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What is the intuitive definition of a subsequence?

  1. Jul 9, 2009 #1
    In Rudin's "Principles of Mathematical Analysis" he gives the strict definition for a subsequence as follows:

    Given a subsequence {pn}, consider a sequence {nk}of positive integers, such that n1<n2<n3<... Then the sequence {pni} is called a subsequence of {pn}. If {pni} converges, its limit is called a subsequential limit of {pn}.

    Just underneath this definition, he says:

    It is clear that {pn} converges to p if and only if every subsequence of {pn} converges to p.

    From this definition alone, the answer is clear. Given any numerical sequence, basically if i start at any nth term and proceed infinitely (assuming that is what Rudin meant by "...") then I have a subsequence. But there are other definitions later in the book which recall this one and make it seem like a subsequence is any sort of segment of a sequence which can have a finite end and beginning, for instance, the following theorem which comes only one page after the previous definition:

    The subsequential limits of a sequence {pn} in a metric space X form a closed subset of X.

    To me, this implies that there are (or at least, could be) multiple subsequential limits. However, the previous definition and the remark just after imply there is only one. Clearly I am misunderstanding something here. Does anyone have the insight I seem to be lacking? Thanks in advance.
  2. jcsd
  3. Jul 9, 2009 #2
    That's correct. A sequence can have multiple subsequential limits if it doesn't converge. The previous remark only concerned convergent sequences. Remember, you can choose any members of the sequence you want to be part of the subsequence, as long as they appear in the same order as in the original sequence. For example, you can have a subsequence consisting of every fifth term in the original sequence. Any method of selection is fine, as long as you don't mess with the order.
  4. Jul 9, 2009 #3


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    Hi Newtime! :smile:
    If by "proceed infinitely" you mean n, n+1, n+2, … then no.

    If the original sequence is 1 0 1 0 1 0 1 0 …,

    then the subsequence 1 1 1 1 1 … converges to 1 (obviously! :rolleyes:) but the subsequence 0 0 0 0 0 … converges to 0. :wink:
  5. Jul 9, 2009 #4

    Thank you both for the replies, it does help clear up some confusion already. However, one thing about Rudin's remark after the definition troubles me: if the sequence {pn} converges to p if and onyl if every subsequential limit is also p, then how can a convergent sequence, like say {10+[tex]\frac{1}{n}[/tex]} n[tex]\in[/tex]Z (it might just be my computer messing up but the tex doesn't come through, the sequence is {10+1/n} for any integer n) converge to 10 as n approaches infinity if I can choose an arbitrary, finite subsequence, say {11, 10.5, 10.25} which obviously does not converge to 10. I think this is the link in thinking I'm missing, because everything else you both said makes sense and that's what I had thought but that one remark after that definition has got be confused still.
  6. Jul 9, 2009 #5
    Forgot to mention that the subsequence needs to contain infinitely many terms from the original sequence. You can't have a finite subsequence.
  7. Jul 9, 2009 #6
    Ahh ok this helps with that previous example I just gave then. That makes much more sense. So in the theorem I cited in my original post about every subsequential limit forming a closed subset, that wasn't talking about a convergent sequence necessarily, just a sequence, but if the sequence was convergent then the set of all subsequential limits would contain one point?
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