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I have been reading through Herbert Endertons introductory book on set theory as I have stumbled upon a claim that baffled me through the day.

As anyone I need sleep so I ask here for help.

Namely as I was reading about order types the author makes a claim that $\bar1 + \bar\omega = \bar\omega$ but that $\bar\omega + \bar1 = \bar\omega^+$

Where $\bar1$ stands for order type for $\langle1,\epsilon_1\rangle$ and $\bar\omega$ stands for order type of $\langle\omega,\epsilon_\omega\rangle$

I have some how conviced myself that first claim is true but for second one I can not even start to understand it.Can someone give a proof of the claim,or even better an intuitive answer?

Along with the answer for this question an example of addition of some arbitrary order types(along with through description of steps) would

Andrés E. Caicedo
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TheCoolDrop
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  • $\omega + 1$ essentially just looks like $\mathbb{N}$ but with one extra element (let's call it 'infinity') which is greater than all the others. This is a well-order. – Kris Sep 04 '14 at 22:54
  • Is this due to the fact that well ordered sets can not be isomorphic to their segments?I do not really see where you are going,is it not true that addition of finite number of elements to infinite set does not change its cardinality? – TheCoolDrop Sep 04 '14 at 22:55
  • I think this has been asked at least twice before. It will help a lot if you clarify what exactly you're not getting. – Asaf Karagila Sep 04 '14 at 22:57
  • http://math.stackexchange.com/questions/98415/i-want-to-know-why-omega-neq-omega1 and http://math.stackexchange.com/questions/913960/are-1%CF%89-and-%CF%891-isomorphic are relevant. But if you explain what you're missing here, we can say with more accuracy whether those are duplicate or not. – Asaf Karagila Sep 04 '14 at 22:58
  • In case the confusion is notational, Enderton is using the superscript $+$ not to denote the next cardinal, but simply the next ordinal. You are correct that $\omega+1$ is still countable. – Andrés E. Caicedo Sep 04 '14 at 22:59
  • @Vanio: There can be different order types with the same cardinality! (There is a bijection between them, but it need not be order preserving.) – Nate Eldredge Sep 04 '14 at 22:59
  • I do not understand why the first examples results in order type of omega,while second one results in order type of its successor. – TheCoolDrop Sep 04 '14 at 23:00
  • I wish I could award each of you points but all I can do is vote up your comments,if anyone of you is willing to turn this into a full blown answer I will accept it.Thank you very much – TheCoolDrop Sep 04 '14 at 23:09

1 Answers1

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Ok so here goes the answer that I came up with.

Let $\bar1 = it\langle1,\epsilon_1\rangle$ and $\bar\omega=it\langle\omega,\epsilon_\omega\rangle$

As by definition of addition of order types $\bar1 + \bar\omega$ we require $\langle A,R\rangle\in\bar\omega$ and $\langle B,S \rangle\in\bar1$ where A and B are disjoint sets.

We now define :

$$R\oplus S = R\cup S \cup (A\times B)$$

It is easily provable that defined set is linear ordering on $A\cup B$

Now notice that since B is equinumerous to 1 it must contain only one element,which is maximum in $A \cup B$ by the $R \oplus S$ ordering

Now let $\alpha$ be the single element in B we can now see that $\langle \omega ,\epsilon_\omega \rangle \cong \langle seg\;\alpha,R\rangle$ thus it can not be isomorphic to $\langle A \cup B , R \oplus S\rangle $ thus to find isomorphism we consider the next largest ordinal namely succesor of omega and thus we see it satisfies since it contains and aditional element to pair with our element $\alpha$

Piece of cake right?

TheCoolDrop
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