# Question on definition of continuity

The following is the definition for continuity at $$x_0$$

$$\forall \quad\varepsilon > 0\ \quad \exists \delta > 0\ \quad \text{s.t. } \quad |x - x_0| < \delta \implies |f(x) - f(x_0)| < \varepsilon$$

But suppose that that $$f(x_0)$$ is undefined, i.e. there is a "hole" at $$f(x_0)$$, the definition still holds for continuity since it does not care about what is happening at $$f(x_0)$$, only around it. Isn't this a problem?

• A function is not continuous at $x_0$ if it is not defined at $x_0.$ It seems like the definition makes that true - $|f(x)-f(x_0)|<\epsilon$ is not true if $f(x_0)$ is not defined. Nov 21, 2019 at 19:21
• Continuity at x_0 in the domain. We don't discuss continuity where it is not defined. Nov 21, 2019 at 19:22
• @ThomasAndrews You mean if its not defined at $f(x_0)$. What if we have an asymptote? Nov 21, 2019 at 19:30

Your definition is missing something. A function $$f: X\to Y$$ is continuous at $$x_0\in X$$ if $$\dots$$. That $$x_0$$ is in the domain of the function (i.e. that the function is defined at $$x_0$$) is part of the definition.
• Understood it when I thought of relations, i.e. $x$ has to have a pair $y$. Nov 21, 2019 at 19:50
$$\forall \ \varepsilon > 0\ \ \exists \delta > 0\ \text{s.t. } \ |x_1 - x_0| < \delta \ \land\ |x_2 - x_0| < \delta \implies |f(x_1) - f(x_2)| < \varepsilon$$
Yes, of course. The only thing that matters for continuity are those points at which the function is defined. What occurs outside the domain is irrelevant. If $$f(x_0)$$ is not defined, then we do not care about it.