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In set theory, $1+ω$ is defined as the ordinal number of ordinal sum of sets $\{a\}$ and ℕ. Also $ω+1$ is defined as the ordinal number of ordinal sum of sets $\Bbb N$ and $\{a\}$. You know what the ordinal sum means: the ordinal sum of well-ordered sets $A$ and $B$ is the set $A \cup B$ ordered as you know. We know that the sets $\{a\} \cup ℕ$ and $ℕ \cup \{a\}$ are equipotent and well-ordered. And also we know by a theorem in set theory, any two equipotent well-ordered set are isomorphic (similar).

Doesn't this argument contradict with the fact that in set theory, the ordinal numbers $1+ω$ and $ω+1$ are not equal to each other?

Pinter mentioned in his book following theorem and he proved it: "Theorem Let A and B be well-ordered classes; exactly one of the following three cases must be hold:

i) A is isomorphic with B.

ii) A is isomorphic with an initial segment of B.

iii) B is isomorphic with an initial segment of A.".

And he concluded " well-ordered classes do not differ from one another except in their size". I remind you the "isomorphism" concept is equivalent to "similar" concept; and it is a one to one correspondence between two sets and order-preserving.

  • Please do not add irrelevant tags. [tag:relations] is meaningless here. This is an elementary question, and [tag:set-theory] is not for those. – Andrés E. Caicedo Aug 30 '14 at 15:03
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    I do not know that is what the ordinal sum means. You need $A,B$ disjoint for that definition to work. And it is not true that two equipotent well-ordered sets are isomorphic. It is the reason that we define ordinals as something different from cardinals. So it's not clear what "theorem of set theory" you are referring to. – Thomas Andrews Aug 30 '14 at 15:14
  • @ThomasAndrews thank you. I brought this theorem from "Set theory Pinter" . –  Aug 30 '14 at 15:26
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    Well, either you misunderstood the theorem, or, highly unlikely, Pinter's "Set Theory" is just wrong. Everything above appears to be paraphrased, not the exact statement from a Pinter. – Thomas Andrews Aug 30 '14 at 15:29
  • @ThomasAndrews I think this theorem is well-known in set theory and it is correct. –  Aug 30 '14 at 15:35
  • Related: http://math.stackexchange.com/questions/98415/i-want-to-know-why-omega-neq-omega1/ – Asaf Karagila Aug 30 '14 at 15:36
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    Sorry, there is no theorem in set theory saying that two equipotent well-ordered sets are order isomorphic. Maybe you are thinking of the finite case? – Cheerful Parsnip Aug 30 '14 at 15:50
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    Since it is not true, and you've just proven that it is not true with your example, which is the fundamental example known to all set theorists of two different ordinals with the same cardinal, maybe, just maybe, you are misreading Pinter. @aminliverpool – Thomas Andrews Aug 30 '14 at 16:01
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    You would help us help you if you quoted the exact theorem from Pinter. – Thomas Andrews Aug 30 '14 at 16:03
  • I edited my question and added the exact form of the theorem from Pinter's book. –  Aug 31 '14 at 05:34
  • Please quote precisely the definition of ordinal sum. – Andrés E. Caicedo Aug 31 '14 at 05:43
  • @AndresCaicedo Definition of ordinal sum: Let A and B be disjoint, well-ordered sets. A ⊕ B, called the ordinal sum of A and B, is the set A ⋃ B ordered as follows. If x,y ∈ A ⋃ B then x <or= y if and only if i) x ∈ A and y ∈ A and x <or= y in A, or ii) x ∈ B and y ∈ B and x <or= y in B, or iii) x ∈ A and y ∈ B. –  Aug 31 '14 at 06:02
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    By the way, please use the "Enter" key to create linebreaks an paragraphs. Your question is impossible to read as it is. – Asaf Karagila Aug 31 '14 at 06:02
  • @AsafKaragila I used the "Enter" key but the continue of my quesion came exactly after my previous sentences without any linebreaks. –  Aug 31 '14 at 06:10
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    Two consecutive linebreaks create a new paragraph. – Asaf Karagila Aug 31 '14 at 06:11
  • @AsafKaragila thank you for your instruction. –  Aug 31 '14 at 06:22

4 Answers4

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The term "isomorphism" means similar structure. But it doesn't specify the structure. The structure needs to be inferred from context, or stated explicitly in the text.

If the context is just sets, then isomorphisms are not required to preserved any structure, in which case $\Bbb N$ and $\Bbb Q$ are isomorphic, since both are countably infinite.

If the context is ordered sets, then the isomorphism must also preserve the order. Not just be a bijection. In this context $\Bbb N$ and $\Bbb Q$ are of course not isomorphic. And also $\omega+1$ and $\omega$ (or $1+\omega$, which is isomorphic to $\omega$ in this context).

To see why two structures are not isomorphic, we can sometimes look at their "inherent properties", things like being a dense set or having a minimum or maximum, are things which must be preserved by isomorphism of ordered sets. So if one ordered set has a maximal element, and another does not then there is no way they are isomorphic.

Now you can easily see that while $1+\omega$ and $\omega+1$ is isomorphic as sets; they are not isomorphic as ordered sets, since $1+\omega$ does not have a maximal element, whereas $\omega+1$ does.

EDIT:

Note that the theorem Pinter quotes does not require that $1+\omega$ and $\omega+1$ are isomorphic. It is still possible, and that much is in fact true, that $1+\omega$ is isomorphic to a proper initial segment of $\omega+1$.

Moreover, a Cantor-Bernstein "like" theorem for well-ordered sets is in fact true: If $A$ and $B$ are two well-ordered sets such that there are order-embeddings from $A$ into $B$ and from $B$ into $A$, then $A$ and $B$ are isomorphic.

But this doesn't contradict my above answer either. Simply because there is no order-preserving injection from $\omega+1$ into $\omega$.

Asaf Karagila
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  • Thank you for your answer. I agree with you, but it contradicts with the theorem which I mentioned. –  Aug 31 '14 at 05:37
  • I edited my question and added the exact form of the theorem from Pinter's book. –  Aug 31 '14 at 05:38
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    You do see three possibilities in this theorem, right? Isomorphism is just one of them. – Asaf Karagila Aug 31 '14 at 05:40
  • For the other two cases we have not the same cardinality. –  Aug 31 '14 at 05:47
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    I'm confused by your comment. But you seem to be confused by other things. So let's work on that first. Why do you think that $\omega+1$ does not have a proper initial segment which has the same cardinality as $\omega+1$? What is the cardinality of $\omega+1$? – Asaf Karagila Aug 31 '14 at 05:51
  • The cardinality of ω and ω + 1 are the same. Please notice to Pinter's conclusion: "well-orederd classes do not differ from one another except in their size." . –  Aug 31 '14 at 06:15
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    I don't have Pinter's book to check that you are not misunderstanding some broader context here, or if this is indeed a mistake. So I can't help you any further than that. Let me leave you with a final piece of advice, once you have definitions you can verify claims about them. In particular you can quite easily verify that $\omega$ and $\omega+1$ are not isomorphic and have the same cardinality. The term "size" here is a bit unclear, and I cannot comment further on that topic without looking at the book. – Asaf Karagila Aug 31 '14 at 06:26
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    @aminliverpool Pinter says "roughly speaking" at this point of the book. And yes, the issue is that what is meant here by "size" isn't cardinality. For example, $\omega$ is a "smaller size" than $\omega+1$ because one is an initial segment of the other. – Andrew Dudzik Aug 31 '14 at 07:28
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There is no contradiction here. While it's certainly true that $\omega+1$ and $1+\omega$ have the same number of elements, and there's therefore a bijection between their elements, isomorphism of wellorders is more demanding.

Let $(A,\leq_A)$ and $(B,\leq_B)$ be wellorders. They are isomorphic iff there's a bijection $f:A\to B$ between them such that $f(x)\leq_B f(y) \iff x\leq_A y$. Being in bijection and also being a bijection on sets that have wellorders is a necessary condition, but far from a sufficient one, for isomorphism of ordinals; in particular, there is no bijection $f$ between the two ordinal sums you mention that satisfies the condition above.

Malice Vidrine
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  • Thank you. I edited my question and added the exact form of the theorem from Pinter's book. –  Aug 31 '14 at 05:40
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A well-ordering exists on both, but the ordinal orderings on $1+\omega$ and $\omega+1$ are different. The former has no maximal element, the latter has one.

Ian
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  • Thank you. Do you agree that any two equipotent well-ordered sets are similar to each other ? –  Aug 30 '14 at 14:54
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    I'm not an expert, but it seems like an isomorphism in this context should be a bijection that preserves order. In $\omega+1$, to what can you map your element $a$ in such a way as to be order-preserving? It seems to me that they are not isomorphic. – paw88789 Aug 30 '14 at 14:56
  • I brought this theorem from "Set Theory Pinter". –  Aug 30 '14 at 14:59
  • $\mathbb{N}\cup{a}$, without further information, isn't necessarily wellordered. One needs to know what happens to the relation $\leq$ when $a$ is added. – Malice Vidrine Aug 30 '14 at 15:05
  • @MaliceVidrine thank you. In the definition of "ordinal sum" the order was inputed. –  Aug 30 '14 at 15:09
  • If $A$ is well-ordered and $B$ is equipotent to $A$, then $B$ inherits a well-order from $A$ by exploiting a bijection. ($a \leq_B b$ if and only if $f(a) \leq_A f(b)$). One may do this with $A=1+\omega$ and $B=\omega+1$, but doing so will not give the ordinal orderings on both sets. – Ian Aug 30 '14 at 15:11
  • "Inputed"? (Not trying to be pedantic, actually trying to figure out your comment.) – Malice Vidrine Aug 30 '14 at 15:17
  • @MaliceVidrine Are you familiar with "ordinal sum" ? –  Aug 30 '14 at 15:21
  • @Ian I cant understand what you mean. –  Aug 30 '14 at 15:24
  • Yes, I'm quite familiar with ordinal arithmetic. That wasn't the part that made your comment opaque. Though if you're talking about concatenation of wellorders you may not want to write it as an ordinary union. – Malice Vidrine Aug 30 '14 at 15:25
  • @MaliceVidrine Is it possible for you to explain more about your comment ? –  Aug 30 '14 at 15:31
  • I edited my question and added the exact form of the theorem from Pinter's book. –  Aug 31 '14 at 05:39
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$\omega+1$ and $1+\omega=\omega$ have the same cardinality and thus there's a bijection $f$ from $\omega+1$ onto $\omega$. $\in$ is the canonical well-ordering on $\omega+1$ and so via $f$ you'll get a well-ordering $<_f$ on $\omega$, namely $\alpha <_f \beta$ iff $f^{-1}(\alpha) \in f^{-1}(\beta)$ for $\alpha,\beta\in\omega$. However, this well-ordering is not the same as the well-ordering $\in$ on $\omega$. Likewise, you can "project" the well-ordering $\in$ from $\omega$ to $\omega+1$ via $f^{-1}$ but what you'll get will not be $\in$ on $\omega+1$.

Frunobulax
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  • Thank you. I cant understand what you mean. Did you mean the theorem which I brought is incorrect ? I edited my question and added the exact form of the theorem from Pinter's book. –  Aug 31 '14 at 05:45