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I am trying to prove the following claim. I did not find the claim in a book, but I believe it to be true.

Claim: Consider a finite set $P\neq \emptyset$ and subsets $S_1, S_2, S_3 \subset P$, each of size $s$. Suppose $s > \frac{2}{3} |P|$. Prove that $S_1 \cap S_2 \cap S_3 \neq \emptyset$.

Here is my proof.

Proof: Consider the elements of $S_1$ and $S_2$. There must be at minimum $\left\lfloor \frac{2|P|}{3} \right\rfloor + 1$ elements in each set. Therefore $$\begin{align*} |P - S_1| &= |P| - s \\ &\le |P| - \left\lfloor \frac{2|P|}{3} \right\rfloor - 1 \\ &= \left\lceil \frac{|P|}{3} \right\rceil - 1 \end{align*}$$ Thus $S_1$ and $S_2$ can have at most $\left\lceil \frac{|P|}{3} \right\rceil - 1$ distinct elements. The other elements must belong to their intersection. $$\begin{align*} |S_1 \cap S_2| &\ge \left\lfloor \frac{2|P|}{3} \right\rfloor + 1 - \left(\left\lceil \frac{|P|}{3} \right\rceil - 1\right)\\ &\ge \left\lfloor \frac{2|P|}{3} \right\rfloor - \left\lfloor \frac{|P|}{3} \right\rfloor + 1\\ &= \left\lfloor \frac{|P| + 1}{3} \right\rfloor + 1 \end{align*}$$

Thus, $$\begin{align*} |S_1 \cap S_2 \cap S_3| &= |S_1 \cap S_2| + |S_3| - |(S_1 \cap S_2) \cup S_3| \\ &\ge \left\lfloor \frac{|P| + 1}{3} \right\rfloor + 1 + \left\lfloor \frac{2|P|}{3} \right\rfloor + 1 - |P|\\ &>0 \end{align*}$$

I am looking for verification, criticism, and for alternative proofs.

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Here is how I would prove it.

Count the total membership count, that is, the total number of times it happens that $p\in S_i$, for any point $p\in P$ and any of the sets $S_i$. Since each set $S_i$ is more than 2/3 of the points in $P$, this total membership count is strictly more than $3\cdot \frac 23|P|$. Thus, the total membership count is strictly more than $2|P|$.

But if no point $p$ is in all three sets, then every point is in at most two $S_i$, and so the total membership count would be at most $2|P|$, a contradiction.

This way of arguing shows the stronger result: you only need one of the sets to have strictly more than 2/3 of the points, if the other two sets have at least 2/3 of the points.

The argument generalizes to $k$ sets, provided that each of them has fraction $\frac{k-1}k$ of the total, with at least one having strictly more than this. In the case of two sets, this is the simple assertion that if $A$ has at least half the points and $B$ has more than half, then $A\cap B$ is not empty.

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