Let $E_1, ..., E_n$ be non empty finite sets. Show that the matrix $A = (A_{ij})_{1 \leq i, j \leq n}$ defined by $A_{ij} = \dfrac{|E_i \cap E_j|}{|E_i \cup E_j|}$, is positive semi-definite.
This is part 5 of Exercise 1 in http://members.cbio.mines-paristech.fr/~jvert/svn/kernelcourse/homework/2019mva/hw.pdf .
I start with the definition but it doesn't seem very promising. Any hint, please?
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The following proof is inspired by my answer to math.stackexchange question https://math.stackexchange.com/q/1340405/, which in turn is inspired by the probabilistic method from extremal combinatorics. It is fully elementary and almost combinatorial ("almost" because it involves a simple limit argument at one point).
We begin with notations:
Let $\mathbb{N}$ be the set $\left\{ 0,1,2,\ldots\right\} $.
For each $n\in\mathbb{N}$, let $\left[ n\right] $ denote the set $\left\{ 1,2,\ldots,n\right\} $.
We also recall the following simple fact (Lemma 5 in my answer to math.stackexchange question https://math.stackexchange.com/q/1340405/):
Lemma 1. Let $Q$ be a finite totally ordered set. Let $J$ be a subset of $Q$. Let $r\in J$. Let $S$ be the set of all permutations of $Q$. Then, \begin{equation} \left\vert \left\{ \sigma\in S\ \mid\ \sigma\left( r\right) >\sigma\left( j\right) \text{ for all }j\in J\setminus\left\{ r\right\} \right\} \right\vert =\dfrac{\left\vert S\right\vert }{\left\vert J\right\vert }. \end{equation}
Now, I shall prove a first positive definiteness statement:
Theorem 2. Let $n\in\mathbb{N}$. Let $E_{1},E_{2},\ldots,E_{n}$ be $n$ finite sets. Let $x_{1},x_{2},\ldots,x_{n}$ be $n$ real numbers. Then, \begin{align} \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v} }{\left\vert E_{u}\cup E_{v}\right\vert +1}\geq0. \end{align}
Proof of Theorem 2. Fix some object $r$ that belongs to none of the sets $E_{1},E_{2},\ldots,E_{n}$. Let $Q$ be the set $E_{1}\cup E_{2}\cup\cdots\cup E_{n}\cup\left\{ r\right\} $. This is a finite set (since the sets $E_{1},E_{2},\ldots,E_{n},\left\{ r\right\} $ are finite). We fix some total order on $Q$.
Let $S$ be the set of all permutations of $Q$. This $S$ is a finite nonempty set (of size $\left\vert Q\right\vert !$). Thus, $\left\vert S\right\vert $ is a positive integer.
If $u\in\left[ n\right] $ and $\sigma\in S$, then we say that $\sigma$ is $u$-friendly if we have $\left( \sigma\left( r\right) >\sigma\left( j\right) \text{ for all }j\in E_{u}\right) $. Now, we claim the following:
For any $u\in\left[ n\right] $ and $v\in\left[ n\right] $, we have \begin{equation} \left\vert \left\{ \sigma\in S\ \mid\ \sigma\text{ is }u\text{-friendly and }v\text{-friendly}\right\} \right\vert =\dfrac{\left\vert S\right\vert }{\left\vert E_{u}\cup E_{v}\right\vert +1}. \label{darij1.pf.t2.1} \tag{1} \end{equation}
[Proof of \eqref{darij1.pf.t2.1}: Fix $u\in\left[ n\right] $ and $v\in\left[ n\right] $. Define a subset $J$ of $Q$ by $J=E_{u}\cup E_{v} \cup\left\{ r\right\} $. (This is well-defined, since the definition of $Q$ shows that $E_{u}$, $E_{v}$ and $\left\{ r\right\} $ are subsets of $Q$.)
The object $r$ belongs to none of the sets $E_{1},E_{2},\ldots,E_{n}$. Thus, in particular, $r$ belongs neither to $E_{u}$ nor to $E_{v}$. In other words, $r\notin E_{u}\cup E_{v}$. This yields \begin{align} E_{u}\cup E_{v}=\underbrace{\left( E_{u}\cup E_{v}\cup\left\{ r\right\} \right) }_{=J}\setminus\left\{ r\right\} =J\setminus\left\{ r\right\} . \end{align} Also, $r\in J$; thus, $\left\vert J\setminus\left\{ r\right\} \right\vert =\left\vert J\right\vert -1$. Now, from $E_{u}\cup E_{v}=J\setminus\left\{ r\right\} $, we obtain $\left\vert E_{u}\cup E_{v}\right\vert =\left\vert J\setminus\left\{ r\right\} \right\vert =\left\vert J\right\vert -1$, so that $\left\vert J\right\vert =\left\vert E_{u}\cup E_{v}\right\vert +1$.
For each $\sigma\in S$, we have the following chain of logical equivalences: \begin{align*} & \ \left( \sigma\text{ is }u\text{-friendly and }v\text{-friendly}\right) \\ & \Longleftrightarrow\ \underbrace{\left( \sigma\text{ is }u\text{-friendly} \right) }_{\substack{\Longleftrightarrow\ \left( \sigma\left( r\right) >\sigma\left( j\right) \text{ for all }j\in E_{u}\right) \\\text{(by the definition of "}u\text{-friendly")}}}\wedge\underbrace{\left( \sigma\text{ is }v\text{-friendly}\right) }_{\substack{\Longleftrightarrow\ \left( \sigma\left( r\right) >\sigma\left( j\right) \text{ for all }j\in E_{v}\right) \\\text{(by the definition of "}v\text{-friendly")}}}\\ & \Longleftrightarrow\ \left( \sigma\left( r\right) >\sigma\left( j\right) \text{ for all }j\in E_{u}\right) \wedge\left( \sigma\left( r\right) >\sigma\left( j\right) \text{ for all }j\in E_{v}\right) \\ & \Longleftrightarrow\ \left( \sigma\left( r\right) >\sigma\left( j\right) \text{ for all }j\in\underbrace{E_{u}\cup E_{v}}_{=J\setminus \left\{ r\right\} }\right) \\ & \Longleftrightarrow\ \left( \sigma\left( r\right) >\sigma\left( j\right) \text{ for all }j\in J\setminus\left\{ r\right\} \right) . \end{align*} Hence, \begin{align*} & \left\vert \left\{ \sigma\in S\ \mid\ \sigma\text{ is }u\text{-friendly and }v\text{-friendly}\right\} \right\vert \\ & =\left\vert \left\{ \sigma\in S\ \mid\ \sigma\left( r\right) >\sigma\left( j\right) \text{ for all }j\in J\setminus\left\{ r\right\} \right\} \right\vert \\ & =\dfrac{\left\vert S\right\vert }{\left\vert J\right\vert }\qquad\left( \text{by Lemma 1}\right) \\ & =\dfrac{\left\vert S\right\vert }{\left\vert E_{u}\cup E_{v}\right\vert +1}\qquad\left( \text{since }\left\vert J\right\vert =\left\vert E_{u}\cup E_{v}\right\vert +1\right) . \end{align*} This proves \eqref{darij1.pf.t2.1}.]
Now, for each $\sigma\in S$, we have \begin{align*} & \left( \sum_{\substack{u\in\left[ n\right] ;\\\sigma\text{ is }u\text{-friendly}}}x_{u}\right) ^{2}\\ & =\left( \sum_{\substack{u\in\left[ n\right] ;\\\sigma\text{ is }u\text{-friendly}}}x_{u}\right) \left( \sum_{\substack{u\in\left[ n\right] ;\\\sigma\text{ is }u\text{-friendly}}}x_{u}\right)\\ & =\left( \sum_{\substack{u\in\left[ n\right] ;\\\sigma\text{ is }u\text{-friendly} }}x_{u}\right) \left( \sum_{\substack{v\in\left[ n\right] ;\\\sigma\text{ is }v\text{-friendly}}}x_{v}\right) \\ & \qquad\left( \begin{array} [c]{c} \text{here, we have renamed the summation}\\ \text{index }u\text{ as }v\text{ in the second sum} \end{array} \right) \\ & =\sum_{\substack{u\in\left[ n\right] ;\\\sigma\text{ is }u\text{-friendly} }}\sum_{\substack{v\in\left[ n\right] ;\\\sigma\text{ is }v\text{-friendly} }}x_{u}x_{v}. \end{align*} Summing up these equalities over all $\sigma\in S$, we obtain \begin{align*} & \sum_{\sigma\in S}\left( \sum_{\substack{u\in\left[ n\right] ;\\\sigma\text{ is }u\text{-friendly}}}x_{u}\right) ^{2}\\ & =\underbrace{\sum_{\sigma\in S}\sum_{\substack{u\in\left[ n\right] ;\\\sigma\text{ is }u\text{-friendly}}}\sum_{\substack{v\in\left[ n\right] ;\\\sigma\text{ is }v\text{-friendly}}}}_{=\sum\limits_{u\in\left[ n\right] } \sum\limits_{v\in\left[ n\right] }\sum\limits_{\substack{\sigma\in S;\\\sigma\text{ is }u\text{-friendly}\\\text{and }v\text{-friendly}}}}x_{u}x_{v}\\ & =\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\underbrace{\sum _{\substack{\sigma\in S;\\\sigma\text{ is }u\text{-friendly}\\\text{and }v\text{-friendly}}}x_{u}x_{v}}_{=\left\vert \left\{ \sigma\in S\ \mid \ \sigma\text{ is }u\text{-friendly and }v\text{-friendly}\right\} \right\vert \cdot x_{u}x_{v}}\\ & =\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] } \underbrace{\left\vert \left\{ \sigma\in S\ \mid\ \sigma\text{ is }u\text{-friendly and }v\text{-friendly}\right\} \right\vert } _{\substack{=\dfrac{\left\vert S\right\vert }{\left\vert E_{u}\cup E_{v}\right\vert +1}\\\text{(by \eqref{darij1.pf.t2.1})}}}\cdot x_{u}x_{v}\\ & =\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{\left\vert S\right\vert }{\left\vert E_{u}\cup E_{v}\right\vert +1}\cdot x_{u}x_{v}\\ & =\left\vert S\right\vert \cdot\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert +1}. \end{align*} Hence, \begin{align} \left\vert S\right\vert \cdot\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert +1} =\sum_{\sigma\in S}\underbrace{\left( \sum_{\substack{u\in\left[ n\right] ;\\\sigma\text{ is }u\text{-friendly}}}x_{u}\right) ^{2}}_{\substack{\geq 0\\\text{(since squares are nonnegative)}}}\geq0. \end{align} We can divide this inequality by $\left\vert S\right\vert $ (since $\left\vert S\right\vert $ is a positive integer). We thus obtain \begin{align} \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v} }{\left\vert E_{u}\cup E_{v}\right\vert +1}\geq0. \end{align} This proves Theorem 2. $\blacksquare$
Now, we apply the tensor power trick. The first step is to replace the $1$ in Theorem 2 by a small rational number $1/m$:
Corollary 3. Let $n\in\mathbb{N}$. Let $E_{1},E_{2},\ldots,E_{n}$ be $n$ finite sets. Let $x_{1},x_{2},\ldots,x_{n}$ be $n$ real numbers. Let $m$ be a positive integer. Then, \begin{align} \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v} }{\left\vert E_{u}\cup E_{v}\right\vert +1/m}\geq0. \end{align}
Proof of Corollary 3. For each $i\in\left[ n\right] $, we define a finite set $B_{i}$ by $B_{i}=E_{i}\times\left\{ 1,2,\ldots,m\right\} $. Then, for every $u\in\left[ n\right] $ and $v\in\left[ n\right] $, we have \begin{align*} & \underbrace{B_{u}}_{\substack{=E_{u}\times\left\{ 1,2,\ldots,m\right\} \\\text{(by the definition of }B_{u}\text{)}}}\cup\underbrace{B_{v} }_{\substack{=E_{v}\times\left\{ 1,2,\ldots,m\right\} \\\text{(by the definition of }B_{v}\text{)}}}\\ & =\left( E_{u}\times\left\{ 1,2,\ldots,m\right\} \right) \cup\left( E_{v}\times\left\{ 1,2,\ldots,m\right\} \right) \\ & =\left( E_{u}\cup E_{v}\right) \times\left\{ 1,2,\ldots,m\right\} \end{align*} and therefore \begin{align} \left\vert B_{u}\cup B_{v}\right\vert & =\left\vert \left( E_{u}\cup E_{v}\right) \times\left\{ 1,2,\ldots,m\right\} \right\vert =\left\vert E_{u}\cup E_{v}\right\vert \cdot\underbrace{\left\vert \left\{ 1,2,\ldots ,m\right\} \right\vert }_{=m}\nonumber\\ & =\left\vert E_{u}\cup E_{v}\right\vert \cdot m. \label{darij1.pf.c3.1} \tag{2} \end{align} But Theorem 2 (applied to $B_{i}$ instead of $E_{i}$) yields \begin{equation} \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v} }{\left\vert B_{u}\cup B_{v}\right\vert +1}\geq0. \label{darij1.pf.c3.2} \tag{3} \end{equation} For each $u\in\left[ n\right] $ and $v\in\left[ n\right] $, we have \begin{align*} \dfrac{x_{u}x_{v}}{\left\vert B_{u}\cup B_{v}\right\vert +1} & =\dfrac {x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert \cdot m+1}\qquad\left( \text{by \eqref{darij1.pf.c3.1}}\right) \\ & =\dfrac{1}{m}\cdot\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert +1/m}. \end{align*} Adding up these equalities for all $u\in\left[ n\right] $ and $v\in\left[ n\right] $, we obtain \begin{align*} \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v} }{\left\vert B_{u}\cup B_{v}\right\vert +1} & =\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{1}{m}\cdot\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert +1/m}\\ & =\dfrac{1}{m}\cdot\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert +1/m}. \end{align*} Thus, \eqref{darij1.pf.c3.2} rewrites as \begin{align} \dfrac{1}{m}\cdot\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert +1/m}\geq0. \end{align} We can multiply this inequality by $m$ (since $m$ is positive), and thus obtain \begin{align} \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v} }{\left\vert E_{u}\cup E_{v}\right\vert +1/m}\geq0. \end{align} This proves Corollary 3. $\blacksquare$
The second part of the tensor power trick is to let $m\rightarrow\infty$:
Theorem 4. Let $n\in\mathbb{N}$. Let $E_{1},E_{2},\ldots,E_{n}$ be $n$ nonempty finite sets. Let $x_{1},x_{2},\ldots,x_{n}$ be $n$ real numbers. Then, \begin{align} \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v} }{\left\vert E_{u}\cup E_{v}\right\vert }\geq0. \end{align}
Proof of Theorem 4. First of all, we notice that $E_{u}\cup E_{v}$ is a nonempty finite set whenever $u\in\left[ n\right] $ and $v\in\left[ n\right] $ (since $E_{1},E_{2},\ldots,E_{n}$ are nonempty finite sets), and thus its size $\left\vert E_{u}\cup E_{v}\right\vert $ is a positive integer. Thus, all the fractions $\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert }$ in Theorem 4 are well-defined.
Now, consider a positive integer $m$ going to infinity. Then, $\lim \limits_{m\rightarrow\infty}\dfrac{r}{q+1/m}=\dfrac{r}{q}$ for every real $r$ and every positive real $q$. Hence, \begin{align} \lim\limits_{m\rightarrow\infty}\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert +1/m}=\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup E_{v} \right\vert } \end{align} for every $u\in\left[ n\right] $ and $v\in\left[ n\right] $. Adding up these equalities for all $u\in\left[ n\right] $ and $v\in\left[ n\right] $, we obtain \begin{align} \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\lim \limits_{m\rightarrow\infty}\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert +1/m}=\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert }. \end{align} Hence, \begin{align*} \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v} }{\left\vert E_{u}\cup E_{v}\right\vert } & =\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\lim\limits_{m\rightarrow\infty}\dfrac {x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert +1/m}\\ & =\lim\limits_{m\rightarrow\infty}\underbrace{\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert +1/m}}_{\substack{\geq0\\\text{(by Corollary 3)}}}\geq \lim\limits_{m\rightarrow\infty}0=0. \end{align*} This proves Theorem 4. $\blacksquare$
Now, let us recall the Iverson bracket notation:
Definition. We shall use the Iverson bracket notation: If $\mathcal{A}$ is any statement, then $\left[ \mathcal{A}\right] $ stands for the integer $ \begin{cases} 1, & \text{if $\mathcal{A}$ is true;}\\ 0, & \text{if $\mathcal{A}$ is false} \end{cases} $ (which is also known as the truth value of $\mathcal{A}$). For instance, $\left[ 1+1=2\right] =1$ and $\left[ 1+1=1\right] =0$.
Corollary 5. Let $n\in\mathbb{N}$. Let $E_{1},E_{2},\ldots,E_{n}$ be $n$ nonempty finite sets. Let $e$ be any object. Let $x_{1},x_{2},\ldots,x_{n}$ be $n$ real numbers. Then, \begin{align} \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{\left[ e\in E_{u}\cap E_{v}\right] }{\left\vert E_{u}\cup E_{v}\right\vert }x_{u} x_{v}\geq0. \end{align}
Proof of Corollary 5. For every $u\in\left[ n\right] $ and $v\in\left[ n\right] $, we have \begin{align*} \left[ e\in E_{u}\cap E_{v}\right] & =\left[ e\in E_{u}\text{ and }e\in E_{v}\right] \\ & \qquad\left( \text{since }e\in E_{u}\cap E_{v}\text{ holds if and only if }\left( e\in E_{u}\text{ and }e\in E_{v}\right) \right) \\ & =\left[ e\in E_{u}\right] \cdot\left[ e\in E_{v}\right] \end{align*} (because the rule $\left[ \mathcal{A}\text{ and }\mathcal{B}\right] =\left[ \mathcal{A}\right] \cdot\left[ \mathcal{B}\right] $ holds for any two statements $\mathcal{A}$ and $\mathcal{B}$) and therefore \begin{align*} \dfrac{\left[ e\in E_{u}\cap E_{v}\right] }{\left\vert E_{u}\cup E_{v}\right\vert }x_{u}x_{v} & =\dfrac{\left[ e\in E_{u}\right] \cdot\left[ e\in E_{v}\right] }{\left\vert E_{u}\cup E_{v}\right\vert } x_{u}x_{v}\\ & =\dfrac{\left( \left[ e\in E_{u}\right] x_{u}\right) \cdot\left( \left[ e\in E_{v}\right] x_{v}\right) }{\left\vert E_{u}\cup E_{v} \right\vert }. \end{align*} Adding up these equalities for all $u\in\left[ n\right] $ and $v\in\left[ n\right] $, we obtain \begin{align*} & \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{\left[ e\in E_{u}\cap E_{v}\right] }{\left\vert E_{u}\cup E_{v}\right\vert } x_{u}x_{v}\\ & =\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{\left( \left[ e\in E_{u}\right] x_{u}\right) \cdot\left( \left[ e\in E_{v}\right] x_{v}\right) }{\left\vert E_{u}\cup E_{v}\right\vert }\geq0 \end{align*} (by Theorem 4, applied to $\left[ e\in E_{i}\right] x_{i}$ instead of $x_{i}$). This proves Corollary 5. $\blacksquare$
We next recall a classical fact (I refer to it as "counting by roll call"):
Lemma 6. Let $Q$ be a finite set. Let $F$ be a subset of $Q$. Then, \begin{align} \left\vert F\right\vert =\sum_{e\in Q}\left[ e\in F\right] . \end{align}
Proof of Lemma 6. The sum $\sum_{e\in Q}\left[ e\in F\right] $ has exactly $\left\vert F\right\vert $ many addends equal to $1$ (namely, all the addends corresponding to $e\in F$), while all its remaining addends are $0$ and thus do not influence its value. Hence, $\sum_{e\in Q}\left[ e\in F\right] =\left\vert F\right\vert \cdot1=\left\vert F\right\vert $. This proves Lemma 6. $\blacksquare$
Theorem 7. Let $n\in\mathbb{N}$. Let $E_{1},E_{2},\ldots,E_{n}$ be $n$ nonempty finite sets. Let $x_{1},x_{2},\ldots,x_{n}$ be $n$ real numbers. Then, \begin{align} \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{\left\vert E_{u}\cap E_{v}\right\vert }{\left\vert E_{u}\cup E_{v}\right\vert }x_{u} x_{v}\geq0. \end{align}
Proof of Theorem 7. Let $Q$ be the set $E_{1}\cup E_{2}\cup\cdots\cup E_{n} $. This is a finite set (since the sets $E_{1},E_{2},\ldots,E_{n}$ are finite).
Fix $u\in\left[ n\right] $ and $v\in\left[ n\right] $. Then, $E_{u}\cap E_{v}$ is a subset of $Q$ (since $Q=E_{1}\cup E_{2}\cup\cdots\cup E_{n}$). Hence, Lemma 6 (applied to $F=E_{u}\cap E_{v}$) yields \begin{align} \left\vert E_{u}\cap E_{v}\right\vert =\sum_{e\in Q}\left[ e\in E_{u}\cap E_{v}\right] . \end{align} Thus, \begin{align} \dfrac{\left\vert E_{u}\cap E_{v}\right\vert }{\left\vert E_{u}\cup E_{v}\right\vert }x_{u}x_{v} & =\dfrac{\sum_{e\in Q}\left[ e\in E_{u}\cap E_{v}\right] }{\left\vert E_{u}\cup E_{v}\right\vert }x_{u}x_{v}\nonumber\\ & =\sum_{e\in Q}\dfrac{\left[ e\in E_{u}\cap E_{v}\right] }{\left\vert E_{u}\cup E_{v}\right\vert }x_{u}x_{v}. \label{darij1.pf.t7.1} \tag{4} \end{align}
Now, forget that we fixed $u$ and $v$. We have \begin{align*} & \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\underbrace{\dfrac {\left\vert E_{u}\cap E_{v}\right\vert }{\left\vert E_{u}\cup E_{v}\right\vert }x_{u}x_{v}}_{\substack{=\sum_{e\in Q}\dfrac{\left[ e\in E_{u}\cap E_{v}\right] }{\left\vert E_{u}\cup E_{v}\right\vert }x_{u}x_{v}\\\text{(by \eqref{darij1.pf.t7.1})}}}\\ & =\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\sum_{e\in Q}\dfrac{\left[ e\in E_{u}\cap E_{v}\right] }{\left\vert E_{u}\cup E_{v}\right\vert }x_{u}x_{v}=\sum_{e\in Q}\underbrace{\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{\left[ e\in E_{u}\cap E_{v}\right] }{\left\vert E_{u}\cup E_{v}\right\vert }x_{u}x_{v} }_{\substack{\geq0\\\text{(by Corollary 5)}}}\\ & \geq\sum_{e\in Q}0=0. \end{align*} This proves Theorem 7. $\blacksquare$
We are now ready for the original question:
Corollary 8. Let $n\in\mathbb{N}$. Let $E_{1},E_{2},\ldots,E_{n}$ be $n$ nonempty finite sets. Let $A$ be the $n\times n$-matrix $\left( \dfrac{\left\vert E_{u}\cap E_{v}\right\vert }{\left\vert E_{u}\cup E_{v}\right\vert }\right) _{u,v\in\left[ n\right] }\in\mathbb{R}^{n\times n}$. Then, $A$ is positive semidefinite.
Proof of Corollary 8. For every $u\in\left[ n\right] $ and $v\in\left[ n\right] $, we have $\dfrac{\left\vert E_{u}\cap E_{v}\right\vert }{\left\vert E_{u}\cup E_{v}\right\vert }=\dfrac{\left\vert E_{v}\cap E_{u}\right\vert }{\left\vert E_{v}\cup E_{u}\right\vert }$ (since $E_{u}\cap E_{v}=E_{v}\cap E_{u}$ and $E_{u}\cup E_{v}=E_{v}\cup E_{u}$). Thus, the matrix $A$ is symmetric. Hence, in order to prove that $A$ is positive semidefinite, it suffices to show that $x^{T}Ax\geq0$ for any vector $x\in\mathbb{R}^{n}$. So let us do this.
Fix $x\in\mathbb{R}^{n}$. We must show that $x^{T}Ax\geq0$. Write the vector $x\in\mathbb{R}^{n}$ in the form $x=\left( x_{1},x_{2},\ldots,x_{n}\right) ^{T}$ for some real numbers $x_{1},x_{2},\ldots,x_{n}$. Then, the definition of $A$ yields \begin{align} x^{T}Ax=\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] } \dfrac{\left\vert E_{u}\cap E_{v}\right\vert }{\left\vert E_{u}\cup E_{v}\right\vert }x_{u}x_{v}\geq0 \end{align} (by Theorem 7). This completes our proof of Corollary 8. $\blacksquare$
Assume the sets $E_i$ are all non-empty. By deleting rows and columns of $A$ if needed one can reduce to the case where the $E_i$ are all distinct.
Write $$A_{ij}= \frac{|E_i\cap E_j|}{|E_i\cup E_j|} = \frac{|E_i\cap E_j|}{|E_i|+|E_j|-|E_i\cap E_j|} = \frac{e_{ij}}{e_i+e_j-e_{ij}} $$ where $e_i$ is shorthand for $|E_i|$ and so on. Then $$A_{ij}= \frac{e_{ij}}{e_i+e_j} \left( 1 + r_{ij} + r_{ij}^2 + r_{ij}^3 + \cdots\right)\tag{*}$$ where $r_{ij}= e_{ij}/(e_i+e_j).$ The distinctness hypothesis implies $|r_{ij}|<1$ and so the convergence of the geometric series in (*).
The matrix with entries $e_{ij}$ is a Gram matrix and hence positive semidefinite. The matrix with entries $1/(e_i+e_j) = \int_0^\infty e^ {-x|E_i|} e^{-x|E_j|} \,dx$ is also a Gram matrix so it too is psd. Schur's product theorem tells us that their elementwise product $r_{ij}$ is thus also psd. By Schur again, all the summand matrices in (*) are psd. So the sum, $A$, is psd.
Examples (where all the $E_i$ are equal, so all $A_{ij}=1$, say) show that positive semi definite cannot be replaced by positive definite as in an earlier form of the problem statement. I suppose that strict positive definiteness holds in the reduced case, but I don't see an argument. Numerical experiments show that the matrix $(r_{ij})$ is not strictly positive definite when (for example) the $E_i$ are all the 2-element subsets of $[4]$. The same experiments however do show $A$ strictly positive definite.