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Let $(A_i)_{i=1}^{n}$ be $n$ different sets. Say $Z := \bigcup_{i=0}^{n}A_i$.
Q1: Is it true that if $|Z| \gt n$ then you can find $x \in Z$ such that the $(A_i-{x)}$ are still all different?
Q2: If $|Z|\gt n-1$ is a sufficient hypothesis, is this tight?

Remark : if Q1 is true then of course you can find $X \subset Z $ such that $|Z-X|=n$ and all $A_i-X$ are different.
The motivation (alternative formulation) is: working in the alphabet ${ \{ 0,1 \}}$:
If you have $n$ different words each of size $n+1$, can you find $i \in [1,n]$ so that removing the $i$ th letter of each word still produces $n$ different words?

Let $(A_i)_{i=1}^{n}$ be $n$ different sets. Say $Z := \bigcup_{i=1}^{n}A_i$.
Q1: Is it true that if $|Z| \gt n$ then you can find $x \in Z$ such that the $(A_i-{x)}$ are still all different?
Q2: If $|Z|\gt n-1$ is a sufficient hypothesis, is this tight?

Remark : if Q1 is true then of course you can find $X \subset Z $ such that $|Z-X|=n$ and all $A_i-X$ are different.
The motivation (alternative formulation) is: working in the alphabet ${ \{ 0,1 \}}$:
If you have $n$ different words each of size $n+1$, can you find $i \in [1,n]$ so that removing the $i$ th letter of each word still produces $n$ different words?

Let $(A_i)_{i=1}^{n}$ be $n$ different sets. Say $Z := \bigcup_{i=0}^{n}A_i$.
Q1: Is it true that if $|Z| \gt n$ then you can find $x \in Z$ such that the $(A_i-{x)}$ are still all different?
Q2: If $|Z|\gt n-1$ is a sufficient hypothesis, is this tight?

Remark : if Q1 is true then of course you can find $X \subset Z $ such that $|Z-X|=n$ and all $A_i-X$ are different.
An alternative formulation is:
If you have $n$ words of size $n+1$ on alphabet ${ \{ 0,1 \}}$. Can you find $i \in [1,n]$ so that the $i$ th letter of each words still produce $n+1$ different words?

Let $(A_i)_{i=1}^{n}$ be $n$ different sets. Say $Z := \bigcup_{i=0}^{n}A_i$.
Q1: Is it true that if $|Z| \gt n$ then you can find $x \in Z$ such that the $(A_i-{x)}$ are still all different?
Q2: If $|Z|\gt n-1$ is a sufficient hypothesis, is this tight?

Remark : if Q1 is true then of course you can find $X \subset Z $ such that $|Z-X|=n$ and all $A_i-X$ are different.
The motivation (alternative formulation) is: working in the alphabet ${ \{ 0,1 \}}$:
If you have $n$ different words each of size $n+1$, can you find $i \in [1,n]$ so that removing the $i$ th letter of each word still produces $n$ different words?

Let $(A_i)_{i=1}^{n}$ be $n$ different sets. Say $Z := \bigcup_{i=0}^{n}A_i$.
Q1: Is it true that if $|Z| \gt n$ then you can find $x \in Z$ such that the $(A_i-{x)}$ are still all different?
Q2: If $|Z|\gt n-1$ is a sufficient hypothesis, is this tight?

Remark : if Q1 is true then of course you can find $X \subset Z $ such that $|Z-X|=n$ and all $A_i-X$ are different.
An alternative formulation is:
If you have $n$ words of size $n+1$ on alphabet ${ \{ 0,1 \}}$. Can you find $i \in [1,n=1]$ so that the $i$th letter of each words still produce $n+1$ different words?

Let $(A_i)_{i=1}^{n}$ be $n$ different sets. Say $Z := \bigcup_{i=0}^{n}A_i$.
Q1: Is it true that if $|Z| \gt n$ then you can find $x \in Z$ such that the $(A_i-{x)}$ are still all different?
Q2: If $|Z|\gt n-1$ is a sufficient hypothesis, is this tight?

Remark : if Q1 is true then of course you can find $X \subset Z $ such that $|Z-X|=n$ and all $A_i-X$ are different.
An alternative formulation is:
If you have $n$ words of size $n+1$ on alphabet ${ \{ 0,1 \}}$. Can you find $i \in [1,n]$ so that the $i$ th letter of each words still produce $n+1$ different words?