# Set of natural numbers, for which the reunion of the sums of all subsets is a set

My math is weak so I'll try to clarify my question with an example. Consider the set $\{1,2,4\}$ for which the powerset $\{ \{\}, \{1\}, \{2\}, \{4\}, \{1,2\}, \{1,4\}, \{2,4\}, \{1,2,4\} \}$.

Now, if I sum the subsets I get the resulting set $\{ 0, 1, 2, 4, 3, 5, 6, 7 \}$.

How would I write the definition of a function that produces such a set?

Update

To be clear, the function should be able to return the set $\{1,2,4\}$ which holds the property. But I'd like to be able to return a set of $n$ length with that property.

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For any set $B$ of natural numbers, let $f(B)$ be the set whose elements are sums of subsets of $B$. So $f:\mathcal{P}(\mathbb{N})\rightarrow\mathcal{P}(\mathbb{N})$. What you desire is to find an inverse of this function, however, the function is not surjective, so its inverse won't be defined on all of $\mathcal{P}(\mathbb{N})$.
For example, suppose we begin with the set $A=\{0,1,2\}$ and suppose that there were some set $B$ such that $f(B)=A$. By looking at sets of cardinality $1$, it follows that $B\subset A$, but notice that no subset of $B$ of $A$ gives $f(B)=A$.
If we only look at $f$ onto its image, we would need to show it is injective to show it's inverse in that case is well-defined. Immediately, we can see that if we consider $0$ a natural number, then $f(\{0,1\})=f(\{1\})$ so the map is not injective.
Even if we don't include $0$ as a natural number, notice that $$f(\{1,2,3,6\})=f(\{1,2,4,5\})=\{1,2,3,4,5,6,7,8,9,10,11,12\})$$ so the map is still not injective, and we cannot hope to find its inverse.