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As an extension to the title question:
If a sequence is bounded below (by say 0) but not above (infinity), does the Bolzano-Weierstrass theorem apply? (i.e. does it have a convergent subsequence?).

I think the answer is yes, since the subsequence can consist of those elements that tend towards zero but I'm not sure since I'm just getting familiar with the Bolzano-Weierstrass theorem.

Would like to kindly check my understanding. Thank you!

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Your title and questions do not seem to be the same –  Henry Mar 12 '12 at 8:13
    
@Henry: Apologies for this Henry. I've edited the question body to better reflect the title and the helpful answers provided by Marc and Aryabhata - it's really helped me a lot. –  xlm Mar 12 '12 at 11:38

2 Answers 2

up vote 1 down vote accepted

No, consider the sequence $x_n = n$.

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Ah, true. From what I gather does this depend on the specifics of the sequence? i.e. can't conclude converge/diverge subsequence from simply 1 bound alone. For example one that alternates between n and 1/n. Then a subsequence could just take all those that are of the form 1/n? –  xlm Mar 12 '12 at 8:02
    
@xlm: Yes, you need the specifics of the sequence. The example you gave does have a convergent subsequence. –  Aryabhata Mar 12 '12 at 8:04
    
Thanks so much! This really solidifies my understanding! –  xlm Mar 12 '12 at 8:06
    
@xlm: You might want to realize that there are sequences which are neither bounded above nor below and still have a convergent subsequence. Bolzano-Weierstrass is about forcing the existence of a convergent subsequence, but there is little that can prevent a convergent subsequence from existing "voluntarily". –  Marc van Leeuwen Mar 12 '12 at 8:18
    
@MarcvanLeeuwen: Very interesting point, thank you for adding this. Is there an example of such a sequence or link that I could examine to further my understanding? Thanks again! –  xlm Mar 12 '12 at 11:32

If your question is "If a sequence is bounded below but not above, does it necessarily have a convergent subsequence?", then the answer is "no", as was indicated. However if your question is your title "Does a convergent subsequence require being bounded below and above?" then the answer is "yes" in either of the following two interpretations: (1) being convergent does require being bounded below and above (only finitely many terms can be at distance more than $1$ from the point of convergence), and also (2) you need the hypothesis of being bounded both above and below to ensure the existence of a convergent subsequence (as the given answers show). You probably meant question (2), but it would be good to choose your title so as to correspond to your actual question (and also in such a way that the answer is not the opposite to that of your actual question).

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Apologies for the confusing title. I didn't realise how different those questions really were from each other. Nonetheless your kind insight and dissection helps me grasp this topic better. Will be more diligent with titles in the future. Thank you! –  xlm Mar 12 '12 at 11:42

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