(I have modified this question after trying to develop an answer for the second part based on the post of spaceisdarkgreen. I have posted my new answer for this part for feedback to see if the answer is correct now. The answer is also detailed since the point is to help me learn.)

I have been reading the book "Introduction to Set Theory" by Jech and Hrbacek. I have come to a pair of exercises that I am having trouble with, and it seems that I do not fully understand some of the concepts introduced in the chapter and section associated with the exercises. The chapter is on the arithmetic of cardinal numbers and the section is on singular and regular cardinals. I am more interested in what I am missing or am getting wrong than just a correct answer.


Here I use $\cf(x)$ to refer to the cofinality of $x$, and $\Ord$ to refer to the "set" of all ordinal numbers. $\boldsymbol{N}$ denotes the natural numbers.

The first exercise statement goes :

Prove $\cf(\aleph_{\omega}) = \cf(\aleph_{\omega + \omega}) = \omega$.

The second exercise statement goes :

Prove $\cf(\aleph_{\omega_{1}}) = \omega_{1}$ and $\cf(\omega_{2}) = \omega_{2}$.

Now I'm going to include some theorems and definitions from the book :

Theorem 7.1.9 :

a.) $\omega_{\alpha}$ is an infinite initial ordinal number for each $\alpha$. b.) If $\Omega$ is an infinite initial ordinal number, then $\Omega = \omega_{\alpha}$ for some $\alpha$.

Definition 9.2.1 :

An infinite cardinal $\kappa$ is called singular if there exists an increasing transfinite sequence $\langle \alpha_{\nu} \; \mid \; \nu < \vartheta \rangle$ of ordinals $\alpha_{\nu} < \kappa$ whose length $\vartheta$ is a limit ordinal less than $\kappa$, and $\kappa = \lim_{\nu \rightarrow \vartheta} \alpha_{\nu}$. An infinite cardinal that is not singular is called regular.

They also give an unlabeled definition of a limit.

Definition X

Let $\langle \alpha_{\nu} \; \mid \; \nu < \vartheta\rangle$ be a transfinite sequence of ordinal numbers of length $\vartheta$. We say that the sequence is increasing if $\alpha_{\nu} < \alpha_{\mu}$ whenever $\nu < \mu < \vartheta$. If $\vartheta$ is a limit ordinal number and if $\langle \alpha_{\nu} \; \mid \; \nu < \vartheta\rangle$ is an increasing sequence of ordinals, we define : \begin{equation} \alpha = \lim_{\nu \rightarrow \vartheta} = \sup\left( \{ \alpha_{\nu} \; \mid \; \nu < \vartheta \} \right) \end{equation} and call $\alpha$ the limit of the increasing sequence.

Another definition :

Definition 9.2.7 :

If $\alpha$ is a limit ordinal, then the cofinality of $\alpha$, $\cf(\alpha)$, is the least ordinal number $\vartheta$ such that $\alpha$ is the limit of an increasing sequence of ordinals of length $\vartheta$.

Theorem 9.2.4 :

Every successor cardinal $\aleph_{\alpha + 1}$ is a regular cardinal.

Here is my current solution for the first exercise :

We know that $\omega$ is a limit ordinal since : \begin{equation} \not \exists \gamma \in \Ord \text{ s.t. } \omega = \gamma + 1 \end{equation} We know by theorem 7.1.9 (a.) that $\omega_{\omega}$ is an infinite initial ordinal number. So $\omega_{\omega}$ is the cardinality of an infinite well-ordered set. This means $\omega_{\omega}$ is a cardinal number. So we have : \begin{equation} \aleph_{\omega} = \omega_{\omega} \end{equation} Since $\omega$ is a limit ordinal and $\aleph_{\omega}$ is a cardinal, then $\aleph_{\omega}$ is a limit cardinal. We can see : \begin{equation} \aleph_{\omega} = \lim_{n \rightarrow \omega} \aleph_{n} = \sup\left( \{ \aleph_{n} \; \mid \; n < \omega \} \right) \end{equation} We see that the length of $\langle \aleph_{n} \; \mid \; n < \omega \rangle$ is $\omega$ and $\aleph_{n} < \aleph_{\omega} \; \forall n < \omega$. We know $\aleph_{0} = \omega$ and $\aleph_{\alpha} > \omega \; \forall \alpha > 0$.

So $\aleph_{\omega} > \omega$. By definition 9.2.1 this means $\aleph_{\omega}$ is a singular cardinal.

So by definition 9.2.7 : \begin{equation} \cf(\aleph_{\omega}) < \aleph_{\omega} \end{equation} We have already established that a sequence of length $\omega$ (with each sequence element less than $\aleph_{\omega}$) exists with limit $\aleph_{\omega}$. If there is no shorter such sequence with this limit, then : \begin{equation} \cf(\aleph_{\omega}) = \omega \end{equation} Need to prove that no shorter sequence exists.

According to definition X, if $\aleph_{\omega}$ is the limit of an increasing sequence, then the upper limit of the indices must be a limit ordinal. In other words : \begin{equation} \lim_{\nu \rightarrow \vartheta} = \aleph_{\omega} \Rightarrow \vartheta \text{ is a limit ordinal } \end{equation} There is no smaller limit ordinal than $\omega$, so we must have : \begin{equation} \cf(\aleph_{\omega}) = \omega \end{equation} We see : \begin{equation} \lim_{n \rightarrow \omega} \aleph_{\omega + n} = \aleph_{\omega + \omega} \end{equation} so that a suitable sequence of length $\omega$ exists such that $\aleph_{\omega + \omega}$ is its limit. By the same argument given for $\aleph_{\omega}$, no suitable sequence of shorter length exists. $\checkmark$ So : \begin{equation} \cf(\aleph_{\omega + \omega}) = \omega \; \checkmark \end{equation}

We know : \begin{equation} \omega_{1} = \sup\left( \{ \beta \in \Ord \; \mid \; \beta < \omega_{1}\right) \end{equation} Define an increasing sequence : \begin{equation} F = \langle \aleph_{\nu} \; \mid \; \nu < \omega_{1} \rangle \end{equation} We see : \begin{equation} \aleph_{\nu} < \aleph_{\omega_{1}} \; \forall \nu < \omega_{1} \end{equation} and : \begin{equation} \lim_{\nu \rightarrow \omega_{1}} \aleph_{\nu} = \aleph_{\omega_{1}} \end{equation} We see that the length of $F$ is $\omega_{1}$. Now need to show that there exists no limit ordinal $\lambda < \omega_{1}$ s.t. there is a sequence : \begin{equation} G = \langle \alpha_{\nu} \; \mid \; \nu < \lambda \rangle \end{equation} such that : \begin{equation} \lim_{\nu \rightarrow \lambda} \alpha_{\nu} = \aleph_{\omega_{1}} \end{equation}

Let $\beta_{\nu}$ be the least ordinal such that $\aleph_{\beta_{\nu}} \geq \alpha_{\nu}$ for any given $\nu < \lambda$.

What does this mean ? We see : \begin{equation} \alpha_{\nu} = \aleph_{0} \Rightarrow \aleph_{\beta_{\nu}} = \aleph_{0} \Rightarrow \beta_{\nu} = 0 \end{equation} and : \begin{equation} \alpha_{\nu} = \aleph_{\omega} \Rightarrow \aleph_{\beta_{\nu}} = \aleph_{\omega} \Rightarrow \beta_{\nu} = \omega \end{equation} and : \begin{equation} \alpha_{\nu} = \aleph_{\omega + 4} \Rightarrow \aleph_{\beta_{\nu}} = \aleph_{\omega + 4} \Rightarrow \beta_{\nu} = \omega + 4 \end{equation} When would we have : \begin{equation} \aleph_{\beta_{\nu}} \neq \alpha_{\nu} \end{equation} Obviously : \begin{equation} \alpha_{\nu} \in \boldsymbol{N} \Rightarrow \beta_{\nu} = 0 \end{equation} We see : \begin{equation} \alpha_{\nu} = \aleph_{\gamma} \text{ for some } \gamma \in \Ord \Rightarrow \beta_{\nu} = \gamma \end{equation} But : \begin{equation} \alpha_{\nu} = \aleph_{\gamma} + k \text{ for some } \gamma \in \Ord \text{ and } k \in \boldsymbol{N} \Rightarrow \beta_{\nu} = \gamma + 1 \end{equation} We know : \begin{align} \alpha_{\nu} \geq \aleph_{0} & \Rightarrow \alpha_{\nu} = \aleph_{\gamma} + k \text{ for some } \gamma \in \text{Ord} \text{ and } k \in \boldsymbol{N} \\ & \Rightarrow \beta_{\nu} = \begin{cases} \gamma & \text{ if } k = 0 \\ \gamma + 1 & \text{ if } k \neq 0 \end{cases} \end{align} We know $\exists \mu < \lambda$ such that $\alpha_{\nu} \geq \aleph_{0} \; \forall \nu \geq \mu$. So : \begin{equation} \alpha_{\nu} = \aleph_{\gamma} + k \Leftrightarrow \beta_{\nu} = \begin{cases} \gamma & \text{ if } k = 0 \\ \gamma + 1 & \text{ if } k \neq 0 \end{cases} \; \; \forall \nu \geq \mu \end{equation} Define a new sequence : \begin{equation} \hat{G} = \langle \hat{\alpha}_{\nu} \; \mid \; \mu \leq \nu < \lambda \rangle \end{equation} Where : \begin{equation} \hat{\alpha}_{\nu} = \alpha_{\nu} \; \forall \nu \in \Ord \text{ such that } \mu \leq \nu < \lambda \end{equation} We see : \begin{equation} \lim_{\nu \rightarrow \lambda} \alpha_{\nu} = \theta \Leftrightarrow \lim_{\nu \rightarrow \lambda} \hat{\alpha}_{\nu} = \theta \end{equation} Define : \begin{equation} \hat{\alpha}_{\nu} = \aleph_{\gamma(\nu)} + k(\nu) \; \forall \nu < \lambda \end{equation} Where : \begin{align} \gamma & : \Ord \rightarrow \Ord\\ k & : \Ord \rightarrow \boldsymbol{N} \end{align} We see : \begin{align} \lim_{\nu \rightarrow \lambda} \alpha_{\nu} = \aleph_{\omega_{1}} & \Rightarrow \lim_{\nu \rightarrow \lambda} \hat{\alpha}_{\nu} = \aleph_{\omega_{1}} \\ & \Rightarrow \lim_{\nu \rightarrow \lambda} \left[ \aleph_{\gamma(\nu)} + k(\nu) \right] = \aleph_{\omega_{1}} \\ & \Rightarrow \sup\left( \{ \aleph_{\gamma(\nu)} + k(\nu) \; \mid \; \nu < \lambda \} \right) = \aleph_{\omega_{1}} \\ & \Rightarrow \sup\left( \{ \aleph_{\gamma(\nu) + 1} \; \mid \; \nu < \lambda \} \right) = \aleph_{\omega_{1}} \\ & \Rightarrow \sup\left( \{ \aleph_{\beta_{\nu}} \; \mid \; \nu < \lambda \} \right) = \aleph_{\omega_{1}} \\ & \Rightarrow \lim_{\nu \rightarrow \lambda} \aleph_{\beta_{\nu}} = \aleph_{\omega_{1}} \\ & \Rightarrow \lim_{\nu \rightarrow \lambda} \beta_{\nu} = \omega_{1} \end{align} and since $\lambda < \omega_{1}$ this implies that $\omega_{1}$ is singular. But by theorem 9.2.4, $\omega_{1} = \aleph_{1} = \aleph_{0+1}$ is a regular cardinal. So we have a contradiction. $\checkmark$

This means no such sequence of length $\lambda$ exists, so that : \begin{equation} \cf(\aleph_{\omega_{1}}) = \omega_{1} \; \checkmark \end{equation}

  • 1
    $\begingroup$ I think you're overthinking the first one quite a bit; consider — what does it mean for a cofinality to be less than $\omega$? As soon as you show there's an $\omega$-sequence whose limit is $\aleph_\omega$ then you're done. $\endgroup$ Commented Oct 11, 2021 at 20:16
  • $\begingroup$ In general if $0\ne x=\cup x\in On$ then $cf(\aleph_x)=cf(x).$ $\endgroup$ Commented Oct 12, 2021 at 4:04

4 Answers 4


$\DeclareMathOperator{cf}{cf}$A cofinal map $f : \alpha \to \beta$ is an increasing map such that $\lim_{\gamma \to \alpha} f(\gamma) = \beta$.

First, note that $\cf(s(\alpha)) = 1$ for all $\alpha$. Conversely, if $\cf(\alpha) = 1$ then $\alpha$ is a successor ordinal. This implies that the only finite cofinalities are $0 = \cf(0)$ and $1$.

Thus, a nonzero limit cardinal must have infinite cofinality.

Furthermore, note that the map $\beta \mapsto \aleph_\beta$ is a cofinal map for all nonzero limit ordinals $\beta$. Therefore, $\cf(\aleph_\beta) \leq \cf(\beta)$ for all nonzero limit $\beta$.

Conversely, if $\beta$ is a nonzero limit ordinal and $f : \gamma \to \aleph_\beta$ is a cofinal map, then define $f' : \gamma \to \beta$ by $f'(\tau) = $ the smallest $\alpha \in \beta$ such that $f(\tau) \in \aleph_\alpha$. Then $f'$ is a cofinal map. So we see that $\cf(\beta) \leq \cf(\aleph_\beta)$ for all nonzero limit $\beta$.

So for all nonzero limit $\beta$, we have $\cf(\beta) = \cf(\aleph_\beta)$.

The other result we need here is that $\tau^+$ is regular for all infinite cardinals $\tau$. For if we have a sequence $\{u_\theta\}_{\theta < \gamma}$ where $\gamma < \tau^+$, then $|u_\theta| \leq \tau$ for all $\theta < \gamma$. And $\gamma \leq \tau$. So $|\sum\limits_{\theta < \gamma} u_\theta| \leq |\sum\limits_{\theta < \gamma} \tau| \leq |\tau \times \tau| = \tau < \tau^+$.

In particular, we have $\cf(\aleph_\omega) = \cf(\omega) = \omega$.

We also have $\cf(\aleph_{\omega + \omega}) = \cf(\omega + \omega)$. Then we see that $\cf(\omega + \omega) = \omega$, since the sequence $n \mapsto \omega + n$ is cofinal.

We have $\cf(\aleph_{\omega_1}) = \cf(\omega_1) = \cf(\omega_0^+) = \omega_0^+ = \omega_1$.

We have $\cf(\omega_2) = \cf(\omega_1^+) = \omega_1^+ = \omega_2$.


As a suggestion for the second: consider the sequence of ordinals $\gamma_\nu = \mathrm{card}(\alpha_\nu)$. Then each $\gamma_\nu$ is some aleph by the axiom of choice, so we can then define the sequence $\beta_\nu$ by $\gamma_\nu = \aleph_{\beta_\nu}$. What can you say about the sequence $\beta_\nu$?

  • 2
    $\begingroup$ You don't need the AoC to give each ordinal a cardinality. Given any ordinal, the set of equipotent ordinals has a least element, and that's the cardinal number you're looking for. $\endgroup$
    – Arthur
    Commented Oct 11, 2021 at 20:37

Yes, your argument for $\aleph_\omega$ and $\aleph_{\omega+\omega}$ is correct, although perhaps a bit inefficient.

For the part you are stuck on concerning $\aleph_{\omega_1},$ if $\lim_{\nu\to \lambda}\alpha_{\nu}=\aleph_{\omega_1},$ where $\alpha_\nu$ is increasing, let $\beta_\nu$ be the least ordinal such that $\aleph_{\beta_\nu}\ge\alpha_\nu.$ So $\lim_{\nu\to \lambda} \aleph_{\beta_\nu}=\aleph_{\omega_1}$... but this means $\lim_{\nu\to \lambda}\beta_\nu=\omega_1.$ So if $\lambda<\omega_1,$ then $\omega_1$ would be singular, but this is false since it's a successor cardinal (assuming AC... countable choice suffices in this instance).

A slightly more general version of the argument yields the very important theorem that for any limit ordinal $\nu,$ $\operatorname{cf}(\aleph_\nu)=\operatorname{cf}(\nu).$


Its actually simple. $\aleph_n$ is cofinal in $\aleph_{\omega}$. Proving that $cf(\aleph_{\omega})=\omega$.

Similarly, $\aleph_{\omega +n}$ is cofinal in $\aleph_{\omega+\omega}$.

For $\aleph_{\omega_1}$,we have that $\aleph_{\alpha}$ for $\alpha <\omega_1$ is cofinal, which shows that $$cf(\aleph_{\omega_1})=cf(\omega_1)=\omega_1$$


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