Is $(On,\in)$ + $(On, \in)$ a well-order?

40 Views Asked by At

the definition of a well-order we were given is that it is a left-narrow, well-founded linear order. Therefore $(On,\in)$ is a well-order. Furthermore we were said that the sum of two well-orders (which is basically glueing them together one after another) is also a well-order, but as I see it $(On,\in)$ + $(On, \in)$ is not left-narrow anymore.

Is there another definition or am I missing something else? Thanks in advance

1

There are 1 best solutions below

1
On BEST ANSWER

In the $\mathsf{ZFC}$ context, unless otherwise specified all mathematical structures - topological spaces, linear orders, rings, groups, etc. - are sets. $(On,\in)$ is therefore not even a linear ordering since it's too big: it's a proper class. Meanwhile, you can prove that the sum of two well-orderings is again a well-ordering (a key point being that the union of two sets is always a set).

However, it sounds like your text is using a slightly odd approach here. "Left-narrow" appears to mean that every proper initial segment should be set-sized; under this definition, $(On,\in)$ is a well-ordering. This is not standard per the above, but if this is what your text is doing, then yes, $(On,\in)+(On,\in)$ is not a well-ordering anymore; indeed, $(On,\in)+{\bf 1}$ isn't a well-ordering! (Which is a good sign that we shouldn't include it.)


It's worth noting that we can talk - to a certain extent anyways - about "class-sized versions" of many kinds of mathematical structures. The issue (as long as we don't go to some class theory to simplify things) is that $\mathsf{ZFC}$ can't quantify over classes, so things like "Given any two class-well-orderings, one is isomorphic to an initial segment of the other" can't even be expressed in $\mathsf{ZFC}$, let alone proved.

Now the situation isn't entirely negative. We can talk about specific examples. Going back to the OP, the class $On\times\{0,1\}$ is definable by a formula $\psi$, and the relation $$(\alpha,i)\trianglelefteq (\beta,j)\quad\iff\quad i<j\mbox{ or }[i=j\mbox{ and }\alpha\le\beta]$$ on that class is definable by a different formula $\varphi$; intuitively, the pair $(\psi,\varphi)$ define "$On+On$." We can then prove, in $\mathsf{ZFC}$, that this class-sized "structure" has no infinite descending chains - that is, it's a class-well-ordering. With more work we can quantify over class-well-orderings whose defining formulas have bounded complexity in a precise sense.

But all of this is significantly messier than the "sets-only" situation, and most of the time is worth avoiding.