Theorem 6.5.1 (Oddtown Theorem).label Let $[n]\define \curl{1,2,...,n}$ and $\F\suf\cali{P}([n])$ be a family such that $\abs{A}\congruent{1}{2}$ for all $A\in\F$ and $\abs{A\cap B}\congruent{0}{2}$ for all $A\neq B\in \F$. Then $\abs{F}\leq n$.
Proof. For each $A\in \F$ consider the indicator associated with $A$, $\indi{A}\in \bb{F}^{n}_{2}$ defined by
\begin{align*}\indi{A}(i)\define \begin{cases}1&i\in A\\ 0&i\nin A\end{cases}\end{align*}
Observe that $\indi{A}\cd \indi{B}=\abs{A\cap B}$ so the hypothesis is equivalent to
(1)
$\indi{A}\cd \indi{A}\congruent{1}{2}$
(2)
$\indi{A}\cd\indi{B}\congruent{0}{2}$
for all $A\neq B\in\F$. If there is a linear relation over $\bb{F}_{2}$ then
\begin{align*}0=\sums{}{A\in\cali{F}}c_{A}\indi{A}\end{align*}
Using (2) we have for each $B\in\F$
\begin{align*}0=\parens{\sums{}{A\in\cali{F}}c_A\indi{A}}\cd\indi{B}=\parens{\sums{}{A\in\cali{F}}c_A(\indi{A}\cd\indi{B})}=c_{B}\indi{B}\cd\indi{B}=c_{B}\end{align*}
showing that
\begin{align*}\abs{\F}\leq \dim_{\bb{F}_2}(\bb{F}_{2}^{n})=n\end{align*}
We shall now provide a proof over $\Q$ or any field of characteristic $0$ (as $\Q$ embeds there). For each $A\in \F$ consider the indicator associated with $A$, $\indi{A}\in \Q^{n}$ defined by
\begin{align*}\indi{A}(k)\define \begin{cases}1&k\in A\\ 0&K\nin A\end{cases}\end{align*}
Observe that $\indi{A}\cd \indi{B}=\abs{A\cap B}$ so the hypothesis is equivalent to
(1)
$\indi{A}\cd \indi{A}\congruent{1}{2}$
(2)
$\indi{A}\cd\indi{B}\congruent{0}{2}$
for all $A\neq B\in\F$. Let $m=\abs{\cali{F}}$ and write $\cali{F}=\curl{A_1,...,A_m}$. Let the matrix $M$ whose $k^{th}$ column is $\indi{A_k}$ with the canonical identification of $\indi{A}\bij (\indi{A}(\el))\in \Q^{n}$ where $\el\in[n]$ then $X\define M^{T}M$ is the matrix such that $X_{ij}=\abs{A_i\cap A_j}$. By (1) and (2)
\begin{align*}\det(X)\equiv\det(I_{k})\congruent{1}{2}\end{align*}
in particular $\det(X)\neq 0$. We conclude
\begin{align*}\abs{\cali{F}}=k=\text{rank}(X)\leq \text{rank}(M)\leq n\end{align*}
$\square$