@asahay: (it is not really a secret) -- owl-sowa.blogspot.com/2015/11/…

@JukkaKohonen: I mean that 40 probably is the largest number $n$ for which all the 0's in $n!$ appear at the end.

Using the heuristic that the digits of $n!$ are just random, followed by some zeroes at the end, you can calculate without too much trouble that the answer to your question is: probably yes. My (very crude) estimate puts it at a better than 99% chance. I don't know whether this heuristic is any good, and I doubt this way of thinking will lead to a real answer . . . but there you have it.

@bof: Yes, that's pretty much the idea.

@bof: Hopefully my answer clears things up. It's a good question whether we can realize $K_{\aleph_1} \cup K_{\aleph_1}$ with a really "simple" family of sets (instead of using some kind of recursion).

I think $|V| \leq \aleph_1$ should always be possible (building the graph recursively, using the fact that $\mathcal P(\omega) / \mathrm{fin}$ is countably saturated). So (if I'm right) a yes answer should at least be consistent. But I don't see how to get all graphs of size $\leq 2^{\aleph_0}$ when CH fails.

Yes, that's right. The model described in that corollary has no condensation from $\kappa^\omega$ to any Polish space when $\aleph_1 < \kappa < \mathfrak{c}$. Also, using the results from the newer paper, it seems that if $\mathrm{cf}(\kappa) > \omega$, then $\omega^\omega$ is a condensation of $\kappa^\omega$ if and only if there is a partition of $[0,1]$ into $\kappa$ Borel sets. I don't see how to extend this to include singular cardinals with cofinality $\omega$.

I think you're right. For $\kappa < \aleph_\omega$, the problems are equivalent, but possibly not for $\kappa \geq \aleph_\omega$. Recently, I've extended some of the stuff Arnie and I did to cardinals past $\aleph_\omega$, using the non-existence of $0^\dagger$. (See arxiv.org/pdf/2101.10088.pdf.) It might be the case that "$0^\dagger$ does not exist" makes these problems equivalent for $\kappa \geq \aleph_\omega$ with $\mathrm{cf}(\kappa) > \omega$. But I'm not sure without thinking more about it.

You can get $\mathrm{cov}(\mathcal N) = \mathfrak{c} = \aleph_3$, together with a partition of $\mathbb R$ into $\aleph_2$ Borel sets, as follows. First force $MA+\mathfrak{c} = \aleph_2$, and then add $\aleph_3$ random reals. Adding the random reals makes $\mathrm{cov}(\mathcal N) = \mathfrak{c} = \aleph_3$, and by an argument due to Stern/Kunen, adding any number of random reals to a model of $MA+\mathfrak{c} = \aleph_2$ will leave you with a partition of $\mathbb R$ into $\aleph_2$ closed sets.

The model from our paper has $\mathrm{cov}(\mathcal N) = \aleph_1$. The forcing used is an iteration of length $\omega_1$, and it adds Cohen reals at every stage. This means that no matter what the ground model is, we'll end up with $\mathrm{cov}(\mathcal N) = \aleph_1$. Since size-$\aleph_1$ partitions are guaranteed in the extension anyway, we could change the length of the iteration to $\omega_2$ and (maybe changing a few other things too) end up with $\mathrm{cov}(\mathcal N) = \aleph_2$. But a new technique would be needed to do any better than that.

"Eventually $BB(n) > n^2$ . . . " Yes, that's putting it mildly. :)