there is no $B\in{\mathcal F}$ with $A\prec B$, by ${\text{\rm MAX}}({\mathcal F})$. Note that if ${\mathcal F}$ is compact every element in ${\mathcal F}$ can be extended to a maximal element in ${\mathcal F}$.

For ${\mathcal F}\subset[{\mathbb{N}}]^{<\omega}$ and $A\in [{\mathbb{N}}]^{<\omega}$ we define $${\mathcal F}(A)=\{ B\in[{\mathbb{N}}]^{<\omega}: A< B, A\cup B\in {\mathcal F}\}.$$ Note that if ${\mathcal F}$ is compact, spreading, closed under restrictions, or hereditary, so is ${\mathcal F}(A)$.

If ${\mathcal F}\subset [{\mathbb{N}}]^{<\omega}$ is compact, we denote by ${\text{{\rm CB}}}({\mathcal F})$ its [*Cantor Bendixson*]{} index, which is defined as follows. We first define the [*derivative*]{} of ${\mathcal F}$ by $$\begin{aligned}
{\mathcal F}'&=\Big\{A\in{\mathcal F}:  \exists (A_n) \subset {\mathcal F}\setminus\{A\}  \quad A_n\to_{n\to\infty } A\Big\} \\
      &=\Big\{A\in{\mathcal F}: \exists (B_n)\subset {\mathcal F}(A)\setminus\{\emptyset\} \quad B_n\to_{n\to\infty} \emptyset\Big\}
      ={\mathcal F}\setminus \{ A\in{\mathcal F}: A\text{ is isolated in }{\mathcal F}\}.\end{aligned}$$ Note that if ${\mathcal F}$ is hereditary then ${\mathcal F}'$ is hereditary, but also note that if ${\mathcal F}$ is only closed under restrictions, this is not necessarily true for ${\mathcal F}'$. Indeed, consider for example $${\mathcal F}=\big\{\emptyset, \{1\},\{1,2\},\{1,2,n\},\, n>2\big\} .$$ Every maximal element $A$ of ${\mathcal F}$ is not in ${\mathcal F}'$ and if ${\mathcal F}$ is spreading, then $${\mathcal F}'={\mathcal F}\setminus {\text{\rm MAX}}({\mathcal F}).$$ For $A\in[{\mathbb{N}}]^{<\omega}$ 