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

• I
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

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

mfb
Mentor
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?

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

mfb
Mentor
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.

Happiness
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?

mfb
Mentor
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.

wrobel
every function defined on a topological space that consists of a single point is continuous.

Last edited:
PeroK
Homework Helper
Gold Member
2020 Award
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.

Happiness
mfb
Mentor
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.