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Suppose we have a set $S$ of real numbers,such that

$ n \neq \Pi_{m \in S-{n}} m^{r_m}$....... $\forall n\in S$ where $r_{m}$ is a rational number....
The set of prime number is an example of such a ser...now my question is that...Is such a set necessarily countable?

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    $\begingroup$ No, take a non enumerable set of algebraically independent real numbers (such a set exists since $\mathbb R$ would be enumerbale otherwise). $\endgroup$ May 17, 2011 at 16:50
  • $\begingroup$ Roland, right, I hadn't seen your comment when I posted my answer. $\endgroup$ May 17, 2011 at 17:09

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Since we are talking about rational exponents, I assume it makes sense to restrict to positive real numbers. If I got it right the question is whether a set $S$ where no element is a product of some others to rational powers is necessarily countable.

One can show non-constructively that the answer is no, that is, that there exist uncountable sets with this property. This is because the condition on $S$ is equivalent to $\{\log x: x\in S\}$ being linearly independent over the rational numbers.

Since there is a basis for $\mathbb{R}$ as a vector space over $\mathbb{Q}$, and such a basis is necessarily uncountable, there is an uncountable set $S$ with the given property.

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