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 May 24th, 2017, 07:52 PM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 242 Thanks: 4 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. Thank you  May 24th, 2017, 08:16 PM   #2
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 Originally Posted by Lalitha183 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, $\lim \limits_{n \to \infty} \dfrac 1 n = 0 \in \mathbb{Q}$ May 24th, 2017, 08:54 PM   #3
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 Originally Posted by romsek 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 Bolzanoâ€“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 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 09:00 PM. May 24th, 2017, 09:33 PM   #4
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 Originally Posted by Maschke 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 Bolzanoâ€“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. May 25th, 2017, 05:43 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,696 Thanks: 2681 Math Focus: Mainly analysis and algebra 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 05:46 AM. May 25th, 2017, 02:22 PM   #6
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Quote:
 Originally Posted by Maschke We don't have boundedness here.
Arbitrary distinct bounds can be chosen so as to define a bounded subsequence. Tags rational, sequence, terms Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post sf7 Algebra 8 July 1st, 2014 02:39 AM Singularity Calculus 0 June 15th, 2014 03:24 PM proglote Calculus 2 April 9th, 2012 10:56 AM scherz0 Number Theory 5 April 19th, 2011 02:00 PM reddd Algebra 3 May 21st, 2010 03:08 PM

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