Non-summable subsets of $[\omega]^{<\omega}$

Question

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 $\sum_{n\in a} \varphi(n) \neq \sum_{n\in b} \varphi(b)$.

Answer 1

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. $\Box$