Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

1 Answer 1

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 $f$.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.