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I have a little problem and confusion. I will write something that is maybe wrong.

Let $\alpha$ and $\beta$ two ordinals and $f$ a bijection between $\alpha$ and $\beta$. $f$ is not necessary a isomorphism between $(\alpha,<)$ and $(\beta,<)$ (if it is then we have $\alpha=\beta$). But $f$ is an isomorphism between $(\alpha,\subseteq)$ and $(\beta,\subseteq)$ (each element of $\alpha$ is a subset of $\alpha$ ...). So a bijection between ordinals can be seen as an isomorphism between ordinals...


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While every element of $\alpha$ is a subset of $\alpha$, not every subset of $\alpha$ is an element of $\alpha$. So $(\alpha,\subseteq)$ is not order isomorphic to $(\alpha,\lt)$. – Arturo Magidin Jun 7 '12 at 20:41
Arturo, I think that we interpret $(\alpha,\subseteq)$ differently. – Asaf Karagila Jun 7 '12 at 20:44
@Asaf: I think so; maybe I was thinking of $(\mathcal{P}(\alpha),\subseteq)$... Because otherwise I don't see how one can justify that there is an isomorphism between what is denoted as $(\alpha,\subseteq)$ and $(\beta,\subseteq)$: if $x,y\in\alpha$, $x\lt y$, and $f(x)\not\lt f(y)$ (which must exist if $f$ is not an order isomorphism), then $x\subseteq y$ but $f(x)\not\subseteq f(y)$... – Arturo Magidin Jun 7 '12 at 20:44
Sounds likely. I see $\beta$ as a particular set which happens to be linearly ordered by $\in$ and from transitivity also by $\subseteq$. – Asaf Karagila Jun 7 '12 at 20:47
@Marc: You are confusing $f$ with the direct image function $\overline{f}\colon\mathcal{P}(\alpha)\to\mathcal{P}(\beta)$. Uou are not using the direct image function to map $\alpha$ to $\beta$, because if you try, it may not even be well defined. Look at the example in my answer. – Arturo Magidin Jun 7 '12 at 21:05
up vote 2 down vote accepted

Let's consider an explicit example. Take $\beta=\omega$, and $\alpha=\omega\cup\{\omega\}$. Define $f\colon\alpha\to\beta$ as $f(n)=n+1$ if $n\in\omega$, and $f(\omega)=0$.

The direct image function $\overline{f}\colon\mathcal{P}(\alpha)\to\mathcal{P}(\beta)$ determined by $f$ from $\alpha$ to $\beta$ does not agree with $f$ itself: $f(\omega)=0$, but the direct image function has $\overline{f}(\{0,1,\ldots,\}) = \{1,2,3\ldots\}\neq 0$. In fact, the values of the direct image function are not elements of $\beta$: for instance, $\overline{f}(\{0\}) = \{1\}$, which is not an element of $\beta$ (since it is not an ordinal).

So the direct image function does not give a bijection between $\alpha$ and $\beta$: it doesn't even map $\alpha$ to $\beta$; so the fact that $x\subseteq y$ implies $\overline{f}(x) \subseteq \overline{f}(y)$ is not really relevant here: you are not using the direct image function. You are using your original $f$ on the elements of $\alpha$ and $\beta$, it's just that you are looking at $\alpha$ as ordered via $\subseteq$ instead of ordered via $\in$. And if $x,y\in\alpha$ are such that $x\lt y$ but $f(x)\not\lt f(y)$, then you have $x\subseteq y$ but you still don't have $f(x)$ (the element of $\beta$) contained in $f(y)$.

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so, in what I wrote in… , I define the $B^i_\delta$ as the direct image of $\alpha_i$ ? – Marc Moretti Jun 8 '12 at 7:06
I want to mean that when I write $\phi(\alpha_i)$, in fact it is the direct image of $\alpha_i$ and not the element $\phi(\alpha_i)\in\delta\times\delta$. An other thing, somebody tells me that a bijection is an isomorphism of sets : but isn't an isomorphism of subsets ? ... – Marc Moretti Jun 8 '12 at 11:22
@Marc: complete, elegant answer ever. – Babak S. Jun 8 '12 at 16:53
@Marc: I have moved your answer to a comment on Arturo's answer. In general, answers are for answers to the question at hand rather than for comments. Comments on answers should go on those answers. (Some new users cannot comment because of rep thresholds and/or because they accidentally created new accounts but this doesn't seem to be a problem in your case.) – Qiaochu Yuan Jun 9 '12 at 0:06

The term bijection between $A$ and $B$ usually refers to (merely) any function from $A$ to $B$ which is both injective and surjective.

When you have additional structures, for example the structure of a linear ordering in the case of ordinals, there is a concept called a homomorphism of that structure. Look in a book on model theory for the general defition, but in the case of linear ordering $(A,<)$ and $(B, <)$, $f$ is a homomorphism if for all $a, a' \in A$, if $a <_A a'$, then $f(a) <_B f(a')$. The term isomorphism, then refers to a bijective homomorphism of linear structures.

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Note that $(\beta,\subseteq)$ is isomorphic to the usual order on $\beta$ since we can reconstruct the well order from the inclusion and we can clearly reconstruct the inclusion from the usual order.

Therefore your claim is wrong. It is true, though, that if $\alpha$ and $\beta$ are equipotent then their power sets are isomorphic when ordered by inclusion.

Note that $\beta$, as an ordinal and in particular as a transitive set, is a subset of its power set and it forms a chain in $(\cal P(\beta),\subseteq)$. However it is a very particular chain, it is a well-ordered chain of order type $\beta$. The power set has other chains as well, of every order type below $|\beta|^+$ to be exact - and more.

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