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I Continuity of a function implies its existence in the neighbourhood?

  1. Jun 16, 2016 #1
    Suppose ##f(x)## is continuous at ##x=c##. Does this imply that ##f(x)## exists in an open neighbourhood of ##c##?

    I believe it does. If ##f(x)## is continuous then ##\lim_{x\to c}f(x)## exists. But if ##f(x)## is undefined for some values of ##x## in the ##\delta##-neighbourhood of ##c##, then we cannot say ##|f(x) - L|<\epsilon## for ##0<|x-c|<\delta##, contradicting that ##\lim_{x\to c}f(x)## exists. So there must exist a ##\delta>0## such that ##f(x)## is defined for all values of ##x## in the ##\delta##-neighbourhood.

    https://en.wikipedia.org/wiki/(ε,_δ)-definition_of_limit
    Screen Shot 2016-06-16 at 8.42.34 pm.png

    If so, does this mean that it is enough to just say "suppose ##f(x)## is continuous in an open subset ##U## of ##\mathbb{R}##" instead of "suppose ##f(x)## exists and is continuous in an open subset ##U## of ##\mathbb{R}##"? (which is what is stated below)

    http://calculus.subwiki.org/wiki/Clairaut's_theorem_on_equality_of_mixed_partials
    Screen Shot 2016-06-16 at 8.54.29 pm.png
     
  2. jcsd
  3. Jun 16, 2016 #2

    mfb

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    Why not?
     
  4. Jun 16, 2016 #3
    Suppose ##f(0.1)## is undefined. Then ##|f(0.1)-L|=|##undefined##-L|## and so it cannot be less than ##\epsilon##.
     
  5. Jun 16, 2016 #4

    mfb

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    But you never want to consider that expression in the first place, because you evaluate eps/delta only for x-values where the function is defined.
     
  6. Jun 16, 2016 #5
    Is

    ##|##undefined##-L|<\epsilon##

    vacuously true?
     
  7. Jun 16, 2016 #6

    mfb

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    No, the problem does not even come up. The equation has to be satisfied only for all x in the domain of the function which are closer than delta to the given point.
     
  8. Jun 16, 2016 #7
    every function defined on a topological space that consists of a single point is continuous.
     
    Last edited: Jun 16, 2016
  9. Jun 16, 2016 #8

    PeroK

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    The definition of continuity includes the condition that ##x## be in the domain of the function. If you were to take a function defined on ##\mathbb{Q}## such that ##f(x) = 1## for all rationals, then that function (defined on the rationals) is continuous. You cannot consider hypothetical function values for points not in the domain.

    Another good example is that ##f(x) = 1/x## is continuous on ##\mathbb{R}-\{0\}##. This is interesting as you will find references to the "discontinuity" of ##1/x## at ##x=0##. What that means technically is that you cannot extend ##1/x## to be a continuous function on ##\mathbb{R}##. But, technically, ##1/x## is a continuous function: there is no point (in its domain) at which it is discontinuous.
     
  10. Jun 16, 2016 #9

    mfb

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    As another example, if you consider f(x) as function over the real numbers and study the continuity at x=0, you don't even think about plugging in imaginary numbers for x. Why? Because those imaginary numbers are not part of the domain of the function.
     
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