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Equivalent continuity definition

  1. Oct 17, 2013 #1
    For every interval [ f(a)-e, (fa)+e ] there exists an interval [ f(a-d), f(a+d) ] such that [ f(a)-e, (fa)+e ] includes [ f(a-d), f(a+d) ]

    is this definition equivalent to the epsilon-delta definition?
  2. jcsd
  3. Oct 18, 2013 #2


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    Hi Jan,

    Yes, it is almost a literal translation. Try writing out the limit definition, then note that if ##\epsilon > 0##, "for all x such that ##|x - a| < \epsilon##" means "for all x in ##[a - \epsilon, a + \epsilon]##"; and ##x \in A \implies x \in B## means that "B is a subset of A", in other words, "A includes B".
  4. Oct 18, 2013 #3
    Not quite. Where you put [itex][f(a-\delta),f(a+\delta)][/itex], you would need to put [tex]f([a-\delta,a+\delta])[/tex] instead.

    Under your definition, any function [itex]f:\mathbb R\to \mathbb R[/itex] with the property that [itex]f(-23)=f(0)=f(23)[/itex] would be continuous at [itex]a=0[/itex] (by setting [itex]\delta=23[/itex]).
  5. Oct 18, 2013 #4
    Quite a point, but mightn't f([a-d, a+d]) also be a point equal to f(a) if f(a-d) = f(a) = f(a+d)? f(a-d) = f(a) = f(a+d) implies the existence of a d for every e, so that if we use this definition of continuity every symmetric function is continuous around its centre of symmetry a.

    The problem f([a-d, a+d]) certainly fixes is that it requires all the points in the interval to be part of the range.
  6. Oct 18, 2013 #5
    [itex]f([a-\delta,a+\delta])[/itex] is only a singleton if [itex]f[/itex] is constant on the whole interval [itex][a-\delta,a+\delta][/itex], a much stronger restriction than just [itex]f(a-\delta)=f(a)=f(a+\delta)[/itex].
  7. Oct 18, 2013 #6
    You're right, I made an error of interpretation.
  8. Oct 19, 2013 #7


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    The intervals must be open: [itex](a - \epsilon, a + \epsilon)[/itex].
  9. Oct 19, 2013 #8
    But the given definition with closed neighbourhoods is still equivalent.
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