Number of subsets required to satisfy condition

Question

Suppose $S$ is a set with $n > 1$ elements and $A_1 ,...,A_m$ are subsets of $S$ with the following property: if $x$, $y$ $\in$ $S$ and $x$ $\neq$ $y$, then there exists $i$ $\in$ $ \{1,...,m\} $ such that,

either $x$ $\in$ $A_i$ and $y$ $\notin$ $A_i$ , or $y$ $\in$ $A_i$ and $x$ $\notin$ $A_i$ . Then the following necessarily holds:

A) $n$ $=$ $2^m$

B) $n$ $\leq$ $2^m$

C) $n\gt2^m$

Here's what I tried:

There are a total of $2^n$ subsets of $S$.

$2^{(n-2)}$ is therefore the number of sets which contain neither $x$ nor $y$, including the null set.

$2^{(n-2)}$ must also be the number of sets which contain both $x$ and $y$.

Thus the maximum number of subsets which do not satisfy our property is:

$2^{(n-2)}$ $+$ $2^{(n-2)}$ $=$ $2^{(n-1)}$

So $m$ must be $\geq$ $2^{(n-1)} + 1$

I think the answer should then be B) $n \leq 2^{(m)}$

Are my reasoning and method correct? Is there a better method of solving this question?

Answer 1

I get (b) too, by this reasoning:

The condition can be rephrased as: There are not two different $x,y\in S$ that are in exactly the same $A_i$s.

Or, in yet other words, the function $$ f: S\to \mathcal P(\{1,2,\ldots,m\}) : x \mapsto \{ i\mid x\in A_i \} $$ is injective. This tells us that $|S|\le 2^m$.

The only question is now whether (b) can be strengthened to (a), but that is not the case because we can easily find an example with, for example, $n=m=2$: $$ S=\{1,2\} \qquad\qquad A_1=\{1\} \qquad A_2 = \{2\} $$ However, this example doesn't satisfy your claim $m\ge 2^{n-1}+1$, so your reasoning can't be right.