Prove that the closure of Q is equal to Rby Tetrinity Tags: closure, rational numbers, real numbers 

#1
Feb2311, 01:34 PM

P: 6

1. The problem statement, all variables and given/known data
As the title states, the problem asks to prove that the closure of the set of rational numbers is equal to the set of real numbers. The problem includes the standard definition of the rationals as {p/q  q ≠ 0, p,q ∈ Z} and also states that the closure of a set X ⊂ R is equal to the set of all its limit points. Our lecturer suggested the following method:
2. Relevant equations I'm not sure if there's anything particularly useful here. My definition of R states that Q is a subset of R, it is ordered, the standard arithmetic operations work and it is complete (i.e. every bounded set S ⊂ R has a suprenum sup(S) ∈ R and infinum inf(S) ∈ R). It also states that every real number can be defined as a convergent sequence of rationals. 3. The attempt at a solution I'm going to use the notation cl(X) to represent the closure of the set X. The first step is pretty easy: Q ⊂ R ⇒ cl(Q) ⊂ cl(R) ⇒ cl(Q) ⊂ R (the closure of N, Z and R are already known to us at this point) My problem lies with the second step. I know that the closure of a set X is the set of all limit points of X, and that a is a limit point to X if a is an adherent point to X\{a}. I'm just not really sure how to begin to prove the second step, and I've racked my brains to the point where I'm probably missing something obvious. Any help with this, be it using this method or a different one, would be appreciated. Thanks. 



#2
Feb2311, 01:43 PM

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P: 16,703

Well, it isn't obvious. Some even think of this as an axiom of R. Thus something that cannot be proven, but that is assumed. So whatever proof you'll find, it will be quite informal. (unless you're working with Dedekind cuts, but I don't assume that this is the case here).
To prove that Q is dense in R, we must show that for every x in R, there exists a sequence [tex](q_n)_n[/tex] in Q, that converges to x. To find the sequence, take x and take as q_{n} the first n decimals of x. This is a rational numher. For example, pi is the limit of following sequence [tex]3,~3.1,~3.14,~3.141,~3.1415,...[/tex] I'll leave it to you to formalize the argument... 



#3
Feb2311, 02:02 PM

P: 828

Couldn't he do something like this: The closure of a set S is defined as the intersection of all closed sets containing S. Now show that the only closed set containing Q is R. Then the intersection of these sets is R. 



#4
Feb2311, 02:06 PM

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P: 16,703

Prove that the closure of Q is equal to R 



#5
Feb2311, 02:27 PM

P: 828





#7
Feb2311, 02:40 PM

P: 6

I can find such a sequence by setting q_{n} = a_{1}a_{2}a_{3}...a_{k1}.a_{k}a_{k+1}...a_{n} where a_{i} is the i^{th} digit of x. Then if we assume that q_{n} is known, we can find a_{n+1} by setting it such that q_{n+1} < x and setting it to be 1 higher causes q_{n+1} ≥ x. This can be done for any x ∈ R. Does this complete the proof that the closure of Q is R? I know it's not amazingly clear, but I can't think of any other way to represent the decimal. And as a sidenote: wrong gender. :) 



#8
Feb2311, 02:46 PM

P: 828





#9
Feb2311, 02:47 PM

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P: 16,703





#10
Feb2311, 02:51 PM

P: 828





#11
Feb2311, 02:53 PM

PF Gold
P: 1,930

Hmm... 



#12
Feb2311, 02:58 PM

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P: 16,703





#13
Feb2311, 03:27 PM

P: 6

I think I may need to stick to the definition we've been given. I lost marks on a previous assignment for using a limit before they were formally introduced in the course, with no indication as to whether my proof was actually correct or not. It still seems a bit unfair, but it might be best to avoid that situation again! Interestingly, I came across another proof of the density of Q in R while reading through my notes just now. Let x, y ∈ R and x < y. We can always find n ∈ N such that n > 1/(yx) Now let m be the smallest integer such that m/n > x (in other words, [m1]/n ≤ x) Then x < m/n = [1 + (m1)]/n = 1/n + (m1)/n We know 1/n < y  x, and (m1)/n ≤ x Therefore 1/n + (m1)/n < (yx) + x = y Therefore x < m/n < y (i.e. there is a rational number between any two reals) Would that proof be better to use, since it avoids having to further prove q_{n} converges to x? I'd been reading through my notes to remind myself of how to show a sequence converges to a limit... 



#14
Feb2311, 03:32 PM

P: 828

Let y < x be reals. Then x  y > 0. We can pick an n such that n > xy => 1/n < 1/(xy). Pick the largest m st m/n < y. Then (m+1)/n > y. Since 1\n < xy, and m/n < x we have that (m+1)/n < x. Hence y < (m+1)/n < x. I think the problem here is that our books have presented information in different orders and differntly. For example, the thing I just proved was an early result in the book I use. 



#15
Feb2311, 03:34 PM

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#16
Feb2311, 03:48 PM

P: 6

Thanks for all of your help! 


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