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Consider the collection of all infinite products of numbers $x_n$ with $0< x_n< 1$ for every $n$. Even if it might be tempting to say that any infinite product of this kind diverges to zero, it is well known that an infinite product of real numbers between zero and one could converge to some real number grater that zero. As an example one could consider the following product: \begin{equation} \prod_{n> 1} \left(1 - \frac{1}{n^2}\right) \end{equation} Now consider the set $\tilde{\Gamma}$ defined as: \begin{equation} \tilde{\Gamma}=\left\{\left\{x_n\right\}_{n>1}|\prod_{n> 1} x_n>0\right\} \end{equation} Now the question is: is the measure of the set $\tilde{\Gamma}$ equal to zero? Or, alternatively, is the set: \begin{equation} \Gamma=\left\{\left\{x_n\right\}_{n>1}|\prod_{n> 1} x_n=0\right\} \end{equation} dense in $\left[0,1\right]^{\infty}$?

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  • $\begingroup$ Note that the set of sequences converging to any value has measure $0$ in the set of all sequences, and the set $\tilde\Gamma$ is a subset of the set of sequences converging to $1$. $\endgroup$ – Don Thousand Feb 23 at 15:26
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    $\begingroup$ Which topology is there on $[0,1]^{\Bbb N}$? $\endgroup$ – Gae. S. Feb 23 at 15:28

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