Every set shattered by $S_j(\mathcal{F})$ is shattered by $\mathcal{F}$

Suppose first that $j\mathrm{\notin }T$. Then every trace of $S_j(\mathcal{F})$ on $T$ is also a trace of $\mathcal{F}$ on $T$, because removing or keeping $j$ changes nothing on the set $T$. Since $T$ is shattered by $S_j(\mathcal{F})$, it is shattered by $\mathcal{F}$ as well.

Now suppose that $j\in T$, and let $A\subseteq T$ be arbitrary. Since $T$ is shattered by $S_j(\mathcal{F})$, both $A$ and $A\cup \{j\}$ occur as traces of shifted sets. The second clause of [F1] implies that whenever $A$ appears as a trace with $j$ removed, the original family already contains a set realising $A$ or a set realising $A\cup \{j\}$; applying this to the two traces $A$ and $A\cup \{j\}$ shows that $\mathcal{F}$ realises both. Hence every subset of $T$ is a trace of $\mathcal{F}$, so $T$ is shattered by $\mathcal{F}$.

The two cases cover all possibilities for $j$, so no new shattered set is created by the down-shift.