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 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 $X_n=(q_n-\frac{1}{2^n}, q_n+\frac{1}{2^n}$. Will $\displaystyle \bigcup_{n=1}^{\infty} X_n=\mathbb{R}$? 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 $q_n\in\mathbb{Q}$ then the union need not equal the real numbers (what if q_n = 0 for all n?). If it means $\bigcup_nq_n=\mathbb{Q}$, 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 $r$ is a real number, take a sequence of rationals $q_k$ converging to $r$. So for each $k$ You can look at 2 intervals. The interval $X_k$ and the interval $Y_k=(q_k-\frac{1}{2^{k_t}}, q_k+\frac{1}{2^{k_t}})$, the latter guaranteeing to contain $r$ (passing to a suitable subsequence if necessary). Perhaps now argue indirectly that $r$ must be in one of the $X_k$.
January 27th, 2011, 07:44 AM   #4
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Re: Union Of Sets

Quote:
 Originally Posted by DrSteve Perhaps now argue indirectly that $r$ must be in one of the $X_k$.
But need it be? Suppose r is pi and the q_i are 3, 3.1, 3.14, 3.141, and so forth, and that they appear at indexes 100, 10000, 1000000, etc. Then (3-2^-100, 3+2^-100) U (3.1-2^-10000, 3.1+2^-10000) U ... does not seem to contain pi.

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 $\mathbb{Q}$ is countably infinite, it is possible to write $\mathbb{Q}=\{q_1,q_2,...\}$, then from this for each $n \ge 1$ let $X_n=(q_n-\frac{1}{2^n}, q_n+\frac{1}{2^n})$ I hope this helps towards the solution a little bit.
January 27th, 2011, 01:20 PM   #6
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Re: Union Of Sets

Quote:
Originally Posted by CRGreathouse
Quote:
 Originally Posted by DrSteve Perhaps now argue indirectly that $r$ must be in one of the $X_k$.
But need it be? Suppose r is pi and the q_i are 3, 3.1, 3.14, 3.141, and so forth, and that they appear at indexes 100, 10000, 1000000, etc. Then (3-2^-100, 3+2^-100) U (3.1-2^-10000, 3.1+2^-10000) U ... does not seem to contain pi.

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?
It looks like you just proved that there is an enumeration of the rationals in which the target theorem is false. It's not clear to me if that's what the question is however? Is the question asking if it's true for all enumerations of the rationals, or is it asking if there is an enumeration of the rationals for which it's true?

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^(n-1) 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 $r \in \mathbb{R}$ we can find a rational that is within epsilon of it. So $|q-r|<\epsilon$ which becomes $-\epsilon < q-r < \epsilon$ then through rearrangment this becomes $-q-\epsilon < -r < -q+\epsilon$ and then get to $q+\epsilon > r > q-\epsilon$ So from this, our interval $(q-\epsilon, q+\epsilon)$ So let $\epsilon=\frac{1}{2^i}$ and I believe that this proves the claim. For any $r \in \mathbb{R}$, it can be found in an interval $(q_i-\frac{1}{2^1}, q_i+\frac{1}{2^1})$. Does this seem reasonable?
January 28th, 2011, 10:47 AM   #10
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Re: Union Of Sets

Quote:
 Originally Posted by gettingit4 Does this seem reasonable?
No, since how do you know that number will appear so early in the sequence? Isn't it possible that for every rational you choose it appears too late on the list to include the real in its interval?

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