Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
up vote 3 down vote accepted

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 $\omega_1$.

share|cite|improve this answer

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.

share|cite|improve this answer
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 4:15
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 9 '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 5:40

Your Answer


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.