Can Ordered Sum of Sets Be Well-Defined in All Cases?

[Office_Shredder]

If M_1 and M_2 are ordered sets, the ordered sum M_1+M_2 is the set M_1\cupM_2 with the ordering defined as:

If a,b \epsilon M_1 or a,b \epsilon M_2 then order them as they would be in the original orderings. If a \epsilon M_1 and b \epsilon M_2 then a<b

The question then is if a \epsilon M_1 and a \epsilon M_2, then we get a< a which is impossible. In general, it seems you'll get a is less than and greater than some elements, which means M_1+M_2 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)

 

[Coin]

If a,b \epsilon M_1 or a,b \epsilon M_2 then order them as they would be in the original orderings. If a \epsilon M_1 and b \epsilon M_2 then a<b

The question then is if a \epsilon M_1 and a \epsilon M_2, then we get a< a which is impossible.

It seems like there's a loophole here since if a \epsilon M_1 and a \epsilon M_2, then you should be interpreting the ordering there under the first clause, not the second clause (since a,a \epsilon M_1).

Alternately how is "M_1_2" defined? If this is a union, then shouldn't the a = a case never come up since something cannot be a member of a set "more than once"?

Otherwise maybe whoever you're getting this from just made a mistake in their wording...

In general, it seems you'll get a is less than and greater than some elements, which means M_1+M_2 isn't really ordered at all

Well, this again comes back to the wording being kind of confusing. Let's say you have a, b where a,b \epsilon M_1 and also a,b \epsilon M_2. And let's say by M_1's ordering a < b, and by M_2's ordering b < a. What do you do here?

However if you can somehow resolve this case, for example if the original wording gives one some excuse to declare that M_1's ordering takes precedence, then I think (a < b ?) will always be unambiguous.

 

[Office_Shredder]

Right, I see what you're talking about with the case a,a \epsilon M_1 cutting you off from 'seeing' the a \epsilon M_1 and a \epsilon m_2 case. But you still have trouble if b \epsilon M_1, c \epsilon M_2 and a<b in M_1 c<a in M_2 then a<b<c<a which means it's not transitive.

I have to apologize, when I wrote M_{12} it was just poorly writing M_1 \cup M_2 so it didn't come out right. But even though a can only be an element of that set once, the way the ordering is defined it still works out fishily, unless it's modified to be if a is in M_2 and not M_1 then it gets ordered as if it was in M_2.

EDIT: This isn't well defined, as if M_1=M_2=N then the order type of \omega + \omega = \omega which certainly isn't true if M_1=N M_2=Z_- where the negative integers are ordered by their absolute value