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Let $\lambda$ a cardinal and $\delta<\lambda^+$. I want to proof there exists a increasing chain $$\{A^i_\delta : i< cf(\lambda)\}\subseteq[\delta\times\delta]^{<\lambda}$$ converging to $\delta\times\delta$.

If $\delta<\lambda$ then $|\delta\times\delta|=|\delta|<\lambda$ and let $A^i_\delta=\delta\times\delta$ for all $i< cf(\lambda)$ and it's ok.

If $\lambda\leq\delta<\lambda^+$ then $|\delta|=\lambda$. Choose a (strictly) increasing sequence $\langle\alpha_\xi : \xi< cf(\lambda)\rangle$ so that $\sup(\alpha_\xi)=\lambda$ and let $\phi$ a bijection between $\lambda$ and $\delta\times\delta$. Let $B^i_\delta=\phi(\alpha_i)\subset \delta\times\delta$ for all $i< cf(\lambda)$. Then $|B^i_\delta|<\lambda$. But it is not necessary an increasing sequence so define by induction $A^i_\delta$ as follows : $$\begin{align*} A^0_\delta&=B^0_\delta\\ A^1_\delta&=A^0_\delta\cup B^1_\delta\\ &\vdots\\ A^{i+1}_\delta&=A^i_\delta\cup B^{i+1}_\delta&\qquad\text{successor}\\ A^i_\delta&=\bigcup_{j$$|\bigcup A^i_\delta|\leq\sum_{j So I want to see that the last $sup$ is less than $\lambda$ (because $i ?). My argument is this one : if $\sup|A^i_\delta|=\lambda$ then, as the sequence of the $|A^i_\delta|$ is increasing, we would have a cofinal sequence in $\lambda$ of length $ which is not possible. Is it ok ? Thanks.

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    Why doesn't the "$A_\delta^i=\delta\times\delta$ for all $i$" solution _always_ work? Is there an unstated additional assumption that you have to satisfy? (Hmm.. perhaps I'm misunderstanding what you mean by $[~\cdot~]^{<\lambda}$?)2012-06-07
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    $[A]^\lambda$ means the subset of $A$ of cardinality less than $\lambda$.2012-06-07
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    Okay. Next question: How can the $B_\delta^i$ fail to be an increasing sequence? The $\alpha_i$'s are increasing subsets of $\lambda$, so surely their images under the bijection $\phi$ are also strictly increasing.2012-06-07
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    the bijection is not necessary an isomorphism ...2012-06-07
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    A bijection _is_ an isomorphism of sets. There's no other structure to require an "isomorphism" to preserve. The bijection $\lambda\to\delta\times\delta$ automatically lifts to an order isomorphism $\mathcal P(\lambda) \to \mathcal P(\delta\times\delta)$.2012-06-07
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    okay. So an $\alpha_i$ can be seen as an element of $\lambda$ or a subset of $\lambda$ ? For the proof, I need to consider it as a subset .2012-06-07
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/3698/discussion-between-marc-moretti-and-henning-makholm)2012-06-07
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    Yes -- since an ordinal is the set of all ordinals less than it, every element of $\lambda$ is also a subset of it (and $\alpha\le\beta \Leftrightarrow \alpha\subseteq\beta$).2012-06-07

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