}{\binom{n}{k_i}} \\
  &= h_{n}(G_{n})\\
  &=\pi_n({\mathcal{H}}).\end{aligned}$$ The sequence $\pi_n({\mathcal{H}})$ is non-negative and decreasing; therefore it converges.

For a fixed set $R:=\{k_1,k_2,\ldots, k_r\}$ (with $k_1<k_2<\cdots < k_r$), an [*$R$-flag*]{} is an $R$-graph containing exactly one edge of each size. The chain $C^R$ is a special $R$-flag with the edge set $${{\rm E}}(C^R)=\{[k_1], [k_2],\ldots, [k_r]\}.$$

\[p1\] For any hypergraph $H$, the following statements hold.

1.  $|R(H)|-1\leq \pi_n(H)\leq |R(H)|$.

2.  For subgraph $H'$ of $H$, we have $\pi(H')\leq \pi(H)- |R(H)|+|R(H')|$.

3.  For any $R$-flag $L$ on $m$ vertices and any $n\geq m$, we have $\pi_n(L)=|R|-1$.

[**Proof:**]{} Pick any maximal proper subset $R'$ of $R(H)$. Consider the complete graph $K_n^{R'}$. Since $K_n^{R'}$ misses one type of edge in $R(H)\setminus R'$, it does not contain $H$ as a subgraph. Thus $$\pi_n(H)\geq h_n(K_n^{R'})=|R'|=|R(H)|-1.$$ The upper bound is due to the fact $h_n(K_n^{R(H)})=|R(H)|$.

Proof of item 2 is similar. Let $S=R(H')$ and $G^S_n$ be an extremal hypergraph for $\pi_n(H')$. Extend $G^S_n$ to $G^{R(H)}_n$ by adding all the edges with cardinalities in $R(H)\setminus S$. The resulting graph $G^{R(H)}_n$ is $H$-free. We have $$\pi_n(H)\geq \pi_n(G^{R(H)}_n)=\pi_{n}(G^S_n)+ |R(H)|-|S|= |R(H)|-|R(H')|+ \pi_n(H').$$ Taking the limit as $n$ goes to infinity, we have $$\pi(H)\geq |R(H)|-|R(H')|+ \pi(H').$$

Finally, for item 3, consider $L$-free hypergraph $G_n^R$. Pick 