$\aleph_\omega$ 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?
 A: 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:

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*Link

*http://journals.impan.pl/cgi-bin/doi?fm205-1-3
A: 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... 
A: 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 $.  
A: 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 $\mathcal F$.
