For an ordinal $\alpha$, define $cf(\alpha)$ to be the smallest ordinal $\beta$ such that there is $f:\beta\to\alpha$ such that $\bigcup f(\beta)=\alpha$. Why is this condition equivalent to $(\forall \gamma < \alpha) (\exists \delta < \beta)(f(\delta) > \gamma)$?

The condition $f:\beta\to\alpha$ such that $\bigcup f(\beta)=\alpha$ means $$\forall x[(x\in \alpha)\iff (\exists z\in f(\beta))(x\in z) ]$$ and the condition that $\beta$ is smallest says that if $\gamma $ is another ordinal such that there is some $g:\gamma\to \alpha$ such that $\bigcup g(\gamma)=\alpha$ then $\alpha \leq \gamma$. But I don't really see how the condition $(\forall \gamma < \alpha) (\exists \delta < \beta)(f(\delta > \gamma)$ comes into play.

Also, I don't "feel" this definition, so if anyone knows what the best way to think about it is, I'd be glad to hear that.

  • 3
    $\begingroup$ One way to "feel" a definition is to compute lots of examples. $\endgroup$
    – Lee Mosher
    Mar 16 at 23:11
  • $\begingroup$ When $S\subseteq x\in On,$ we say $S$ is unbounded above in $ x$ iff $\forall y\in x\,\exists s\in S\,(y\le s).$ When $\cup x=x\in On$ then $cf(x)$ is the least ordinal that is order-isomorphic to an unbounded-above subset of $x.$ $\endgroup$ Mar 17 at 21:27

1 Answer 1


I too once struggled to "feel" the definition, so allow me to provide some insight.

Let's first talk about the equivalence in your question. The equivalence is saying that the union of a set of ordinals is exactly the supremum (i.e. the least upper bound) of that set of ordinals. In other words, $\bigcup f(\beta) = \sup_{\gamma < \beta} f(\gamma)$. I provide a proof here.

  • First and foremost, $\bigcup f(\beta)$ is indeed an ordinal. See this answer.

  • To see that $\bigcup f(\beta)$ is an upper bound of $\{f(\gamma) : \gamma < \beta\}$, we observe that for any $\gamma < \beta$, we have $f(\gamma) \subseteq \bigcup f(\beta)$, so $f(\gamma) \leq \bigcup f(\beta)$.

  • Suppose $\delta$ is another upper bound of $\{f(\gamma) : \gamma < \beta\}$. We want to show that $\bigcup f(\beta) \leq \delta$. This is simply because by definition of upper bound, we have that $f(\gamma) \leq \delta$ for all $\beta$, i.e. $f(\gamma) \subseteq \delta$, so $\bigcup f(\beta) \subseteq \delta$ as well.

This then leads to my next point: I think it's easier to see $\operatorname{cf}(\alpha)$ as the smallest ordinal $\beta$ such that there exists a sequence $\{\alpha_\gamma : \gamma < \beta\}$ in which $\sup_\gamma \alpha_\gamma = \alpha$. This is because the union of a set of ordinals is exactly the supremum (i.e. the least upper bound) of that set of ordinals.

Cofinality of $\alpha$ is basically a measure of how fast we can approximate $\alpha$. In other words, it is the shortest path from $0$ to $\alpha$, but we are not allowed to actually reach $\alpha$. From this intuition, we can clearly see a few simple properties:

  • Since we are only concerned with the length and not the order-type, $\operatorname{cf}(\alpha)$ is always a cardinal.

  • The most straightforward way of "climbing" the ordinals below $\alpha$ is to just go through all the ordinals below it, which has length $|\alpha|$. Thus, $\operatorname{cf}(\alpha) \leq |\alpha|$.

  • If $\operatorname{cf}(\alpha) = \kappa$, then any path that "climbs" $\kappa$ should "lift" up to paths that "climb" $\alpha$. But since $\kappa$ is the shortest length that can "climb" $\alpha$, we must have $\operatorname{cf}(\kappa) = \kappa$. In other words, cardinals which are cofinalities must always be regular.

Of course, these are not proofs, but merely serve to provide an intuition to the proofs that you see in various books.

Let's look at a few examples.

  • $\mathrm{cf}(\omega) = \omega$. The only way to approximate $\omega$ from is to keep climbing the natural numbers - the good ol' and honest way.

  • $\mathrm{cf}(\omega_1) = \omega_1$. To see this, we note that we must have $\operatorname{cf}(\omega_1)$ is either $\omega$ or $\omega_1$. If $\operatorname{cf}(\omega_1) = \omega$, then we are saying that we can reach $\omega_1$ by having a countable sequence of ordinals below $\omega_1$ (i.e. countable ordinals) in which supremum is $\omega_1$. This is false as the countable union of countable sets is still countable (under $\mathsf{AC}$).

  • $\operatorname{cf}(\aleph_\omega) = \omega$. This is because we may take a "shortcut" to $\aleph_\omega$ by the sequence $\aleph_0,\aleph_1,\aleph_2,\dots$, which approximates but never reaches $\aleph_\omega$. By a similar reasoning, we may also conclude that $\operatorname{cf}(\aleph_\alpha) = \operatorname{cf}(\alpha)$ for all infinite limit ordinals $\alpha$.

  • $\operatorname{cf}(2^{\aleph_0}) = \; ?$. Well, it turns out that we cannot determine the cofinality of $2^{\aleph_0}$ from just $\mathsf{ZFC}$. By works of Solovay and König, it turns out that for any uncountable regular cardinal $\kappa$, there is a model of $\mathsf{ZFC}$ such that $\operatorname{cf}(2^{\aleph_0}) = \kappa$ (it is fine if you are confused with this example. It will come back to you as you learn more set theory).

So why is cofinality so important? That is because it turns out that many interesting in set theory (especially on large cardinals and combinatorial set theory), the length of such approximation of an ordinal $\alpha$ plays a huge part in whether $\alpha$ satisfy certain set-theoretic properties. For instance, if $\alpha$ is regular, it avoids certain loopholes to proofs that may fail by taking a "shortcut" to $\alpha$.

  • 1
    $\begingroup$ The last bullet point is a bit off. (1) It could be that $\aleph_\omega<2^{\aleph_0}$, but there is no extension satisfying that the cofinality of any ordinal is a singular cardinal. (2) What if you are working in a model where CH holds? The statement seems to suggest that the only values for the continuum have cofinality $\omega_1$. (3) As far as the continuum itself is concerned, this observation is due to Solovay, not Easton. Easton took that observation and expanded it to "all regular cardinals simultaneously", but you only talk about the continuum here. $\endgroup$
    – Asaf Karagila
    Mar 17 at 10:18
  • $\begingroup$ @AsafKaragila Thanks for the comments. I'm not exactly sure why (1) is relevant. and for (3) do you have the reference for Solovay's result? $\endgroup$ Mar 17 at 15:55
  • 1
    $\begingroup$ (1) Read what you wrote. You wrote that if $\kappa\leq2^{\aleph_0}$ is uncountable, then there is an extension in which that is its cofinality. If $2^{\aleph_0}=\aleph_{\omega+1}$, then $\kappa=\aleph_\omega$ is most certainly uncountable. (3) If my memory serves me right, this isn't published, but it's in some historical surveys of the results. I'll see what I can wrestle up. $\endgroup$
    – Asaf Karagila
    Mar 17 at 16:13
  • $\begingroup$ You can find this explained, amongst other places, in Kanamori's Introduction to the Handbook of Set Theory, specifically on section 3.2, line 5 (of the section). $\endgroup$
    – Asaf Karagila
    Mar 17 at 16:19
  • 1
    $\begingroup$ @lopofa it seems like you're a victim of poor notation. I'm sure here $\bigcup f(\beta)$ means $\bigcup_{\gamma < \beta} f(\gamma)$, or $\bigcup f[\beta]$, where $f[\beta]$ denotes the image of $f$ on the set $\beta$. $\endgroup$ Mar 18 at 12:17

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