$T_{\gamma} = [k+1,k+\gamma]$ and $S_{\gamma} = [2,k]\cup \{\gamma\}$. Condition (1) in Definition \[defresnu\] is satisfied if $\gamma\le k-1$. Condition (2) is satisfied iff $k+\gamma>2\gamma+2$, which is equivalent to $\gamma\le k-3$. (Note also that $\gamma=1$ is resistant for $k=3$.) From the discussion above, we also conclude that, for $k>3$, we have $\gamma_{k-2} = {n-k\choose n-k-1}=n-k.$

The following observation is given for the sake of familiarizing the reader with the statement.

The bound in Theorem \[thmfull1\] is always at least as strong as the bound in Theorem \[thm1\] for intersecting ${\mathcal F}\subset {[n]\choose k}$ with diversity $\gamma({\mathcal F})\le {n-4\choose k-3}$.

Let us first compare the statement of Theorem \[thmfull1\] with the statement of Theorem \[thm1\] for $\gamma_l := {n-u-1\choose n-k-1}$ with integer $u$. Such $\gamma_l$ is resistant for any $u\in [3,k]$, and we have $A_{\gamma_l}=[2,u+1]$. Thus, if we substitute such $\gamma_l$ in , then we get the bound $$|{\mathcal F}|\le {n-2\choose n-k}+\ldots +{n-u-1\choose n-k-u+1}+\gamma_l = {n-1\choose n-k}-{n-u-1\choose n-k-u}+{n-u-1\choose n-k-1},$$ which is exactly the bound for $\gamma_l$.

Remark that the inequality , however, gives a stronger conclusion and is valid in weaker assumptions. Indeed, while we know that this bound is sharp for $\gamma({\mathcal F}) = \gamma_l$, Theorem \[thmfull1\] also tells us that 