A function is continuous iff it is continuous at all points in its domain. What confuses me about this definition is that for undefined points in a function, those points are removed from the domain.

Does this mean a function $f(x) = x$ with domain $(-\infty, -10) \cup (3, \infty)$ is continuous even though there's a big blank space between the two pieces?

Does this mean $f(x) = \frac{x-3}{x-3}$ is continuous even though it has a removable singularity at $x=3$?

Would a single point be continuous even though it lacks limits on either side?

Does this mean $f(x) = \tan(x)$ and $f(x) = 1/x$ are continuous even though they have asymptotes where the function goes to infinity?

Does this mean the only functions that are discontinuous are ones with jump discontinuities?

  • $\begingroup$ Your first one is continuous. Your second one is just f(x)=1 which has no singularity. If you are referring to a normal rational function, then this function is continuous for all points except at the singularity. $\endgroup$ – Harry Alli Feb 26 '18 at 0:22
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    $\begingroup$ @HarryAlli The second one is $f(x)=1$ for all $x \neq 3$. At $x=3$ it isn't defined since the denominator is $0$. $\endgroup$ – user525966 Feb 26 '18 at 0:23

It is because we get a little bit loosey-goosey with the concept of domain.

We had too many questions in Algebra class / pre-calculs class were we were asked what is the domain of $f(x).$ i.e. the domain of $f(x) = \frac 1x$ is $\mathbb R - \{0\}.$ But, it is really the responsibility of whoever it is that defines the function to also define the domain. It could be that there are points in the domain where the function is not defined. When we get to these continuity questions, we are often being asked to include points in the domain where the function may, in fact, be undefined.

$f(x) = \frac 1{x}$ is continuous on $(-\infty,0)\cup (0,\infty),$ but it is not continuous on other domains such as $[-1,1]$ or $\mathbb R.$

  • $\begingroup$ Does it even make sense to use a different domain though, since this can include points that aren't even defined? $\endgroup$ – user525966 Feb 26 '18 at 0:30
  • $\begingroup$ If I define my function $f:\mathbb R \to \mathbb R$ then the domain is $\mathbb R$ even if there are points where the function is infact undefined. We say a function is a mapping from the domain to the co-domain where every input has "at most one" output. There could certainly be no output for a given input and still be a valid function. $\endgroup$ – Doug M Feb 26 '18 at 0:33
  • $\begingroup$ So if we don't explicitly give the domain then we generally assume the domain is wherever it's defined -- and that it's probably going to be a continuous function granted no jump or essential discontinuities, but if we explicitly make the domain include undefined points, then it won't be continuous anymore for sure? $\endgroup$ – user525966 Feb 26 '18 at 0:34
  • $\begingroup$ That is exactly how I think about it. $\endgroup$ – Doug M Feb 26 '18 at 0:35
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    $\begingroup$ “It could be that there are points in the domain where the function is not defined.” No, the domain of a function is (by definition!) the set of points where it's defined. It doesn't make sense to talk about “$f(x)=1/x$ on the domain $[-1,1]$”. $\endgroup$ – Hans Lundmark Feb 26 '18 at 6:04

Terence Tao's book Analysis II makes the following definition (p. 420):

Let $(X,d_X)$ be a metric space, and let $(Y,d_y)$ be another metric space, and let $f:X \to Y$ be a function. If $x_0 \in X$, we say that $f$ is continuous at $x_0$ iff for every $\epsilon > 0$, there exists a $\delta > 0$ such that $d_Y(f(x),f(x_0)) < \epsilon$ whenever $d_X(x,x_0) < \delta$. We say that $f$ is continuous iff it is continuous at every point $x \in X$.

Folland gives a definition similar to Tao's.

So, using the definition in Tao or Folland, it is correct to say that the functions you mentioned are continuous. For example, using Tao's definition, it is correct to state that the function $f(x) = \tan(x)$ is continuous. (But the Wikipedia article quoted below warns us that not everyone uses Tao's terminology, so we should be careful that the meaning of this statement is clear in context.)

The answer to your final question is no. Picture the function $f$ defined by $f(x) = \sin(1/x)$ if $x \neq 0$ and $f(0) = 0$.

By the way, we should note that the Wikipedia says the following:

There are several different definitions of continuity of a function. Sometimes a function is said to be continuous if it is continuous at every point in its domain. In this case, the function $f(x) = \tan(x)$, with the domain of all real $x \neq (2n+1)\pi/2$, $n$ any integer, is continuous. Sometimes an exception is made for boundaries of the domain. For example, the graph of the function $f(x) = \sqrt{x}$, with the domain of all non-negative reals, has a left-hand endpoint. In this case only the limit from the right is required to equal the value of the function. Under this definition $f$ is continuous at the boundary $x= 0$ and so for all non-negative arguments. The most common and restrictive definition is that a function is continuous if it is continuous at all real numbers. In this case, the previous two examples are not continuous, but every polynomial function is continuous, as are the sine, cosine, and exponential functions. Care should be exercised in using the word continuous, so that it is clear from the context which meaning of the word is intended.

  • $\begingroup$ Those are essential discontinuities because they oscillate forever, and that's fine, I did forget to mention those, but I'm mostly interested in the other categories, and more broadly, when a function is considered discontinuous $\endgroup$ – user525966 Feb 26 '18 at 0:32
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    $\begingroup$ “The most common and restrictive definition is that a function is continuous if it is continuous at all real numbers.” Really?!? I don't think I've ever seen that. The standard definition is “a function is continuous iff it is continuous at every point in its domain”, so you should really trust Tao more than Wikipedia here. $\endgroup$ – Hans Lundmark Feb 26 '18 at 6:08
  • $\begingroup$ @HansLundmark Thanks, I edited the answer to emphasize the Tao / Folland definition and deemphasize the Wikipedia passage. $\endgroup$ – littleO Feb 26 '18 at 7:13

Usually continuity is defined for those points at which the function is defined and what occurs outside the domain is not relevant.

Anyway note that in some context a function is defined to be continuos if its domain is an interval, and it is continuous at every point of that interval.

Here you can find a good reference from MIT for a full classification Continuity and Discontinuity.

With respect to your examples, following the definition proposed in the reference

  • $f(x) = x$ with domain $(-\infty, -10) \cup (3, \infty)$ is continuous on each interval but it is not a continuos function
  • $f(x) = \frac{x-3}{x-3}$ has a removable discontinuity at $x=3$ which can be eliminated by definining $f(3)=1$
  • a single point $(x_0,y_0)$ is a continuous function since $f(x_0)=y_0$
  • $f(x) = \tan(x)$ and $f(x) = 1/x$ are continuous in the intervals of definition and have infinity dicontinuities
  • Does this mean the only functions that are discontinuous are ones with jump discontinuities? No that's only a special kind of discontinuity
  • $\begingroup$ This one says $\tan(x)$ is not continuous for example? $\endgroup$ – user525966 Feb 26 '18 at 0:29
  • $\begingroup$ @user525966 yesexactly, in the sense that it is not continuos in a whole interval but it is continuos in $(-\pi/2,\pi/2)$ for example. $\endgroup$ – user Feb 26 '18 at 0:35
  • $\begingroup$ I would say the definitions in that document are most nonstandard. (Requiring the domain to be an interval? What? And requiring points of discontinuity to be isolated? What?!? Surely everyone would say that every point is a point of discontinuity for the Dirichlet function $\mathbf{1}_{\mathbb{Q}}$?) $\endgroup$ – Hans Lundmark Feb 26 '18 at 6:17
  • $\begingroup$ And example 7 (“if $f(x)=|x|$ then $f'$ has a jump discontinuity at the origin”) is misleading. Since $f'$ isn't even defined at the origin, one shouldn't even talk about continuity or discontinuity there to begin with (although I confess that I might have committed this sin myself from time to time). Doing so, one contradicts the Corollary to Theorem 5.12 in Rudin's Principles (a derivative can't have jump discontinuites). $\endgroup$ – Hans Lundmark Feb 26 '18 at 6:24
  • $\begingroup$ @HansLundmark Always there are a lot of discussion on this kind of topic, I really think that it is mainly a fact of definitions which depends upon the particular context. Thus it is important to keep in mind that we need to be aware about this as we can have different definitions and related interpretation. $\endgroup$ – user Feb 26 '18 at 8:27

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