Let $[\omega]^{<\omega}$ denote the collection of finite subsets of the integers, and let us call $E\subseteq [\omega]^{<\omega}$ *non-nested* if $a\not\subseteq b$ whenever $a\neq b\in E$.


Is there a non-nested set $E \subseteq [\omega]^{<\omega}$ with the following properties?


1. Every $n\in \omega$ is contained in infinitely many members of $E$, and
2. For every bijection $\varphi:\omega\to\omega$ and for every $a\neq b\in E$ we have $\sum\_{n\in a} \varphi(n) \neq \sum\_{n\in b} \varphi(b)$.



Non-summable subsets of $[\omega]^{<\omega}$
Theorem: There is no such $E$.


Claim: for each $a\in E$ there exists $b\in E$ with $|b\setminus a|\ge 2$.
Proof of Claim: Let $a$ be a counterexample. Then all $b\ne a$ contain exactly one element each that's not in $a$. Let $n\not\in a$. Since $n$ is in infinitely many $b$'s, some $b$'s must be equal, contradiction.


Proof of Theorem:
Fix any $a\in E$, and a corresponding $b$ as in the Claim.


To show that some $\varphi$ exists with $\sum\_{n\in a}\varphi(n)=\sum\_{n\in b}\varphi(n)$, we may assume $a\cap b=\emptyset$ since $a\cap b$ contributes the same to both sides.


Let $a=\{\alpha\_1,\dots,\alpha\_n\}$ and $b=\{\beta\_1,\dots,\beta\_m\}$.


Let $\varphi(\alpha\_i)$ for $i<n$ be distinct and arbitrary.


Let $\varphi(\beta\_i)$ for $i\le m$ be distinct, with each $\varphi(\beta\_i)> \sum\_{j<n}\varphi(\alpha\_j)$.


Then for any $k$ and $t\ne k$ we have
$$\sum\varphi(\beta\_i)\ge \varphi(\beta\_t)+\varphi(\beta\_k) >\varphi(\beta\_t) + \sum\_{j<n}\varphi(\alpha\_j)$$
so $\sum\varphi(\beta\_i)-\sum\varphi(\alpha\_i)> \varphi(\beta\_t)$.
Therefore we are free to define
$$\varphi(\alpha\_n)=\sum\varphi(\beta\_i)-\sum\_{j<n}\varphi(\alpha\_i),$$
and we have defined an injective function from $a\cup b$ to $\omega$. This can now be extended to a bijection of $\omega$ arbitrarily.


This completes the construction of the desired $\varphi$ showing that no such $E$ exists. $\Box$


