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May 24th, 2017, 06:52 PM   #1
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Rational terms sequence

Let $(a_n)$ be a sequence where all rational numbers are terms (and all terms are rational). Then
A) no sub sequence of $(a_n)$ converges.
B) there are uncountably many convergent sub sequences of $(a_n)$.
C) Every limit point of $(a_n)$ is a rational number.
D) no limit point of $(a_n)$ is a rational number.

This question can have multiple answers.
My answers are Option B&D.

Please check and let me know.
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May 24th, 2017, 07:16 PM   #2
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Originally Posted by Lalitha183 View Post
Let $(a_n)$ be a sequence where all rational numbers are terms (and all terms are rational). Then
A) no sub sequence of $(a_n)$ converges.
B) there are uncountably many convergent sub sequences of $(a_n)$.
C) Every limit point of $(a_n)$ is a rational number.
D) no limit point of $(a_n)$ is a rational number.

This question can have multiple answers.
My answers are Option B&D.

Please check and let me know.
Thank you
A) is clearly false

B) is clearly true, consider $m + \dfrac 1 n,~\forall m \in \mathbb{Z}$

I believe (C) is false. The rationals are dense in the reals so for any real number you can construct a sequence of rationals that will converge to it.

D) is clearly false, $\lim \limits_{n \to \infty} \dfrac 1 n = 0 \in \mathbb{Q}$
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May 24th, 2017, 07:54 PM   #3
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A) is clearly false
If I'm understanding the question, the sequence includes all the rationals. That is, it's an enumeration of the rationals. And those are very wild, aren't they?

I am not saying this isn't false ... but I don't see that it's clearly false. I'm not sure I see the proof. Maybe I'm missing something obvious but I'm wondering why some enumeration couldn't jump around so much that no subsequence converges.

The Bolzano–Weierstrass theorem says that every bounded sequence has a convergent subsequence. We don't have boundedness here.

I can see that the jumping around probably can't prevent some sequence from converging. But I think I'd have to work at a proof.

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Originally Posted by romsek View Post
B) is clearly true, consider $m + \dfrac 1 n,~\forall m \in \mathbb{Z}$
If I can't convince myself any subsequence converges, I certainly can't see uncountably many converging. And I don't understand your argument, you only expressed a countable set.

I agree with you about C and D.

I must be missing something obvious.

Last edited by Maschke; May 24th, 2017 at 08:00 PM.
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May 24th, 2017, 08:33 PM   #4
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Originally Posted by Maschke View Post
If I'm understanding the question, the sequence includes all the rationals. That is, it's an enumeration of the rationals. And those are very wild, aren't they?

I am not saying this isn't false ... but I don't see that it's clearly false. I'm not sure I see the proof. Maybe I'm missing something obvious but I'm wondering why some enumeration couldn't jump around so much that no subsequence converges.

The Bolzano–Weierstrass theorem says that every bounded sequence has a convergent subsequence. We don't have boundedness here.

I can see that the jumping around probably can't prevent some sequence from converging. But I think I'd have to work at a proof.



If I can't convince myself any subsequence converges, I certainly can't see uncountably many converging. And I don't understand your argument, you only expressed a countable set.

I agree with you about C and D.

I must be missing something obvious.
I see, it's my error. I didn't take into account that the order of the original sequence must be maintained.

(A) and (B) seem fairly deep questions then.
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May 25th, 2017, 04:43 AM   #5
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C and D talk about limit points of the sequence of all rationals. I'm reasonably sure that such a sequence doesn't converge and that $\limsup a_n$ and $\limsup a_n$ are $\pm \infty$. This is because we must include all arbitrarily large rationals in the sequence. The arbitrarily small ones are also there, of course, but that would just make the sequence oscillate.

The fact that we need all arbitrarily small rationals leads me to suggest that there is guaranteed to be a subsequence that converges to zero. I also suspect that the set of such sequences must be uncountable.

Of course all of this is intuition rather than proof.

Last edited by v8archie; May 25th, 2017 at 04:46 AM.
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May 25th, 2017, 01:22 PM   #6
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We don't have boundedness here.
Arbitrary distinct bounds can be chosen so as to define a bounded subsequence.
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