [tex]a \epsilon M_1[/tex] and [tex]a \epsilon M_2[/tex], then we get [tex]a< a[/tex] which is impossible. In general, it seems you'll get a is less than and greater than some elements, which means [tex]M_1+M_2[/tex] isn't really ordered at all
(I use epsilon as the 'element of' symbol as I couldn't find a more appropriate one in the latex pdfs)