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I understand deltas and epsilon proofs for the most part

  1. Mar 21, 2010 #1
    so 0 < l x-a l < delta and l f(x)-L l < epsilon

    What I don't understand is how come deltas and epsilons can't be greater than or equal to their respective differences?
  2. jcsd
  3. Mar 21, 2010 #2
    Convention? One less line to draw!
  4. Mar 21, 2010 #3


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    Because it is "less than" that is important- and if I can find [itex]\delta[/itex] so that [itex]0< |x- a|\le \delta[/itex] I could always choose [itex]\delta[/itex] just slightly larger and have [itex]0< |x- a|<\delta[/itex].
  5. Mar 21, 2010 #4
    Some writers (though not many) even follow the convention that < means "less than or equal to", and reserve [tex]\lneq[/tex] for strict inequality. More commonly, [tex]\subset,\subseteq[/tex] are used interchangeably for set inclusion, with proper inclusion indicated by [tex]\subsetneq[/tex].

    It's not a bad convention IMO, because, for instance, for a sequence converging to L, you can write [tex]\forall n,a_n<c\Rightarrow L<c[/tex], without worrying that limits don't preserve strict inequality.
  6. Mar 22, 2010 #5
    Dang why is that a good convention? Maybe that just looks weird to me. Anyways I'm perfectly content with just knowing that when I pass off to limits, then I need the non-strict inequality sign. These are rather nitpicky things that I cared way too much about when I started learning analysis. Also, I don't understand the point of getting a nice "less than epsilon" end to an argument, though I admit sometimes it's maybe worth the few extra minutes to finish with "< e" instead of "< e(some ugly factor)".
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