$\mid S_j(\mathcal{F})\mid =\mid \mathcal{F}\mid$, and $w(S_j(\mathcal{F}))\le w(\mathcal{F})$ with equality only when $S_j(\mathcal{F})=\mathcal{F}$

Suppose $s_j(F,\mathcal{F})=s_j(G,\mathcal{F})$ with $F\mathrm{\ne }G$. Then at least one of $F$ or $G$ is shifted. If both were shifted, then $F\setminus \{j\}=G\setminus \{j\}$ and adding $j$ back gives $F=G$, impossible. So exactly one is shifted, say $F$, and then $s_j(F,\mathcal{F})=F\setminus \{j\}=G$ because $G$ is not shifted. But [F1] says precisely that $F\setminus \{j\}\mathrm{\notin }\mathcal{F}$ when $F$ is shifted, a contradiction. Therefore $F\mapsto s_j(F,\mathcal{F})$ is injective.

Since the map is injective, it is a bijection from the finite set $\mathcal{F}$ onto its image $S_j(\mathcal{F})$, so $\mid S_j(\mathcal{F})\mid =\mid \mathcal{F}\mid$.

Every shifted set loses the element $j$ and every unshifted set keeps its size, so $w(S_j(\mathcal{F}))\le w(\mathcal{F})$. Equality holds exactly when no set is shifted, and that is exactly the condition $S_j(\mathcal{F})=\mathcal{F}$.