October 12th, 2019, 12:23 PM  #1 
Senior Member Joined: Dec 2015 From: Earth Posts: 823 Thanks: 113 Math Focus: Elementary Math  Real numbers in (0,1)
Show whether the product of all real numbers in interval (0,1) converges or diverges ?

October 12th, 2019, 01:52 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,852 Thanks: 743 
Trivial: Consider a subset consisting of all the reciprocals of the integers. The product is $\displaystyle\lim_{n\to \infty} \frac{1}{n!} =0$. All other contributions to the product are $\le 1$, making it go to zero faster.
Last edited by skipjack; October 12th, 2019 at 02:13 PM. Reason: typo 
October 14th, 2019, 06:19 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,696 Thanks: 2681 Math Focus: Mainly analysis and algebra 
Here's a problem though. How do you form the product of uncountably many numbers? A product is formed by multiplying numbers, but that operation is only defined as a binary operation. We can chain them together, but that at best only gives us countably many (as in Mathman's subset).

October 14th, 2019, 07:54 PM  #4 
Senior Member Joined: Jun 2019 From: USA Posts: 376 Thanks: 202  That may be a problem in general, but it's not going to affect the outcome of the question at hand. Once you've got down to zero with a countably infinite set of values, the rest aren't going to make it nonzero, whether you can figure out how to multiply them in or not.

October 15th, 2019, 12:58 AM  #5  
Senior Member Joined: Oct 2009 Posts: 905 Thanks: 354  Quote:
$$\prod_{x\in (0,1)} x = L$$ if and only if for each $\varepsilon>0$, there is some finite subset $F_0\subseteq (0,1)$ such that if $F$ is any finite subset of $(0,1)$ with $F_0\subseteq F$, then $$\left \prod_{x\in F} x  L\right<\varepsilon.$$ With this definition, you can indeed prove the product is $0$. Obviously other definitions are possible, but this is a really common one.  

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