Continuity of a function implies its existence in the neighbourhood?

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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
 

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  • #2
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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##
Why not?
 
  • #3
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Why not?
Suppose ##f(0.1)## is undefined. Then ##|f(0.1)-L|=|##undefined##-L|## and so it cannot be less than ##\epsilon##.
 
  • #4
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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.
 
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  • #5
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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.
Is

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

vacuously true?
 
  • #6
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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.
 
  • #7
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every function defined on a topological space that consists of a single point is continuous.
 
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  • #8
PeroK
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Suppose ##f(0.1)## is undefined. Then ##|f(0.1)-L|=|##undefined##-L|## and so it cannot be less than ##\epsilon##.
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.
 
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  • #9
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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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