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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 : $$\begin{align*} A^0_\delta&=B^0_\delta\\ A^1_\delta&=A^0_\delta\cup B^1_\delta\\ &\vdots\\ A^{i+1}_\delta&=A^i_\delta\cup B^{i+1}_\delta&\qquad\text{successor}\\ A^i_\delta&=\bigcup_{j<i}A^j_\delta&\qquad\text{limit case} \end{align*}$$ First, $\bigcup_{i<cf(\lambda)}A^i_\delta=\bigcup B^i_\delta=\bigcup\phi(\alpha_i) =\phi(\bigcup \alpha_i)=\phi(\lambda)=\delta\times\delta$ and all the $A^i_\delta$ are subsets of $\delta\times\delta$. Second, we need to proof that the cardinality of each $A^i_\delta$ is $<\lambda$ : for the successor case, it's ok. But for the limit one, I have some problem : let $i<cf(\lambda)$ and consider $A:=\bigcup_{j<i}A^j_\delta$ with $|A^j_\delta|<\lambda$. Does $|A|<\lambda$ ? We have $$|\bigcup A^i_\delta|\leq\sum_{j<i}|A^j_\delta|\leq|i|.\sup|A^i_\delta|<cf(\lambda).\sup|A^i_\delta|$$ So I want to see that the last $sup$ is less than $\lambda$ (because $i<cf(\lambda)$ ?). 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 $<cf(\lambda)$ 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 $[~\cdot~]^{<\lambda}$?) – Henning Makholm Jun 7 '12 at 16:04
$[A]^\lambda$ means the subset of $A$ of cardinality less than $\lambda$. – Marc Moretti Jun 7 '12 at 16:23
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. – Henning Makholm Jun 7 '12 at 16:34
the bijection is not necessary an isomorphism ... – Marc Moretti Jun 7 '12 at 17:05
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 $\mathcal P(\lambda) \to \mathcal P(\delta\times\delta)$. – Henning Makholm Jun 7 '12 at 17:07

As you said, if $\delta<\lambda$ then we can just take $A_\delta^i=\delta\times\delta$.

Otherwise fix a bijection $f$ from $\lambda$ to $\delta\times\delta$, and let $\langle \xi_i\mid i<\operatorname{cf}(\lambda)\rangle$ be an increase sequence of ordinals, $\xi_i<\lambda$ for all $i$.

For every $i$ define $A_\delta^i=f''\xi_i$ is a set of cardinality $|\xi_i|<\lambda$, so $A_\delta^i\in[\delta\times\delta]^{<\lambda}$, and since for $i<j$ we have $\xi_i\subseteq\xi_j$ it follows that $A_\delta^i\subseteq A_\delta^j$. Lastly, because $\bigcup\xi_i=\lambda$ we have that $$\bigcup A_\delta^i=\bigcup f''\xi_i=f''\bigcup\xi_i=f''\lambda=\delta\times\delta.$$

As wanted.

(Where $f''X=\{f(x)\mid x\in X\}$, the direct image of $X$ under $f$.)

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