= Solution
For each positive <integer> $n$, consider the threshold <set> $A^{(n)}=\{a\in A:a\ge1/n\}$. If it were infinite, one could choose recursively a <sequence> of distinct elements from it. Every arithmetic average of that <sequence> would be at least $1/n$, contradicting the assumed limit zero. Hence every $A^{(n)}$ is a <finite set>.
Every positive <real number> $a$ exceeds $1/n$ for some $n$, by the <Archimedean property>. Therefore
$$
A=\bigcup_{n=1}^{\infty}A^{(n)}.
$$
A <countable union of countable sets> is countable, so $\boxed{A\text{ is countable}}$. In fact, under the stipulated infinitude, $A$ is a <countably infinite set>. Positivity matters: it ensures that these thresholds cover $A$. This is <countability from vanishing averages of distinct sequences>; ordinary <Cesaro convergence of a sequence> of one chosen enumeration would not suffice to control the whole <set>.
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