May 24th, 2017, 06:52 PM  #1 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 236 Thanks: 2  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, 07:16 PM  #2  
Senior Member Joined: Sep 2015 From: USA Posts: 2,530 Thanks: 1390  Quote:
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, 07:54 PM  #3  
Senior Member Joined: Aug 2012 Posts: 2,356 Thanks: 738  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. Quote:
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.  
May 24th, 2017, 08:33 PM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 2,530 Thanks: 1390  Quote:
(A) and (B) seem fairly deep questions then.  
May 25th, 2017, 04:43 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 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 04:46 AM. 
May 25th, 2017, 01:22 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,927 Thanks: 2205  

Tags 
rational, sequence, terms 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Sequence, find its terms  sf7  Algebra  8  July 1st, 2014 01:39 AM 
How many terms for a recurrent, not linear sequence get to 0?  Singularity  Calculus  0  June 15th, 2014 02:24 PM 
Terms in a sequence (own)  proglote  Calculus  2  April 9th, 2012 09:56 AM 
Number of terms in a sequence of primes  scherz0  Number Theory  5  April 19th, 2011 01:00 PM 
Write the first five terms of each sequence...  reddd  Algebra  3  May 21st, 2010 02:08 PM 