# Minimal set with same Schnirelmann Density

Define the Schnirelmann density of a set of integers by $\sigma(A) = \inf_{n \in \mathbb N} \frac{A(n)}{n}$ where $A(n)$ is the number of elements of $A$ which are $\le n$.

I would like a proof that there is a subset $A'$ of $A$ with the same density and the additional property that removing any element of $A'$ would decrease its density. I could not come up with one myself. Thanks for any proof or sources.

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Define $A_0=A$ and for each $i\geq 1$ define inductively $$A_i=\begin{cases} A_{i-1}\setminus\{i\}&\text{ if \sigma(A_{i-1}\setminus\{i\})=\sigma(A)}\\ A_{i-1}&\text{ otherwise.}\end{cases}$$ Define $A'=\cap_{i\geq 0}A_i$. By induction, $\sigma(A_i)= \sigma(A)$ for all $i$, which implies $\sigma(A')= \sigma(A)$. But by construction, removing any element of $A'$ lowers the Schnirelmann density.
More abstractly, $\sigma$ in continuous in the sense that $\sigma(\cap A_i)=\inf \sigma(A_i)$ for any descending chain $A_i$. So the set of subsets $A'\subseteq A$ with $\sigma(A')=\sigma(A)$ is (reverse-)chain-complete. By Zorn's lemma we can pick a minimal element.