I've previously tried to define a meaningful notion of size for sets with this question. It turns out that the notion I tried to define was more then related to the one of measurable cardinals, which can not be proven to exist in ZFC. I'm now here to ask for what I hope to be an weaker version of this notion which was motivated by the previous one.
Let $R$ be a set of size $\mathfrak c=2^{\aleph_0}$ and say $\mathcal F\subset \mathcal P(R)$ is a family of humongous sets if it has the two following properties:
- If $P = \{P_n: n\in\mathbb N\}$ is a countable chain ($P_0\subset P_1\subset P_2\subset\dots$) such that $\displaystyle \bigcup_{n\in\mathbb N} P_n = R$, then $P\cap\mathcal F\ne\varnothing$, i.e., there is an humongous set in $P$.
- If $\{H_n: n\in\mathbb N\}\subset\mathcal F$ is a countable family of humongous sets, then $\left|\displaystyle\bigcap_{n\in\mathbb N} H_n\right| = \mathfrak c$.
I wonder if one can find a family of humongous sets. The difference from my previous question is the change of $P$ from a partition to a chain. With a partition, by taking $R:=\mathbb R$, one could force intervals of arbitrarily small length to be humongous, which would contradict the intersection property. With the change for a chain I hope to make this no longer possible.
No, one cannot find such a family.
[Update: the following is a starker counter-example than the original answer, thanks to an observation in the comments by bof.]
Take $R=\mathbb [0,1]$. Then for each $r\in [0,1]$, there is some $n_r\in\mathbb N$ such that $H_r:=\{r\}\cup\mathbb [0,1]\backslash (r-\frac{1}{n_r},r+\frac{1}{n_r})$ is humongous.
Let $I_r=[0,1]\cap (r-\frac{1}{n_r},r+\frac{1}{n_r})$. By compactness, there is a finite subset $S\subset [0,1]$ such that $\bigcup_{s\in S} I_s=[0,1]$.
But then for each $t\in [0,1]$ we have $t\in I_s$ for some $s$, whereby $t\notin H_s$ unless $t=s$. Therefore $\bigcap_{s\in S} H_s\subseteq S$.