Finding a rational sequence with no subsequence with rational limit

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Implication
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#1
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So I'm given a bunch of sets, and told to find an example of each of the below for said set. The set I'm considering at the moment is the rationals, and I must find:
(i) a convergent sequence in the set whose limit is not in the set;
(ii) a sequence in the set having no subsequence with limit in the set.

The first was pretty simple; I wanted a rational sequence with an irrational limit so I just took the sequence defined iteratively by
x_{n+1}=x_{n}-\dfrac{x_{n} ^{2}-2}{2x_{n}}, \ x_{0}=1
Clearly the sequence is rational but its limit is root 2 - which is irrational - by the Newton-Raphson method.

In the case of the second, I think I wanted a rational sequence with every subsequence having an irrational limit or no limit at all. So I figure I can just take an unbounded, divergent sequence e.g. 2^n. And my question is just whether or not this thinking is correct?

The examples I've been given in the solutions are the same for (i) and (ii); they've just used 3, 3.1, 3.14, 3.141, 3.1415,.... It's rational and converges to pi, and every subsequence converges to pi (I think?).
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Implication
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Okay so I've just spotted that at the bottom of the question it states,

of course, if you find an answer to (i) then this is automatically an answer to (ii) also
Why is this true? Why does a convergent sequence in a set with limit not in said set necessarily have no subsequences with limits in the set?


edit: so i'm guessing all subsequences of a convergent sequence converge to the limit of the original sequence? if so, why?
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Mark13
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(Original post by Implication)
Okay so I've just spotted that at the bottom of the question it states,



Why is this true? Why does a convergent sequence in a set with limit not in said set necessarily have no subsequences with limits in the set?


edit: so i'm guessing all subsequences of a convergent sequence converge to the limit of the original sequence? if so, why?
It follows very quickly from the definition of a limit, and it would be a good exercise to try and prove it.
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Implication
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(Original post by Mark13)
It follows very quickly from the definition of a limit, and it would be a good exercise to try and prove it.
I've got it common-sense wise now, kind of hard to explain since it seems so obviously true now I've understood it. Not sure how I'd go about proving it though, is it trivial?
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james22
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The proof that if a sequence has a limit, L, then every subsequence converges to L follows almost immediately from the definition of the limit.
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Implication
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Got it

ty gentlemen

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already repped you recently james
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matt2k8
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The main key to the proof of subsequences going to the same limit is that if n_k is a strictly increasing sequence of natural numbers, we always have n_k \geq k
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