# Is this an infinite number of discontinuities?

1. Aug 22, 2014

### Hertz

So consider a function $f(x)$ which is continuous for all $x$ except on some finite interval, say $[a, b]$. Imagine, for example, a function which goes to $-\infty$ from the left at $x=a$, is undefined from a to b, and then "comes from" infinity at $x=b$ and is defined and continuous everywhere else. Is this considered an infinite number of discontinuities, or a single discontinuity? My first thought is that it would be considered an infinite number of discontinuities, but I want to be sure

2. Aug 22, 2014

### micromass

Staff Emeritus
A discontinuity only really makes sense on the domain of the function. Since points in $[a,b]$ are not in the domain (the function is not defined there), those points cannot be discontinuities.
It follows that the function is continuous on its domain.

3. Aug 22, 2014

### gopher_p

Edit: Drats, sniped!

Last edited: Aug 22, 2014
4. Aug 22, 2014

### disregardthat

If what you want is a real function which is nowhere continuous only on a finite interval, consider the function $f : \mathbb{R} \to \mathbb{R}$ given by f(x) = 0 on $(-\infty,0)$, f(x) = x if x is rational, 0 if x is irrational on [0,1], and f(x) = 0 on $(1,\infty)$. This would amount to an infinite number of discontinuities.

5. Aug 22, 2014

### Hertz

Ok I see, thank you :)

See, my line of thought was that you could use the definition of continuity to argue that the function is "not continuous" at every point in the interval. I guess 'not continuous' and 'discontinuous' have two different meanings? You cannot speak of discontinuity at a point where the function isn't defined? Ok :)

6. Aug 23, 2014

### disregardthat

not continuous and discontinuous mean the exact same thing. But it doesn't make sense to speak of a function at a place where it isn't defined. That's like asking whether a real function is continuous or discontinuous at the imaginary number i.

7. Aug 23, 2014

### Florentino

As people said, if your function is not defined on the interval [a,b] it can't be discontinuous on the whole interval, it can however be discontinuous on points that are not in the function's domain if they are limit points (every open ball centered around these points contains one point of the domain distinct from itself).

A function that is discontinuous on [a,b], and in this case it should be defined at least on a subset that is dense on [a,b], like (a,b) or $(a,b) \cap \mathbb Q$, will have infinite points of discontinuity, more specifically uncontably infinite.

I assume the domain of your function is $\mathbb R$. So I'll consider consider a function $f:\mathbb R \rightarrow \mathbb R$.

I believe a function like you intended to describe would be something like that:

$f(x)= \frac{1}{x-a} \ \text{if} \ x < a$, $f(x)= 0 \ \text{if} \ x \in \mathbb Q \cap [a,b], \ f(x)=1 \ \text{if} \ x \in (\mathbb R - \mathbb Q) \cap [a,b]$ and $f(x)= \frac{1}{b-x} \ \text{if} \ x>b$.

8. Aug 23, 2014

### HallsofIvy

Staff Emeritus
Or even asking whether a function is continuous or discontinuous at "green"!

9. Aug 23, 2014

### Hertz

If someone were to ask me whether a real function is continuous at i, my response (prior to this topic) would be "no, considering that the function does not fulfill the criteria of continuity at that point, I can conclude that the function is 'not continuous' at that point." I understand what you are saying now, but are my semantics entirely unreasonable? I've spoken of functions where they are not defined many times in my life, such as discussing their limits at these points, and the conversations still 'made sense'. Either way, thanks for the replies . I understand now that continuity only exists where the function is defined. Prior to this I would have considered a vertical asymptote on a real function to be a point of discontinuity, this is where my confusion came from.

10. Aug 23, 2014

### disregardthat

What is the definition of continuity?

A function $f : A \to B$ (with A , B $\subseteq \mathbb{R}$) is continuous at $a \in A$ if for all $\epsilon > 0$, there exists a $\delta > 0$ such that $|f(x)-f(a)| < \epsilon$ for all x such that $|x-a|<\delta$.

So, suppose $a \in \mathbb{R}-A$. What does "f is continuous at a" mean?

Well, actually, continuity is not defined at a, because a function can only be continuous at a point in its domain by the very definition of continuity. So it's quite unreasonable to say that ( not (f is continuous at a) ), since (f is continuous at a) is non-sensical.

11. Aug 23, 2014

### Hertz

I was taught that for a function to be continuous at a point, it's limit at that point must equal the actual function value at that point. Hence, I would conclude that the function $f(x)=x^{-2}$ is not continuous at the point $x=0$ because clearly the 'function value' at that point cannot equal the limit at that point considering that the function is not defined there. I would also conclude that the function is not continuous at GREEN because neither the limit nor the function exist there. Do you understand how my misuse of terminology is not entirely "unreasonable" although it may be incorrect? Or is my English also in need of harsh criticism? Am I misunderstanding the meaning of "reason"?

It's also worth noting that while continuity in this sense is a mathematical term, it carries along with it an intuitive meaning. You can tell by visual inspection whether or not something is continuous because the mathematical definition of continuity is just an formal version of a conceptual idea that we all understood as a first step. Before continuity was given a formal definition, I'm sure there were plenty of mathematicians who considered asymptotes and other such points on a function as "points where the function is not continuous" even though the function itself is not defined there.

12. Aug 23, 2014

### pwsnafu

This is how I teach my students. You can still ask that question, just like you can ask "what is the maximum of the set $(-1,1)$?". And I've had assignments/exams where the question asked whether the function was continuous at a point even though it wasn't defined there. The answer is not a simple yes/no. "The question is invalid because..."

It's not what people expect when you use the term "discontinuity" without quantification. The presumption is that the function is indeed defined there, but the limits don't equal the value. The term "singularity" is more often used for this situation. However, it is a fairly common misuse of terminology that $\frac{\sin x}{x}$ has a removable discontinuity at x=0. "Removable singularity" is the correct term.

None the less "continuous at a point" is different from "continuous function". $f(x)=x^{-2}$ is not continuous at x=2 (under this interpretation), but it is a continuous function.

If you are interested before the definition was finalized mathematicians had assumed that "continuous equals differentiable". And yes they were aware that this was problematic: there are plenty of letters discussing the plucking of guitar stings.

Last edited: Aug 23, 2014