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