# B Continuous Function

1. Aug 31, 2016

### Gurasees

What do we mean when we say a function is continuous on its domain? How is that different from simply saying that a function is continuous?

2. Aug 31, 2016

### Staff: Mentor

A function is always associated with its domain, so if someone says a function is continuous, the implication is that it is continuous at each point in its domain.

3. Aug 31, 2016

### Math_QED

Yes, but a continuous function f: [a,b] ⊂ R→ R is in the following way defined:

f is continuous in (a,b)
f is right continuous in a
f is left continuous in b

So, we cannot truly say that a continuous function is continuous in every point of its domain, as this function is not continuous in a and b.

4. Aug 31, 2016

### Staff: Mentor

I might be wrong, but when the domain of a function is a closed interval (such as f(x) = $sqrt{x}$), we can say that f is continuous on its domain, and it is understood that we mean the continuity is one-sided at 0.

5. Aug 31, 2016

### Math_QED

I agree, but you said that a continuous function is continuous on every point of its domain and obviously, the function
$f: [0,+∞) → ℝ: x → √{x}$ is not continuous at x = 0, although it is a continuous function (see the definition above). It's one-sided continous in x = 0 as you stated yourself. Of course, these are details, and I'm sure you are aware of this fact, but I just wanted to point that out.

6. Aug 31, 2016

### Staff: Mentor

I understand, but I don't think it's necessary to qualify the term "continuous" at an endpoint of the domain.

From Wikpedia (https://en.wikipedia.org/wiki/Continuous_function), FWIW (emphasis added):
At an endpoint of the domain, the approach to c necessarily can only be one-sided.

7. Aug 31, 2016

### Math_QED

Then we are in agreement.

8. Aug 31, 2016

### Ssnow

When I say that a function is continuous, it is understood that it is continuous in its '' natural '' domain. For example If I say that $f(x)=\frac{1}{x}$ is continuous, it is understood that its natural domain is $\mathbb{R}\setminus \{0\}$ ...

9. Aug 31, 2016

### PeroK

That's not right at all. When considering the domain as $[0, +\infty)$ the epsilon-delta definition of continuity holds completely as there are no points less than $0$ under consideration. Right and left sided limits don't come into it.

10. Aug 31, 2016

### Math_QED

I thought that a function is continuous in a point x = a if lim x-> a f(x) = f(a). I haven't seen the epsilon-delta definition yet, but for the function f(x) = √x: lim x-> 0 f(x) does not exist as we can't approach f(x) from the left? Or did I miss something obvious?

EDIT: I read this: http://math.stackexchange.com/questions/637280/limit-of-sqrt-x-as-x-approaches-0 . This cleared a lot up for me. Thanks for pointing that out @PeroK.

Last edited: Aug 31, 2016
11. Aug 31, 2016

### jbriggs444

Consider f(x) = x defined for integer x. Is f continuous?

Under the epsilon-delta definition, this function is continuous at every point in its domain.

12. Sep 1, 2016

### PeroK

More generally, when you consider a function defined on a more general set, you no longer have the temptation to consider points not in its domain.

Because you are so familiar with the real line, you were tempted to consider the function's behaviour for negative x.

If you think of a function defined on, say, the set of invertible matrices, then why would you consider its behaviour outside of this set?

In summary, the domain of a function is fundamentally part of the function's definition. It's a good lesson in pure mathematical thinking to be able to focus on the domain and not see a different function defined on a larger domain.

13. Sep 3, 2016

### Stephen Tashi

You might be correct, but we could subject that to technical analysis. Is there any limitation on x in the condition "$|x - a| < \delta$"? Is the statement " $| f(x) - L | < \epsilon$ " true when $x$ is not in the domain of $f$?

Things might be clearer if we use the topological definition of continuity in terms of open sets. But then we have the question of which toplogy to use. For real valued functions, do we use the standard topology of the real line or do we use the relative topology it induces on the domain of the function?

If we really want to drive ourselves crazy, we can pose a similar question about uniform continuity.

14. Sep 3, 2016

### pwsnafu

The epsilon delta is very clear: you only test points in your domain.
Edit: realised that the link is to limits not continuity. Strike that.

The former gives $[0,\infty)$ as neither open nor closed, so its not a topology for $[0,\infty)$.

15. Sep 3, 2016

### JonnyG

If a function is given then the topology for the domain and codomain must also be given. Then continuity is defined in terms of preimages of open sets being open as well.

Now, regardless of which topology $[a,b]$ is equipped with, continuity makes sense at the end points. A function $f: [a,b] \rightarrow \mathbb{R}$ is continuous at $a$ if for each open set containing $f(a)$, there is an open set $U \ni a$ such that $U \subset f^{-1}\big(f(a)\big)$. A similar definition is given for continuity at $b$. You'll notice that this definition actually makes sense for any point $x \in [a,b]$. So we can talk about continuity at the end points of a closed interval and in fact it is not, in any sense, a sort of "special" continuity. It's the same old definition.

16. Sep 4, 2016

### Stephen Tashi

It's very clear in the Wikipedia, but a point of interest is whether that "precise definition" in the Wikipedia is equivalent to the epsilon-delta definitions in typical calculus books. Perhaps modern day books are more precise than books of a decade ago. If we speak of "the" epsilon-delta definition of limit then what will a calculus student understand that to mean if he consults his book? Answering the original post is going to involve emphasizing that there is a precise epsilon-delta definition as well as some not-so-precise epsilon delta definitions.

Somewhere in that definition, you need a symbol representing the "open set containing $f(a)$".

Last edited: Sep 4, 2016
17. Sep 4, 2016

### JonnyG

Yes, I meant to write that there is a neighborhood $U \ni a$ such that $f(U) \subset \mathrm{the open set containing } f(a)$

I will edit my original post to reflect this.

EDIT: I can't edit my original post anymore. So the topological definition of continuity at a point: $f: X \rightarrow Y$ is continuous at a point $a \in X$ if for every open set $V \ni f(a)$, there is an open set $U \ni a$ such that $f(U) \subset V$

18. Sep 4, 2016

### Stephen Tashi

And it would simplify the life of students of analysis if they learned this definition early in their studies, instead of approaching $\mathbb{R^2}, \mathbb{R^3}$ as completely new vistas. One can make a nice analogy to the epsilon-delta definition by using "E" to denote the open set containing $f(a)$ and "D" to denote the open set containing $a$.

19. Sep 8, 2016

### chiro

Hey Gurasees.

The intuitive idea of continuity is that if you were able to draw a continuous mapping (like a function - but not necessarily) then you could do it by using a marking element (like a pencil) and you would be able to represent the information without taking the marking element off of the paper.

It means that if you can do that then the object you are representing is continuous - if you have to take the marking element off for some reason then it isn't.

That is basic continuity and formally it means saying that any limits from the left and right equal the actual value at that point across all of the points you consider.

If for some reason you need to take the marking element off then it violate the limit conditions and that shows the discontinuity part.