MathOverflow is a question and answer site for professional mathematicians. 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

Consider the following question:

Is there a family $\mathcal{F}$ of subsets of $\aleph_\omega$ that satisfies the following properties?

(1) $|\mathcal{F}|=\aleph_\omega$

(2) For all $A\in \mathcal{F}$, $|A|<\aleph_\omega$

(3) For all $B\subset \aleph_\omega$, if $|B|<\aleph_\omega$, then there exists some $B'\in \mathcal{F}$ such that $B\subset B'$.

I am not sure if there is anything special about $\aleph_\omega$, but this was the example that came up.

Any help?

share|cite|improve this question
up vote 13 down vote accepted

I think the following diagonalization will show that there is no such set $\mathcal{F}$.

Suppose there were such an $\mathcal{F}$. Then we could split it up into $\omega$ many chunks $( \mathcal{F}_i ) _{i \in \omega} $ such that each $\mathcal{F} _i$ had exactly the sets of size $\aleph_i $ or smaller that were in $\mathcal{F}$. Now for each $\mathcal{F}_i$ we will construct a countable set $S_i \subset \aleph _\omega$ such that every set $A \in \mathcal{F} _i$ has only finite intersection with $S_i$. If we can make such an $S_i$, then by unioning together all the $S_i$ for every $i \in \omega$, we will get a countable set which is not contained in any $A \in \mathcal{F}$.

So: for a given $\mathcal{F} _i$, if there are fewer than $\aleph _\omega $ sets in it, then it's easy to make our set $S_i$, since the union of all the sets in $\mathcal{F} _i$ is smaller than $\aleph _\omega $. Now suppose there are $\aleph _\omega$ many sets in $\mathcal{F} _i$. Break $\mathcal{F} _i$ up into $\omega$ many chunks $( \mathcal{G}_j ) _{j \in \omega} $ such that each $ \mathcal{G}_j $ has size $\aleph _j $ and such that if $m < n$ then $\mathcal{G}_m \subset \mathcal{G}_n $. Note that the union of each $ \mathcal{G}_j $ has size less than $\aleph _\omega$. So now we can construct our set $S_i$ as follows: pick the $j$-th element to be something outside the union of $ \mathcal{G}_j $. Then $S_i$ has only finite intersection with any $A \in \mathcal{F}_i $.

share|cite|improve this answer
Thank you. Your argument answers the question. As pointed out my Andreas Blass your answer and his answer are essentially the same. Since your answer came first, I marked it as the accepted answer. – Ioannis Souldatos Aug 1 '11 at 17:03

There is no such family $\mathcal F$. Suppose, toward a contradiction, that you had such an $\mathcal F$ and list it in a sequence of order-type $\aleph_\omega$. For each $n\in\omega$, let $\mathcal F_n$ consist of the first $\aleph_n$ members of the sequence that have cardinality at most $\aleph_n$. Notice that $\mathcal F$ is the union of these subfamilies $\mathcal F_n$. The union of all the sets in $\mathcal F_n$ has cardinality at most $\aleph_n$, so we can choose some $a_n$ that is in $\aleph_\omega$ but not in this union. Then $\{a_n:n\in\omega\}$ is a countable subset of $\aleph_\omega$ not covered by any element of $\mathcal F$.

share|cite|improve this answer
Andy Voellmer's answer, which arrived while I was typing mine, is essentially the same. – Andreas Blass Jul 29 '11 at 21:56
Andreas: Can anything be said beyond the pcf bound? – Andrés E. Caicedo Jul 29 '11 at 22:12
Andres: Not that I can think of now (but I seem to still be a bit jet-lagged, even after 2 days back in the western hemisphere). – Andreas Blass Jul 30 '11 at 0:55

a family F as above of minimal size satisfies $|F|=cov(\aleph_\omega,\aleph_\omega,\aleph_\omega,2)=pp(\aleph_\omega)=cof([\aleph_\omega]^\omega)\ge \aleph_{\omega+1}$.

Edit: following Ali's suggestion, here are more details. $$cov(\lambda,\theta,\kappa,\sigma):=min{ |\mathcal F| : \mathcal F\subseteq [\lambda]^{<\theta} s.t. \forall A\in [\lambda]^{<\kappa}\exists\mathcal A\in[\mathcal F]^{<\sigma}(A\subseteq\bigcup\mathcal A)}$$ The definition is due to Shelah, of course. Note that $cov(\lambda,\kappa,\kappa,2)=cf([\lambda]^{<\kappa},\subseteq)$. The definition of $pp$, may be found in several places; for a crush treatment, see for example: . Let $\lambda$ denote a singular cardinal. It is always the case that $\lambda^+\le pp(\lambda)\le cov(\lambda^+,\lambda,cf(\lambda)^+,2)\le cf([\lambda]^{cf(\lambda)},\subseteq)$. Now, consider the preceding three cardinal invaritans (of $\lambda$). Shelah proved that if $\lambda$ is the least (singular) cardinal for which any of the three is greater than $\lambda^+$, then all three are equal. In particular, the openning equation that I gave (concerning $\aleph_\omega$) holds. See [Sh:E12] for pointers to Shelah's works on ``pp VS. cov''.

Edit2: I see that the definition of $cov$ is rendered incorrectly. The definition may be found here as well: . See (the proof of) Lemma 3.4 from there, and the subsequent works on this subject: 1. 2.

share|cite|improve this answer
@saf: for those of us who are not as conversant as yourself with pcf theory, could you elaborate? i.e., please provide definitions, references, etc. – Ali Enayat Jul 30 '11 at 1:35
Hi Asaf. This covers the doubt I had (in a comment to Andreas's answer). – Andrés E. Caicedo Jul 30 '11 at 6:35
Good to see you here, welcome :-) – Asaf Karagila Jul 30 '11 at 12:59
@saf: thanks for the elaboration; when you get the chance please edit the fourth line of your answer (defining $cov$) since it is not compiling correctly (at least on my machine). – Ali Enayat Jul 30 '11 at 15:28
@saf, Oh, never mind, I guess you referred us to a paper on your second edit for the definition of $cov$. – Ali Enayat Jul 30 '11 at 15:29

This question has been already answered thoroughly. I just wanted to address the OP's comment "I am not sure if there is anything special about $\aleph_\omega$".

Actually, there is nothing special about $\aleph_\omega$ other than the fact that it's a singular cardinal. Let $\kappa$ be a cardinal and let $S(\kappa)$ be the following statement:

There is a family $\mathcal{F} \subset [\kappa]^{<\kappa}$ such that $|\mathcal{F}|=\kappa$ and for every $F \in [\kappa]^{<\kappa}$ there is $G \in \mathcal{F}$ such that $F \subset G$.

Then $S(\kappa)$ holds if and only if $\kappa$ is a regular cardinal.

But things become more complicated if we just consider subsets of $\kappa$ of a fixed cardinality smaller than $\kappa$. For example, let $C(\kappa)$ be the statement:

There is a family $\mathcal{F} \subset [\kappa]^{\aleph_0}$ such that $|\mathcal{F}|=\kappa$ and for every $F \in [\kappa]^{\aleph_0}$ there is $G \in \mathcal{F}$ such that $F \subset G$.

Then $C(\aleph_n)$ is true for every $0< n< \omega$, $C(\aleph_\omega)$ is false for essentially the same reason that $S(\aleph_\omega)$ is false, but the truth value of $C(\aleph_{\omega+1})$ depends on your set theory. Namely, if there is an $\aleph_{\omega+1}$-sized family of countable subsets of $\aleph_\omega$ which is cofinal in $([\aleph_\omega]^\omega, \subseteq)$ then $C(\aleph_{\omega+1})$ is true, while if $cof([\aleph_\omega]^\omega, \subseteq) \geq \aleph_{\omega+2}$ (which is consistent with ZFC, modulo large cardinals) then $C(\aleph_{\omega+1})$ is clearly false...

share|cite|improve this answer

EDIT: As Andreas Blass points out below, this approach doesn't work.

Suppose $\mathcal{F}=\lbrace X_\alpha: \alpha\in\aleph_\omega\rbrace$ were such a collection of sets. Now for $\alpha\in\aleph_\omega$, let $rank(\alpha)=\min\lbrace \beta: \alpha\in X_\beta\rbrace$, and let $\le_W\subseteq\aleph_\omega\times\aleph_\omega$ be a well-ordering of $\aleph_\omega$ with the property that $rank(\alpha)$ < $rank(\beta)\implies \alpha\le_W\beta$. Note that no $X_\alpha$ is cofinal in $\le_W$: since each $X_\alpha$ has size $<\aleph_\omega$, for each $\alpha$ the set $\bigcup_\beta\le\alpha X_\beta$ has size $<\aleph_\omega$, and hence (by the third assumption on $\mathcal{F}$) there is some $\gamma>\alpha$ with $X_\gamma\not\subseteq X_\alpha$; any element of $X_\gamma-X_\alpha$ is then $\le_W$-above each element of $X_\alpha$. Let $C\subseteq \aleph_\omega$ be countable and cofinal in the $\le_W$ ordering. Then clearly $C\not\subseteq X_\alpha$ for any $\alpha\in\aleph_\omega$ since no $X_\alpha$ is cofinal in $\le_W$. But $C$ has cardinality $\aleph_0<\aleph_\omega$; a contradiction.

share|cite|improve this answer
I don't see why $\bigcup_{\beta\leq\alpha}X_\beta$ has size $<\aleph_\omega$. Even the union of the first $\omega$ of the sets $X_\beta$ could have size $\aleph_\omega$ (e.g., if $|X_n|=\aleph_n$ for each $n\in\omega$). I also don't see how you get a countable $C$ cofinal in the $\leq_W$ ordering; why couldn't that ordering have uncountable cofinality? On the other hand, why couldn't the $\leq_W$ ordering have a last element? (Suppose, for example, that $X_0=\omega-\{0\}$, that $X_n=\omega_n-\omega_{n-1}$` for all non-zero $n\in\omega$, and that `$0\in X_\omega$.) – Andreas Blass Jul 30 '11 at 6:03
Good point - this doesn't work at all. I think one might be able to ensure that $\le_W$ to have order type $\aleph_\omega$, but there is no way to ensure that $\bigcup_{\beta\le\alpha}X_\beta$ has size $<\aleph_\omega$. Thanks for pointing this out. – Noah Schweber Jul 30 '11 at 15:53

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.