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Can $\omega_1$ (the first uncountable ordinal) be represented as union of an uncountable collection of cofinal, pairwise disjoint subsets?

2 Answers 2

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For an explicit example of an uncountable partition of $\omega_1$ into cofinal sets, for each $\alpha<\omega_1$ let $A_\alpha = \{ \beta<\omega_1\mid \exists \gamma<\omega_1,\thinspace \beta=\gamma +\omega^\alpha\}$. That is, partition $\omega_1$ by looking at the last summand in the Cantor normal form of each element of $\omega_1$.

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Assuming the axiom of choice, the answer is yes. Under AC there is a bijection from $\omega_1$ to $\omega_1 \times \omega_1$, so we can partition $\omega_1$ into an uncountable collection of uncountable sets. Any uncountable subset of $\omega_1$ is cofinal.

Edit: As Andres Caicedo points out in his comment (thanks!), AC is not needed.

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    @Tim: By recursion. It’s a little easier to describe the bijection $\varphi:\omega_1\to\omega_1\times\omega_1$: given $\varphi\upharpoonright\alpha$ for some \alpha<\omega_1, let $\varphi(\alpha)$ be the $\preceq$-minimal element of $(\omega_1\times\omega_1)\setminus\operatorname{ran}(\varphi\upharpoonright \alpha)$. (Sorry to have been so slow to answer this.)2012-01-11