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

1. Jun 16, 2016

### Happiness

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

2. Jun 16, 2016

### Staff: Mentor

Why not?

3. Jun 16, 2016

### Happiness

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

4. Jun 16, 2016

### Staff: 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.

5. Jun 16, 2016

### Happiness

Is

$|$undefined$-L|<\epsilon$

vacuously true?

6. Jun 16, 2016

### Staff: 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.

7. Jun 16, 2016

### wrobel

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

Last edited: Jun 16, 2016
8. Jun 16, 2016

### PeroK

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.

9. Jun 16, 2016

### Staff: 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.