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If every sequence of distinct elements of a positive-real set $A$ has arithmetic averages tending to zero, then $A$ is a countable set. For each positive integer $n$, the threshold set $\{a\in A:a\ge 1/n\}$ must be finite: an infinite sequence inside it would have averages at least $1/n$. The Archimedean property makes $A$ their union, so the countable union of countable sets theorem applies. The hypothesis concerns every distinct-element sequence, not merely one enumeration.

For each positive integer $n$, consider the threshold set $A^{(n)}=\{a\in A:a\ge 1/n\}$. 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. ThereforeA 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.