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?



$\aleph_\omega$ many subsets of $\aleph_\omega$
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 $. 


