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Here's an argument that for $A$ a finite set of algebraic numbers, $d_A(n)$ decays at most exponentially.

Suppose $A$ is contained in some number field $K$. For $x\in K^\times$, there's a product formula $$ \prod_v |x|_v=1, $$ where $v$ runs over the valuations on $K$ and $|\cdot|_v$ is a suitably normalized absolute value. In order for some absolute value to be small, the product of the others must be large.

For fixed $A$, there is a finite set of places $P$ of $K$ such that $|x|_v\leq 1$ for all $x\in S_A(n)$ whenever $v\not\in P$ ($P$ consists of all the Achimedean places, and the finite places dividing denominators of elements of $A$). It's not hard to see that for fixed $v$, $\max \{|x|_v:x\in S_A(n)\}$ grows at most exponentially in $n$.

In our case, $K$ is embedded in the real numbers. Let $v_0$ be the valuation coming from this embedding. Since the number of $v$ for which $|x|_v>1$ is bounded, we have \begin{gather} d_A(n)=\min \big\{|x|_{v_0}:x\in S_A(n)\backslash\{0\}\big\}=\min\left\{\prod_{v\neq v_0}|x|_v^{-1}\right\} \geq\min\left\{\prod_{v\in P\backslash\{v_0\}} |x|_v^{-1}\right\}. \end{gather} The right hand side decays at most exponentially, so the left hand side does too.

Here's an argument that for $A$ a finite set of algebraic numbers, $d_A(n)$ decays at most exponentially.

Suppose $A$ is contained in some number field $K$. For $x\in K^\times$, there's a product formula $$ \prod_v |x|_v=1, $$ where $v$ runs over the valuations on $K$ and $|\cdot|_v$ is a suitably normalized absolute value. In order for some absolute value to be small, the product of the others must be large.

For fixed $A$, there is a finite set of places $P$ of $K$ such that $|x|_v\leq 1$ for all $x\in S_A(n)$ whenever $v\not\in P$ ($P$ consists of all the Archimedean places, and the non-Archimedean places dividing denominators of elements of $A$). It's not hard to see that for fixed $v$, $\max \{|x|_v:x\in S_A(n)\}$ grows at most exponentially in $n$.

In our case, $K$ is embedded in the real numbers. Let $v_0$ be the valuation coming from this embedding. Since the number of $v$ for which $|x|_v>1$ is bounded, we have \begin{gather} d_A(n)=\min \big\{|x|_{v_0}:x\in S_A(n)\backslash\{0\}\big\}=\min\left\{\prod_{v\neq v_0}|x|_v^{-1}\right\} \geq\min\left\{\prod_{v\in P\backslash\{v_0\}} |x|_v^{-1}\right\}. \end{gather} The right hand side decays at most exponentially, so the left hand side does too.

Here's an argument that for $A$ a finite set of algebraic numbers, $d_A(n)$ grows at most exponentially.

Suppose $A$ is contained in some number field $K$. For $x\in K^\times$, there's a product formula $$ \prod_v |x|_v=1, $$ where $v$ runs over the valuations on $K$ and $|\cdot|_v$ is a suitably normalized absolute value. In order for some absolute value to be small, the product of the others must be large.

For fixed $A$, there is a finite set of places $P$ of $K$ such that $|x|_v\leq 1$ for all $x\in S_A(n)$ whenever $v\not\in P$ ($P$ consists of all the Achimedean places, and the finite places dividing denominators of elements of $A$). It's not hard to see that for fixed $v$, $\max \{|x|_v:x\in S_A(n)\}$ grows at most exponentially in $n$. Since the number of $v$ for which $|x|_v>1$ is bounded, we have \begin{gather} d_A(n)=\max \big\{|x|_{v_0}:x\in S_A(n)\big\}=\max\left\{\prod_{v\neq v_0}|x|_v^{-1}\right\} \geq\max\left\{\prod_{v\in P\backslash\{v_0\}} |x|_v^{-1}\right\}. \end{gather} The right hand side decays at most exponentially, so the left hand side does too.

Here's an argument that for $A$ a finite set of algebraic numbers, $d_A(n)$ decays at most exponentially.

Suppose $A$ is contained in some number field $K$. For $x\in K^\times$, there's a product formula $$ \prod_v |x|_v=1, $$ where $v$ runs over the valuations on $K$ and $|\cdot|_v$ is a suitably normalized absolute value. In order for some absolute value to be small, the product of the others must be large.

For fixed $A$, there is a finite set of places $P$ of $K$ such that $|x|_v\leq 1$ for all $x\in S_A(n)$ whenever $v\not\in P$ ($P$ consists of all the Achimedean places, and the finite places dividing denominators of elements of $A$). It's not hard to see that for fixed $v$, $\max \{|x|_v:x\in S_A(n)\}$ grows at most exponentially in $n$.

In our case, $K$ is embedded in the real numbers. Let $v_0$ be the valuation coming from this embedding. Since the number of $v$ for which $|x|_v>1$ is bounded, we have \begin{gather} d_A(n)=\min \big\{|x|_{v_0}:x\in S_A(n)\backslash\{0\}\big\}=\min\left\{\prod_{v\neq v_0}|x|_v^{-1}\right\} \geq\min\left\{\prod_{v\in P\backslash\{v_0\}} |x|_v^{-1}\right\}. \end{gather} The right hand side decays at most exponentially, so the left hand side does too.

Here's an argument that if $A$ is a finite set of algebraic numbers, then $d_A(n)$ grows at most exponentially.

Suppose $A$ is contained in some number field $K$. For $x\in K^\times$ there's a product formula $$ \prod_v |x|_v=1, $$ where $v$ runs over the valuations on $K$ and $|\cdot|_v$ is a normalized absolute value. In order for some absolute value to be small, the product of the others must be large.

For fixed $A$, there's a finite set of places $S$ of $K$ such that $|x|_v\leq 1$ for all $x\in S_A(n)$ whenever $v\not\in S$ ($S$ consists of all the Achimedean places, and the finite places dividing denominators of elements of $A$). It's not hard to see that for fixed $v$, $\max \{|x|_v:x\in S_A(n)\}$ grows at most exponentially in $n$. Since the number of $v$ for which $|x|_v>1$ is bounded, we get lower bound of exponential decay the minimal absolute value of elements of $S_A(n)$.

Here's an argument that for $A$ a finite set of algebraic numbers, $d_A(n)$ grows at most exponentially.

Suppose $A$ is contained in some number field $K$. For $x\in K^\times$, there's a product formula $$ \prod_v |x|_v=1, $$ where $v$ runs over the valuations on $K$ and $|\cdot|_v$ is a suitably normalized absolute value. In order for some absolute value to be small, the product of the others must be large.

For fixed $A$, there is a finite set of places $P$ of $K$ such that $|x|_v\leq 1$ for all $x\in S_A(n)$ whenever $v\not\in P$ ($P$ consists of all the Achimedean places, and the finite places dividing denominators of elements of $A$). It's not hard to see that for fixed $v$, $\max \{|x|_v:x\in S_A(n)\}$ grows at most exponentially in $n$. Since the number of $v$ for which $|x|_v>1$ is bounded, we have \begin{gather} d_A(n)=\max \big\{|x|_{v_0}:x\in S_A(n)\big\}=\max\left\{\prod_{v\neq v_0}|x|_v^{-1}\right\} \geq\max\left\{\prod_{v\in P\backslash\{v_0\}} |x|_v^{-1}\right\}. \end{gather} The right hand side decays at most exponentially, so the left hand side does too.