# Proving cardinality of intersection of union of subsets

I have a discrete math problem I've stumbled onto during my research, but I'm not sure I have the math background to solve it. The problem is as follows:

There is a set, $C$, of $n$ elements, where $n = 3t+1$ for some integer $t$.

There are $n$ subsets of $C$, called $C_1 ... C_n$. Each subset contains $n-t$ elements of $C$, and the exact same $n-t$ elements never appear in more than $t$ subsets.

Each subset $C_t$ modifies itself by setting its content to the union of $n-t$ of the subsets of $C$ - this union must include $C_t$. Call the modified subsets $C_1'...C_n'$.

Prove that the intersection of all the modified subsets, $C_1' \cap C_2' \cap ... C_n'$, always has a cardinality of at least $n-t$. (or find a counterexample)

Can someone provide help or at least a starting point?

I found a counterexample.

$n = 10, t = 3, n - t = 7$
$C = \{1,2,3,4,5,6,7,8,9,10\}$
$C_1, C_2, C_3 = \{1,2,3,4,5,6,7\}$
$C_4, C_5, C_6 = \{1,2,3,4,5,6,8\}$
$C_7, C_8, C_9 = \{1,2,3,4,5,6,9\}$
$C_{10} = \{1,2,3,4,5,6,10\}$

All $C$ subsets have $\{1,2,3,4,5,6\}$ so it just becomes a matter of getting all $C'$ subsets to have the same 7th item.

$C_1', C_8' = C_1 \cup C_2 \cup C_3 \cup C_7 \cup C_8 \cup C_9 \cup C_{10} = \{...7,9,10\}$ (no 8)
$C_2', C_5' = C_1 \cup C_2 \cup C_3 \cup C_4 \cup C_5 \cup C_6 \cup C_{10} = \{...7,8,10\}$ (no 9)
$C_3', C_6', C_9' = C_1 \cup C_2 \cup C_3 \cup C_4 \cup C_5 \cup C_6 \cup C_7 = \{...7,8,9\}$ (no 10)
$C_4', C_7' = C_4 \cup C_5 \cup C_6 \cup C_7 \cup C_8 \cup C_9 \cup C_{10} = \{...8,9,10\}$ (no 7)
$C_{10}' = C_1 \cup C_2 \cup C_4 \cup C_5 \cup C_7 \cup C_8 \cup C_{10} = \{...7,8,9,10\}$ (all items)

The intersection of $C_1'$ through $C_{10}'$ is $\{1,2,3,4,5,6\}$, which has a cardinality of 6, which is less than $n-t$ (7).