# Why is continuity only defined on its domain?

As mentioned in this question students sometimes struggle with the fact that continuity is only defined at points of the function's domain. For example the function $f:\mathbb R\setminus\{0\} \to \mathbb R: x \mapsto \tfrac 1x$ is continuous although it has a "jump" at $x=0$ (cf. this answer with more details). So:

Why is continuity only defined on the function's domain? What's the benefit? How should a lecturer answer to such a question of a student?

My attempt to answer the question: I would give two arguments:

• When we take the sequence limit definition of continuity $\lim_{n\to\infty} f(x_n) = f\left(\lim_{n\to\infty} x_n\right) = f(x_0)$, then this definition makes only sense when $x_0 = \lim_{n\to\infty} x_n$ is in the domain of $f$.
• The concept students have in mind is "continuous continuation" and not "continuity". Thus, one have to distinguish between both concepts.

What do think about my answer? Have I missed something or are there other good arguments?

Note: This is another follow up question of How can I motivate the formal definition of continuity? I hope that's okay since I ask here for another aspect of continuity. I want to write an introductory article for continuity. That's the reason why I ask all these questions here...

• – quid Feb 20 '16 at 15:47
• I've noticed a fashion (that I don't understand, because I think it makes things much harder for students with little benefit) to define limits at limits points of the domain. However, the definition of continuity doesn't exist outside the domain, because it says 'the limit is equal to the actual value of the function'. – Jessica B Feb 20 '16 at 18:34
• @JessicaB How do you define the derivative without limits at limit points? The function $h \mapsto (f(x_0 + h) -f(x_0))/h$ does not have $0$ in its domain. – quid Feb 20 '16 at 18:49
• @quid, True, but it only requires one point to be missing from an interval, and students already have a reasonable concept of the boundary of an open interval. Also, usually there is time to get the idea of limits not caring about the value at the one point while learning about continuity, before starting differentiation. – Jessica B Feb 20 '16 at 21:03
• @Jessica B: Of course, the idea of defining limits at limit points of the domain (that do not belong to the domain) is a way of distinguishing between the limiting value of the function and the value of the function, but I suspect the MAIN reason is that it's such an important concept/tool in mathematics past elementary calculus and many people may overlook the fact that it's usefulness is not something beginning students will have any experience with (even in the calculus course, unless it's a highly theoretical course by U.S.A. standards). – Dave L Renfro Feb 22 '16 at 20:57

"Is the sine function continuous at Mount Everest?"

This or some modification thereof could be a playful start for a discussion about the fact, which you addressed more technically, that there must be some meaning to the function where it is considered. Put differently where else should continuity be considered?

One reason for the confusion may be that the students mainly think about functions defined on the real numbers with some exceptional points.

An actual example to discuss, beyond the playful start, could be the (real) logarithm function: "Is $\log$ continuous or discontinuous at $-27$?"

Discussing such examples could help to get the concept across that the function must have some agreed upon meaning at the point where continuity is considered.

The specific example of $1/x$ is a bit tricky as the question "is the reciprocal function continuous" does not have that clear-cut an answer.

For this specific function, and related ones, it may make sense to explain that the type of phenomenon one has for $1/x$ around $0$ has a particular name: pole (or more generally isolated singularity).

This may help them to categorize functions. Once $1/x$ becomes continuous on $\mathbb{R} \setminus \{0\}$, with a pole at $0$, it may be less confusing that it should be a continuous function "just like" $x^2 + 3$, since the one has a pole while the other does not.

• great answer! ;-) – Stephan Kulla Feb 20 '16 at 19:44

A somewhat analogous point of view is the one of "continuous extensions". What looks weird in saying $\frac 1 x$ is a continuous function is, of course, what is happening around $0$.

What is happening around $0$ can be summarized by saying:

there exists no continuous function on the whole $\mathbb R$ extending $\frac 1 x$

The whole idea of removable discontinuity is centered on this idea of extending a continuous function to another continuous function on a larger domain. This point of view has its strength in reconciling with the idea of something non continuous going on here...

Of course, in general, a continuous extension may not exist, may exist and not be unique and whatever else. From this point of view $\log$ can be made continuous at $-27$ (e.g. extending it via $\log|x|$), but not at $0$.

One can of course use this point of view to play around a little bit with subtle ideas like "maximal extensions" (where maximal is referred to the domain), gluing of continuous functions and the like.

• This formulation, no continuous extension, captures better than my own what I want to get at. – quid Feb 21 '16 at 16:45