Is a function with an undefined removable discontinuity considered continuous?

Question

I've heard a few times the definition of a continuous function simplified to "Being able to draw the graph of the function without picking our pencil".

A more rigorous definition states that a function is continuous on a domain $(a,b)$ if for all $c$ in the domain $\lim_{x\to c} f(x) = f(c)$.

My professor told us that a function is continuous if it continuous on it's domain, but I seem to doubt that statement because it doesn't respect the "picking pencil" rule.

If we remove a point $p$ on a continuous function where $D=\rm I\!R$ then its new domain is $(-\infty, p) \cup (p,\infty)$.

Therefore this function is continuous along it's domain, so is it continuous?

Answer 1

That "picking pencil" is not a "rule", it's just an intuition, and an incomplete one at that. A function defined on a domain with holes can be continuous. So you really need to amend your pencil thing to say "a function is continuous if, on each interval of its domain of definition, its graph can be drawn without picking up the pencil". And even then, this rule is incomplete, because if the domain is not open then the "interval" thing isn't very good.

You don't do math with just intuition and metaphors. You do math with precise definitions.

Answer 2

That pencil rule only gives an intuitive idea about the meaning of continuity in the case of functions whose domain is an interval, not in the general case. For instance, the function$$\begin{array}{rccc}\iota\colon&\mathbb{R}\setminus\{0\}&\longrightarrow&\mathbb R\\&x&\mapsto&\frac1x\end{array}$$is continuous although, of cource, the pencil rule does not apply.