Given: $2^{n-1}$ subsets of a set with $n$ elements with the property that any three have nonempty intersection, prove that the intersection of all the sets is nonempty.
My Solution: Let $S$ be the family of subsets of the set $A$ with $n$ elements. Because $S$ has $2^{n-1}$ elements, for every subset B of A, either B or its complement $B^c$ is in $S$. $WLOG$, let's assume that subset $B$ is in $S$, this implies that its complement $B^c$ will be in complement of $S$ (which is $S^c$). This will follow for every element in $S$. So now it directly follows that intersection of all the subsets will be an empty subset. $QED$
Putnam and Beyond: Let $S$ = {$A_1, A_2, . . . ,$ $A_{2^{n−1}}$} be the family of subsets of the set $A$ with $n$ elements. Because $S$ has $2^{n−1}$ elements, for any subset $B$ of $A$, either $B$ or its complement $B^c$ is in $S$. (They cannot both be in $S$ by the other hypothesis.) So if $A_i$ and $A_j$ are in $S$, then either ${A_i}∩{A_j}$ is in $S$, or its complement is in $S$. If the complement is in $S$ then ${A_i}∩{A_j}∩({A_i}∩{A_j})^{c}$ is empty, contradicting the fact that the intersection of any three elements of $S$ is nonempty. Hence ${A_i}∩{A_j}∈S$. We will now show by induction on $k$ that the intersection of any $k$ sets in $S$ is nontrivial. We just proved the base case $k=2$. Assume that the property is true for any $k−1$ elements of $S$, and let us prove it for $A_{i1}, A_{i2}, . . . > ,A_{ik}∈S$. By the induction hypothesis, $A_{i1}∩···∩Ai_{k−1}∈S$, and also $A_{ik}∈S$, so $(A_{i1}∩···∩A_{ik−1})∩A_{ik}$ is in $S$. This completes the induction. For $k=2n−1$, we obtain that the intersection of all sets in $S$ is nontrivial.
As it may be seen my solution is fully inspired by that of book but I really couldn't get it and feel that induction is unnecessary in this. Is there some logical error in what I wrote up or not?