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Uniform continuity

  1. Jan 28, 2009 #1
    Hi,
    This may sound lame but I am not able to get the definition of uniform continuous functions past my head.

    by definition:
    A function f with domain D is called uniformly continuous on the domain D if for any eta > 0 there exists a delta > 0 such that: if s, t D and | s - t | < delta then | f(s) - f(t) | < eta. Click here for a graphical explanation.

    I can just choose "delta" that is a large number that will make any 2 points on the curve satisfy this condition. 1/x would be uniform continuous if I simply choose a large enough delta.

    moreover, what is the utility and application of uniform continuous fynctions?

    thanks,
    Sam
     
  2. jcsd
  3. Jan 28, 2009 #2

    quasar987

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    I think what you're missing is that in the definition, the part that says

    "if s, t D and | s - t | < delta then | f(s) - f(t) | < eta"

    actually means

    "if for all s,t in D such that |s-t|<delta, we have | f(s) - f(t) | < eta."

    So it does not suffice that you can find two points of D a distance less than delta apart that satisfy | f(s) - f(t) | < eta, but rather, the definition is saying that all the points in D that are a distance less than delta apart must satisfy | f(s) - f(t) | < eta !
     
  4. Jan 28, 2009 #3

    daniel_i_l

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    For starters, in order to prove that the Darboux integral is defined for any continuous function on a closed interval, Cantors theorem - which states that a continuous function on a closed interval is uniformly continuous - is used and then the uniform continuity is used to prove that the integral is defined.
     
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