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I Epsilon in the limit definition

  1. Jan 13, 2017 #1
    in the limit definition of a sequence, why does epsilon has to be greater than 0 and not greater or equal to 0?

    thanks in advance.
     
  2. jcsd
  3. Jan 13, 2017 #2

    fresh_42

    Staff: Mentor

    It doesn't really matter with sequences of real numbers. You could take both as you can always find another epsilon that is slightly smaller. In general, however, one speaks of open neighborhoods around the limit point as they are the defining element of general (topological) spaces. And open translates to smaller than, whereas smaller or equal includes the boundaries, and as such are closed sets. So the restriction to smaller than is somehow simply consequent, even if not needed (and it's available on the keyboard).
     
  4. Jan 13, 2017 #3

    pwsnafu

    User Avatar
    Science Advisor

    Choose ##\epsilon = 0##, then in order for ##a_n \to L## we need to find a ##N## such that ##a_n = L## for all ##n>N##.
    So under this definition the only sequences that converge are those that are eventually constant.
     
  5. Jan 13, 2017 #4

    fresh_42

    Staff: Mentor

    Yes, you're right. I confused it with the condition ##\,\vert \,a_n -L\,\vert \, < \varepsilon## where you could take ##\leq## instead.
    Of course ##\varepsilon = 0## would make no sense as there would be only constant sequences left over.
     
  6. Jan 13, 2017 #5
    thank to both of you!
     
  7. Jan 13, 2017 #6

    Mark44

    Staff: Mentor

    Consider ##a_n = \frac {n - 1} n, n \ge 1##. It's easy to show that ##\lim_{n \to \infty}a_n = 1##. However, if ##\epsilon = 0##, it's not possible to find a specific number N for which ##|a_n - 1| = 0##, for all ##n \ge N##.

    With ##\epsilon > 0##, all that has to happen is to force the terms in the tail of the sequence arbitrarily close to L, not necessarily making them exactly equal to it.
     
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