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Suppose $A$ is an infinite linear order with cardinality $\kappa$, and take the family $\langle A_{\alpha}: \alpha < \kappa \rangle$, define like this:

$$A=\bigcup_{\alpha < \kappa} A_{\alpha}$$

for all $\alpha < \beta < \kappa$, $A_{\alpha} \subseteq A_{\beta}$. and for all $\alpha < \kappa$, $A_{\alpha}$ is a well order. I need to prove that if $cf \kappa > \omega$ then A is a well order.

I have tried to develop the fact that a well order is a linear order with no infinite descending chains, but I don't know what else to do, if someone could help me or guide me i would really appreciate it.

Asaf Karagila
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Sara
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1 Answers1

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Suppose $A$ is not a well order. Then we can find an infinite descending sequence $a_1 > a_2 > \dots$ in $A$. Each $a_n$ belongs to some $A_{\alpha_n}$ for some $\alpha_n < \kappa$. Since $\kappa$ has uncountable cofinality, there is some $\beta < \kappa$ such that $\alpha_n < \beta$ for every $n$. But now, each $a_n$ belongs to $A_{\beta}$ contradicting the fact that $A_{\beta}$ is well ordered.

Peter
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