Funcoid bases/Failed condition

The following condition:

4. $S$ is an upper set and $S\cap\Gamma$ is a filter on the boolean lattice $\Gamma$ (defined in the chapter "Funcoids are filters" of the book).

was once thought by me to be equivalent to other three conditions. It is not so:

Proposition 4 does not imply 3.

Proof Take $S_0 = \operatorname{up} 1$ (where $1$ signifies the diagonal of some infinite set) and $S_1 = \bigcup_{F \in S_0} \operatorname{up}^{\Gamma} F$. Both $S_0$ and $S_1$ are upper sets. $S_0 \neq S_1$ because $1 \in S_0$ and $1 \notin S_1$. The $S \cap \Gamma$ for both is the filter on $\Gamma$ defining the funcoid $1$.

Thus in assumption of 4 we have not 3 (as otherwise both $S_0 = \operatorname{up} 1$ and $S_1 = \operatorname{up} 1$).