January 26th, 2011, 04:44 PM  #1 
Newbie Joined: Jan 2011 Posts: 26 Thanks: 0  Union Of Sets
Let q be in the rational numbers. Define an interval . Will ? Explain in detail? I am not sure how to go about this. I know there is the density of rationals. I think the answer is yes, but I am not quite sure how to explain it. 
January 26th, 2011, 04:55 PM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Union Of Sets
I'm not sure what you mean by "Let q be in the rational numbers". If it means then the union need not equal the real numbers (what if q_n = 0 for all n?). If it means , then I feel as you do.

January 27th, 2011, 02:07 AM  #3 
Senior Member Joined: Nov 2010 From: Staten Island, NY Posts: 152 Thanks: 0  Re: Union Of Sets
I suspect it's true also (assuming what CR said is what you mean). I would try something like this:: If is a real number, take a sequence of rationals converging to . So for each You can look at 2 intervals. The interval and the interval , the latter guaranteeing to contain (passing to a suitable subsequence if necessary). Perhaps now argue indirectly that must be in one of the . 
January 27th, 2011, 07:44 AM  #4  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Union Of Sets Quote:
Or to put it differently: suppose we have a fixed ordering for the rationals. Go through the order and whenever your target irrational r would otherwise be included in the interval, move the rational at that index to the first index that would exclude r, moving some rational outside [r  2^n, r + 2^n] in its place (WLOG you can choose some integer in {0, 1, ..., n}), and swapping the displaced numbers. This yields a new list that seems to never include r. Did I just prove what I thought wasn't true? Or have I made a mistake?  
January 27th, 2011, 09:16 AM  #5 
Newbie Joined: Jan 2011 Posts: 26 Thanks: 0  Re: Union Of Sets
Alright, so we are also told that since is countably infinite, it is possible to write , then from this for each let I hope this helps towards the solution a little bit. 
January 27th, 2011, 01:20 PM  #6  
Senior Member Joined: Nov 2010 From: Staten Island, NY Posts: 152 Thanks: 0  Re: Union Of Sets Quote:
The first seems to be false. I'm not sure about the second. In particular, your argument shows that my particular suggestion for proof doesn't work.  
January 27th, 2011, 04:59 PM  #7 
Newbie Joined: Dec 2010 Posts: 18 Thanks: 0  Re: Union Of Sets
Hi, the length of the interval X_n is 2^(n1) for each n. Therefore the Lebesgue measure of the union of the X_n never exceeds 2, no matter the enumeration of the rationals. So, the union can never cover the whole real line. Best regards, Thommy 
January 27th, 2011, 05:27 PM  #8 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Union Of Sets
I had wondered about that too, Thommy.

January 28th, 2011, 10:35 AM  #9 
Newbie Joined: Jan 2011 Posts: 26 Thanks: 0  Re: Union Of Sets
I was thinking about it, and I feel that its still true. This is my reasoning. The rationals are dense in the reals that is for any we can find a rational that is within epsilon of it. So which becomes then through rearrangment this becomes and then get to So from this, our interval So let and I believe that this proves the claim. For any , it can be found in an interval . Does this seem reasonable? 
January 28th, 2011, 10:47 AM  #10  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Union Of Sets Quote:
 

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