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So, another one in set theory (I think I am falling inlove with the subject). The question itself as presented:

Given $\Bbb Z$ is ordered by $<'$, where $a<'b$ iff

  1. $a\ge 0, a<b$, or
  2. $b<a<0$, or
  3. $b<0\le a$.

Prove that $\langle \Bbb Z, <'\rangle$ is a well-ordered set, which ordinal number is $\omega +\omega$.

My attempt at it: First I proved it is strictly ordered, then I proved it is well-ordered. Reaching for the second part of the question, I thought about trying to show it is isomorphic (has a one-to-one and onto function that keeps the same order) with the set N that is ordered so that the even numbers come first and the odd numbers come later. I showed that set has that ordinal number, and now I am looking for a one-to-one and onto function that keeps the same order between the two. I even went on to show an unto function that will keep the order will be enough as well, but I can't even think of one as such. The best I came up with was a function from Z onto N so that f(z)=z, z is not negetive and even, f(z)=z-1, z is not negetive and odd, f(z)=-z, z is negetive and odd, f(z)=-z-1, z is negetive and even. While that function is onto, it doesn't keep the order.

Any different approaches to solve this, or hints towards a helpful function will be extremely appreciated!

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  • $\begingroup$ Yes indeed, and it means onto, sorry about that. Thanks for the heads up, will edit. $\endgroup$
    – Studentmath
    Commented Feb 3, 2014 at 2:55
  • $\begingroup$ You might wanna correct the definition of $<'$ too. (Also, typo in the title.) $\endgroup$
    – Asaf Karagila ♦
    Commented Feb 3, 2014 at 2:57
  • $\begingroup$ Oh yes. That wouldn't make sense if it was so. Need more sleep. Cheers and sorry about that. $\endgroup$
    – Studentmath
    Commented Feb 3, 2014 at 3:09

1 Answer 1

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You can construct an isomorphism directly.

$$f(n)=\begin{cases} |n| & n\leq 0\\ \omega+n-1 & 0<n\end{cases}$$

Then you have to show that this injective and surjective and order preserving. The first two are quite easy. The third is not much harder either. In fact, if you do this from the beginning then you have essentially proved that it is a well-order, without going through the definition.

But, if you already proved that this is a well-order, then it is isomorphic to a unique ordinal. And $\omega+\omega$ is the unique limit ordinal that has exactly one limit ordinal smaller than itself. So it is the order type of every well-order without a maximal element, that every element except the minimal and another one, are successors.

So now we have to find successors for every number and show that there is no maximal elements, but that's easy. if $k\leq 0$ then $-k$ is the successor; and if $k>0$ then $k+1$ is its successor; and $0$ and $1$ are exactly those two numbers which are not successors in $<'$.

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    $\begingroup$ You're welcome. I should say that loving set theory is a very good idea! :-) $\endgroup$
    – Asaf Karagila ♦
    Commented Feb 3, 2014 at 0:21
  • $\begingroup$ @bof: Oh, yeah, that seems to be a typo, and perhaps it's better to alert the OP to that. It seems that I glazed over it when I read the question (because it's quite obvious what the writer had meant there, and that this is a typo). $\endgroup$
    – Asaf Karagila ♦
    Commented Feb 3, 2014 at 1:54

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