Can the Rationals be Contained in Open Intervals with Infinitely Small Width?

zhang128
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Homework Statement


Prove that the rationals as a subset of the reals can all be contained in open intervals the sum of whose width is less than any \epsilon > 0.


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The Attempt at a Solution

 
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You aren't playing the game here. What do you think about the problem? You can't leave the Attempt at a Solution completely blank.
 
ups, my bad. I was thinking that since rationals are countable, I just need to make the open interval around each rational such that the total sum is less than \epsilon. Then the question turns to prove the sum of a finite series converges to the epsilon??
 
zhang128 said:
ups, my bad. I was thinking that since rationals are countable, I just need to make the open interval around each rational such that the total sum is less than \epsilon. Then the question turns to prove the sum of a finite series converges to the epsilon??

Much better, thanks. Can you write a series that converges to epsilon? If you can write a series that sums to say, 1, you should be able to write a series that converges to epsilon.
 
hmmm, like epsilon/(n^2+n)??
 
zhang128 said:
hmmm, like epsilon/(n^2+n)??

Sure, that works. I would have said sum epsilon*(1/2)^n. But whatever you like.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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