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