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

Vladimir Reshetnikov
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2 Answers2

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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 $\omega_1$.

Philip Brooker
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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.

Nate Eldredge
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  • Nate, AC is not used here. Provably in ZF, there is a definable function that to each infinite ordinal $\alpha$ associates a bijection between $\alpha$ and $\alpha\times\alpha$. Choice is used to prove a stronger result, namely, that we can pick the $\omega_1$ pieces in the partition to be stationary. – Andrés E. Caicedo Nov 25 '11 at 04:15
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    And the following relation well-orders $\omega_1\times\omega_1$ in type $\omega_1$ and thus indirectly provides the desired bijection. For $\langle\alpha,\beta\rangle,\langle\gamma,\delta\rangle\in\omega_1\times\omega_1$ define $\langle \alpha,\beta\rangle \preceq\langle\gamma,\delta\rangle$ iff one of the following holds: (1) $\max{\alpha,\beta}<\max{\gamma,\delta}$; (2) $\alpha=\gamma$ and $\beta\le\delta<\alpha$; (3) $\beta<\alpha=\delta\ge\gamma$; or (4) $\beta=\delta$ and $\alpha\le\gamma\le\beta$. – Brian M. Scott Nov 26 '11 at 20:01
  • @BrianM.Scott: +1 How does the relation provide a bijection from $ω_1×ω_1$ to $ω_1$? – Tim Jan 09 '12 at 15:57
  • @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.) – Brian M. Scott Jan 11 '12 at 05:40