a cardinality inequality Denote $E=\{1,2,...,n\}$ for some positive integer $n\geq 2$. Denote
\begin{equation*}
E_k=\{\Omega\subset E||\Omega|=k\},\quad k=1,2,...,n.
\end{equation*}
Here, $|\cdot|$ denotes a cardinality of a set.
Let $S_k\subset E_k$ for $k=1,2,...,n$. Suppose that for any $\Omega\in S_k$ with $k<n$ and for any $i\in E\setminus \Omega$, it holds $$(\Omega\cup\{i\})\in S_{k+1}.$$
I guess $\frac{|S_k|}{|E_k|}\leq\frac{|S_{k+1}|}{|E_{k+1}|}$ for any $k<n$? Is my proposition correct? Can anyone help me prove it or give some counter example?
 A: The result follows from a modification of a comment by @OlivierRoche
Out of each $\Omega \in S_k$ we can build $n-k$ elements of $S_{k+1}$ by adjuncting some $i \in E \setminus \Omega$.  
Let us be more explicit.  Consider all triplets of the form $(\Omega, i, \Omega \cup \{i\})$ where $\Omega \in S_k, i \in E \setminus \Omega$.
The number of such triplets $= |S_k| (n - k)$, because for every $\Omega \in S_k$ we can choose $i$ in $(n-k)$ different ways.
In each such triplet, the last member $(\Omega \cup \{i\}) \in S_{k+1}$ by the given property in the OP.
Now, consider some $\Lambda \in S_{k+1}$.  In how many triplets can $\Lambda$ appear (as the last member, obviously)?  I claim that $\Lambda$ can appear in at most $k+1$ such triplets.  This is because the $i$ in the triplet must be $\in \Lambda$, and there are only $k+1$ such choices for $i$.  (This is what I meant when I originally said "$\Lambda$ can only be built from some $\Omega$ in at most $k+1$ different ways.")
Let $N(\Lambda)$ be the number of times $\Lambda$ appears in these triplets, i.e. the no. of ways to build $\Lambda$ using some $\Omega \in S_k$.  Thus we have $N(\Lambda) \le k+1$.
Summing over all $\Lambda \in S_{k+1}$ we have:
$$\sum_{\Lambda \in S_{k+1}} N(\Lambda) = \text{no. of triplets} = |S_k| (n-k)$$
Note that the summation only needs to cover $\Lambda \in S_{k+1}$, because only such $\Lambda$ can appear as the third member in these triplets, by the given property of the OP.  I.e., we do not need to include any $\Lambda' \in E_{k+1} \setminus S_{k+1}$ because such $\Lambda'$ cannot appear in these triplets (i.e. cannot be built from some $\Omega \in S_k$).
Meanwhile, based on $N(\Lambda) \le k+1$ we also have:
$$\sum_{\Lambda \in S_{k+1}} N(\Lambda) \le \sum_{\Lambda \in S_{k+1}} (k+1) = |S_{k+1}| (k+1)$$
$$|S_{k+1}| \ge {n-k \over k+1} |S_k|$$
The result now easily follows from $|E_k| = {n \choose k}$ by expanding into factorials:
$${|E_{k+1}| \over |E_k|} = {{n \choose k+1} \over {n \choose k}} = {n! \over (k+1)! (n-k-1)!}{k! (n-k)! \over n!} = {n-k \over k+1} \le {|S_{k+1}| \over |S_k|}$$
