# Upward generators of $[\omega]^\omega$

If $(P,\leq)$ is a poset and $S\subseteq P$ we let $$\uparrow S = \{p\in P: p\geq s\text{ for some }s\in S\}.$$

Let $([\omega]^\omega,\subseteq)$ denote the collection of infinite subsets of $\omega$, ordered by set inclusion. If $S\subseteq [\omega]^\omega$ has the property that $\uparrow S = [\omega]^\omega$, does this imply $|S|=2^{\aleph_0}$?

• Take an almost disjoint family $A$ in $[\omega]^\omega$ of size $2^{\aleph_0}$. Then $S$ has to contain an element below every element of $A$, and those sets have to be distinct by almost disjointness (intersections finite). – Wojowu Aug 12 '18 at 15:23

## 1 Answer

As Wojowu says, the existence of an almost disjoint family of size continuum gives an affirmative answer. For completeness, here's one way to construct one of those:

• Fix a bijection $b$ from $2^{<\omega}$ to $\omega$.

• For each $f\in 2^{\omega}$, let $X_f=\{b(\sigma): \sigma\prec f\}$.

Since any two distinct elements of $2^\omega$ only agree on a finite initial segment, the set $\{X_f: f\in 2^{\omega}\}$ is an almost disjoint family - and it clearly has size continuum.