Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Subsequence converging

  1. Apr 26, 2008 #1
    1. The problem statement, all variables and given/known data

    Consider the sequence [tex]\left\{ x_{n} \right\}[/tex].

    Then [tex] x_{n}[/tex] is convergent and [tex]\lim x_{n}=a[/tex] if and only if, for every non-trivial convergent subsequence, [tex]x_{n_{i}}[/tex], of [tex]x_{n}[/tex], [tex]\lim x_{n_{i}}=a[/tex].



    2. Relevant equations
    The definition of the limit of a series:
    [tex]\lim {x_{n}} = a \Leftrightarrow [/tex] for every [tex]\epsilon > 0[/tex], there exists [tex] N \in \mathbb{N}[/tex] such that for every [tex]n>N[/tex], [tex]\left| x_{n} - a \right| < \epsilon[/tex].



    3. The attempt at a solution

    Ok, so I easily see how to show that it [tex]\lim {x_{n}} = a [/tex], then every convergent subsequence must also converge to [tex]a[/tex].
    But I'm stuck on how to show the other way.
     
  2. jcsd
  3. Apr 26, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I would say, well isn't a_n a subsequence of itself? But you also said 'non-trivial'. I'm not sure exactly what that means, but can't you split a_n into two 'non-trivial' subsequences, which then converge, but when put together make all of a_n?
     
  4. Apr 26, 2008 #3
    Yeah, I asked my prof about this one. To him, apparently "non-trivial" just means that the subsequence is not equal to the original sequence. I don't think it actually has any bearing on the problem.

    The trick, he said, is that you have to consider every non-trivial (convergent) subsequence.

    I'm not sure I know what that means.
     
  5. Apr 26, 2008 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Ok, then suppose a_n has two convergent subsequences with different limits. Then does a_n have a limit?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook