In a sequence of all rationals, why is every real number a subsequential limit?

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Every real number can be a subsequential limit of a sequence of rational numbers, as the set of all rationals is dense in the reals. While a sequence of all rationals can be ordered, it does not need to be monotonic, allowing for subsequences that converge to any real number. The discussion clarifies that not every sequence of rationals will have every real number as a limit point; rather, it is the complete set of rationals that achieves this. Additionally, it is established that for any real number, a corresponding sequence of rational numbers can be constructed to converge to it. Thus, the density of rationals in the reals is fundamental to understanding subsequential limits.
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If {x} is a sequence of rationals, I understand every real number will be a limit point. However, sequences have an order to them, right? So if this sequence of all rationals is monotonically increasing, then it will converge to infinite and all subsequences will have to converge to infinite. If it is monotonically decreasing, then it will converge to negative infinite and all subsequences will converge to negative infinite. How can the subsequence converge to a real number such as 1, when there exists a number x s.t. |x-1| > ε for any ε>0, such as when x = 1.1, since 1.1 is rational.

Can we ignore all values greater than one when constructing the subsequence? Then if I had a sequence from [0,-5) which converges to -5, can't I get subsequences with multiple subsequential limits, specifically every real number between [0,-5]. Then the upper limit of the sequence is -5, since this is convergence point of the sequence, but the supremum of subsequential limits is 0, since every real will be a subsequential limit, and 0 is the greatest number of the set.
 
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docholliday said:
If {x} is a sequence of rationals, I understand every real number will be a limit point.
Can't be right. What about the sequence 1 1 1 1 1...? You must mean something else.
 
I think the sequence must contain all rationals, as the say {(n+1)/(n+2)} does not have every real number. Since we have every rational we cannot be monotonic. We always have for N<n infinite |a_n-L|<epsilon, because there are infinite a_n and N is finite.
 
docholliday said:
So if this sequence of all rationals is monotonically increasing,

If the first term is x1, then how do you get x1-1 in the sequence?
 
I think you are misunderstanding. "If {x} is a sequence of rationals, every real number will be a limit point" implies that every sequence of rationals has every real number as limit which is not true. There are two true statements that might be meant:
1) The sequence of all rational numbers (since the set of all rationals is countable, they can be ordered into a single sequence, though not in the "usual" ordering) has every real number as a subsequential limit.
This is true because
2) Given any real number there exist a sequence of rational numbers which converges to it.
In fact, one way to define "real numbers" is "the set of equivalence classes of Cauchy sequences of rational numbers where two sequences {an} and {bn} are "equivalent" if and only if the sequence {an- bn} converges to 0. We then say that every sequence in the equivalence class defining a "converge to a".
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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