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I have read a proof of the following theorem in Rudin's Real and Complex Analysis:

Suppose $\{u_n\}$ is a sequence of bounded complex functions on a set S, such that $\sum |u_n(s)|$ I converges uniformly on $S$. Then the product $$ f(s) = \prod_{n=1}^\infty (1+u_n(s)) $$ converges uniformly on $S$, and $f(s_0) = 0$ at some $s_0\in S$ if and only if $u_n(s_0) = - 1$ for some $n$.

Furthermore, if $\{n_1 n_2 , n_3 ,\ldots\}$ is any permutation of $\{1, 2, 3,\ldots\}$, then we also have $$ f(s) = \prod_{i=1}^\infty (1+u_{n_i}(s)) $$

Can anyone provide an example of an infinite product, where the factors does not commute? I can't seem to find one...

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Let $(a_{n})$ be a sequence whose sum converges, but not absolutely.
Then the factors of the product $$\prod_{k=0}^{\infty}2^{a_{n}}=2^{\sum_{k=0}^{\infty} a_{k}}$$ do not commute.

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    $\begingroup$ Ah, neat! I think that answers my question, thanks! $\endgroup$ – user114158 Jun 20 '15 at 13:43
  • $\begingroup$ You're welcome! $\endgroup$ – preferred_anon Jun 20 '15 at 13:43

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