# $\aleph_\omega$ many subsets of $\aleph_\omega$

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?

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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 $A \in \mathcal{F}_i$.

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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 $\mathcal F$.

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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 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: http://papers.assafrinot.com/?num=5 . 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: http://papers.assafrinot.com/?num=1 . See (the proof of) Lemma 3.4 from there, and the subsequent works on this subject: 1. http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.jsl/1164060456 2. http://journals.impan.pl/cgi-bin/doi?fm205-1-3

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@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 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...

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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.

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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