# Math Help - Tricky Continuity Question

1. ## Tricky Continuity Question

If I ever teach a class on real analysis I be sure to put this question on the exam.

TRUE/FALSE
-------------
A function $f$ is continous at $x_0\in \mbox{dom}(f)$ if and only if:
$\lim_{x\to x_0}f(x) = f(x_0)$.

2. Originally Posted by ThePerfectHacker
If I ever teach a class on real analysis I be sure to put this question on the exam.

TRUE/FALSE
-------------
A function $f$ is continous at $x_0\in \mbox{dom}(f)$ if and only if:
$\lim_{x\to x_0}f(x) = f(x_0)$.
Okay, I'll bite. I would say this question is meant to be a "trick question" so I
would guess the correct answer is "FALSE."

But I don't know why?

-Dan

3. Originally Posted by topsquark
Okay, I'll bite. I would say this question is meant to be a "trick question" so I
would guess the correct answer is "FALSE."

But I don't know why?

-Dan
The actual true definition (as appears in analysis books) is as follows:

A function is continous at $x_0\in \mbox{dom}(f)$ iff any sequence $(x_n)$ in $\mbox{dom}(f)$ converging to $x_0$ we have that $\lim f(x_n)=f(x_0)$.

With some work it is possible to prove another equally important theorem (which is equivalent, so you can use this one instead as definition of continuity):

A function is continous at $x_0\in \mbox{dom}(f)$ iff for any $\epsilon >0$ there exists a $\delta>0$ such that for all $|x-x_0|<\delta$ and $x\in \mbox{dom}(f)$ imply that $|f(x)-f(x_0)|<\epsilon$.

Now limits $\lim_{x\to x_0}f(x)$ are only defined on functions whose domain contain an open interval containing $x_0$ except possible at $x_0$.

So whenever a limit is written it is assumed we are talking about a function defined on an open interval.

Note, the function $f:\{0\}\mapsto \mathbb{R}$ is continous! But it is not defined on an open interval containing $0$. So writing $\lim_{x\to 0}f$ makes no sense. Consequently, $\lim_{x\to 0} f(x)=f(0)$ just does not make sense.

However, if a function is defined on an open interval then writing the limit version of continuity is okay.

In fact, I once read through a Calculus book that when talked about continuity it put in paranthesis "(defined on some open interval)" and I always wondered to myself why that is. It was not until I learned some real analysis when I realized why that was necessay. Indeed, most Calculus books do not even write such a statement in paranthesis.

My last remark, is that this is a common mistake. Supprisingly, my complex analysis author made this same error when he was talking about Complex continous functions! So it is really a trick question.

4. Originally Posted by ThePerfectHacker
The actual true definition (as appears in analysis books) is as follows:

A function is continous at $x_0\in \mbox{dom}(f)$ iff any sequence $(x_n)$ in $\mbox{dom}(f)$ converging to $x_0$ we have that $\lim f(x_n)=f(x_0)$.

[snipped]

Now limits $\lim_{x\to x_0}f(x)$ are only defined on functions whose domain contain an open interval containing $x_0$ except possible at $x_0$.
Not true. Since $f(x)$ is being written, it is implicit that we must have $x \in \mbox{dom}(f)$ for $f(x)$ to be defined. Therefore this definition is just shorthand (maybe sloppy shorthand) for your longer definition.

So whenever a limit is written it is assumed we are talking about a function defined on an open interval.

Note, the function $f:\{0\}\mapsto \mathbb{R}$ is continous! But it is not defined on an open interval containing $0$. So writing $\lim_{x\to 0}f$ makes no sense. Consequently, $\lim_{x\to 0} f(x)=f(0)$ just does not make sense.
For the domain $\{0\}$, the only possible sequence is the sequence of zeroes, so $\lim_{x\to 0} f(x)=f(0)$ makes sense and is true.

When you take graduate-level Topology, you will learn the most general definition of continuity of a function: a function is continuous iff for every open set $O$ in the codomain space, $f^{-1}(O)$ is open in the domain space. And you will learn when the special definitions that you've given apply. (See Continuous function (topology).)

In the case of the domain space $\{0\},$ the only open sets are $\{0\}$ and $\emptyset$. Thus $f^{-1}(O)$, which is either $\{0\}$ or $\emptyset,$ is trivially open for any open $O$ of the codomain space.

The neat thing about Topology is that it looks at things in a very abstract way and shows the relationships between a lot of special definitions you learn as an undergraduate. Since you like to look closely at definitions, I think you'll really like Topology.

5. Originally Posted by ThePerfectHacker
The actual true definition (as appears in analysis books) is as follows:

A function is continous at $x_0\in \mbox{dom}(f)$ iff any sequence $(x_n)$ in $\mbox{dom}(f)$ converging to $x_0$ we have that $\lim f(x_n)=f(x_0)$.
.
You will note that with this definition any function who's domain is the
integers is vacuously continuous. This definition says nothing about the
the domain being an open interval or $x_0$ being in an open interval
contained in the domain of $f$.

If the domain contains an isolated point $x_0$ then it is vacously continuous
there as every sequence $x_n; n=1, 2, ..$ which goes to $x_0$ is constant
so $(f(x_n)) \to f(x_0)$as all the $f(x_n) = f(x_0)$](eventualy anyway, it can do something else initialy)

RonL

6. Originally Posted by JakeD
Not true. Since $f(x)$ is being written, it is implicit that we must have $x \in \mbox{dom}(f)$ for $f(x)$ to be defined. Therefore this definition is just shorthand (maybe sloppy shorthand) for your longer definition.
When you say "not true" you mean the answer to my true/false question, i.e. you agree with me.

Originally Posted by CaptainBlank
You will note that with this definition any function who's domain is the
integers is vacuously continuous.
Yes. I did notice that. It used to bother me that: $f:\{0,1\} \mapsto \mathbb{R}$ is a continous function. But I realized it does not matter how you define something as long as you can prove everything that you need from the definition. And this definition worked really well.

7. Originally Posted by ThePerfectHacker
When you say "not true" you mean the answer to my true/false question, i.e. you agree with me.

Yes. I did notice that. It used to bother me that: $f:\{0,1\} \mapsto \mathbb{R}$ is a continous function. But I realized it does not matter how you define something as long as you can prove everything that you need from the definition. And this definition worked really well.
But that means you don't need the "defined on an open interval"!

RonL

8. Originally Posted by CaptainBlank
But that means you don't need the "defined on an open interval"!
You need that term if you want to talk about limits of functions. And hence define continuity as $\lim_{x\to x_0}f(x)=f(x_0)$. (Because you cannot talk about limit not defined on an open interval).

But the the definition I posed first is more general, it does not need it.

For example: $f:\{x_0\} \to \mathbb{R}$. Let me do if with the more general (and correct vesion) of $\epsilon,\delta$.
1)Choose any $\epsilon >0$.
2)Choose $\delta = 1$.
3)All $x\in \mbox{dom}(f)$ such that $|x-x_0|< 1$ is $\{x_0\}$
4)Then all in #3 satisfy $|f(x)-f(x_0)| <\epsilon$ (since $x=x_0$)
5)The function is continous. Q.E.D.

Let us use the same function $f(x)$ and show that (the limit version fails):
$\lim_{x\to x_0}f(x)=f(x_0)$.
Because we need to show,
"For any $\epsilon>0$ there exist a $\delta>0$ so that all $|x-x_0|<\delta$ are in $\mbox{dom}(f)$ and $|f(x)-f(x_0)|<\epsilon$.
1)Choose $\epsilon =1$
2)No matter what $\delta$ we work with interval $I=(x_0-\delta,x_0+\delta)$
3)So by #2 $f$ needs to be defined on $I$ which is impossible.
So it is not continous? But that is wrong. My point is the limit version does not work here. Which is why the good Calculus book put in paranthesis "(defined on open interval)" because that version only works when $f$ is defined on an open interval.

9. Originally Posted by ThePerfectHacker

Let us use the same function $f(x)$ and show that (the limit version fails):
$\lim_{x\to x_0}f(x)=f(x_0)$.
Because we need to show,
"For any $\epsilon>0$ there exist a $\delta>0$ so that all $|x-x_0|<\delta$ are in $\mbox{dom}(f)$ and $|f(x)-f(x_0)|<\epsilon$.
No, that is incorrect. What you need to show is

For any $\epsilon>0$ there exists a $\delta>0$ so that for all $x \in \mbox{dom}(f)$, $|x-x_0|<\delta$ implies $|f(x)-f(x_0)|<\epsilon$.

Since $x \in \mbox{dom}(f) = \{x_0\}$ means $x = x_0$, $|x-x_0|<\delta$ implies $|f(x)-f(x_0)|<\epsilon$ is trivially true and $f$ is continuous.

But to really understand that, you would need to understand the term "inherited topology." Here $\{x_0\}$ is a subspace of $\mathbb{R}$, so we can use the topology on $\mathbb{R}$ defined by the metric $d(x,y) = |x-y|$ to define a topology on $\{x_0\}.$

You'll need to study topology to understand what this means.

10. Originally Posted by JakeD
No, that is incorrect. What you need to show is
That is exactly what I am saying! I am saying it is wrong. (Unless a function is defined on an open interval). And I said exactly what you said in my post before.

11. Originally Posted by ThePerfectHacker
Because we need to show,
"For any $\epsilon>0$ there exist a $\delta>0$ so that all $|x-x_0|<\delta$ are in $\mbox{dom}(f)$ and $|f(x)-f(x_0)|<\epsilon$.
Originally Posted by JakeD
No, that is incorrect. What you need to show is

For any $\epsilon>0$ there exists a $\delta>0$ so that for all $x \in \mbox{dom}(f)$, $|x-x_0|<\delta$ implies $|f(x)-f(x_0)|<\epsilon$.
Originally Posted by ThePerfectHacker
That is exactly what I am saying! I am saying it is wrong. (Unless a function is defined on an open interval). And I said exactly what you said in my post before.
Your statement of what is to be shown and my statement are different. We are not saying the same thing.

Your statement (which I'm saying is incorrect)

"all $|x-x_0|<\delta$ are in $\mbox{dom}(f)$ and $|f(x)-f(x_0)|<\epsilon$"

is not the same as my statement (which I'm saying is correct)

"for all $x \in \mbox{dom}(f)$, $|x-x_0|<\delta$ implies $|f(x)-f(x_0)|<\epsilon$."

12. Originally Posted by JakeD
Your statement of what is to be shown and my statement are different. We are not saying the same thing.

Your statement (which I'm saying is incorrect)

"all $|x-x_0|<\delta$ are in $\mbox{dom}(f)$ and $|f(x)-f(x_0)|<\epsilon$"

is not the same as my statement (which I'm saying is correct)

"for all $x \in \mbox{dom}(f)$, $|x-x_0|<\delta$ implies $|f(x)-f(x_0)|<\epsilon$."
I believe my position and statements are equivalent to JakeD's

RonL

13. I have been very reluctant to join this thread. Because I agree with JakeD, all this could be cleared up with a basic course in topology. I know TPH has used Ken Ross’s book for undergraduate analysis. And I think he has misread Ross’s definition of limit. Ross restricts the approach to be on a set S that is a subset of the domain. There is no requirement for the domain to have an interior about the value.
There is another text that has a beautiful discussion of limit and continuity: Undergraduate Analysis by Serge Lang. It is in Chapter 7 in my 1989 edition. There the issues raised by TPH are addressed head-on.

14. I think the problem is that I do not explain myself sufficiently well. Which creates the effect that I am arguing with you. I am not arguing with you!

Originally Posted by Definition 1
$|x-x_0|<\delta \mbox{ and }x\in \mbox{dom}(f) \mbox{ imply }|f(x)-f(x_0)|<\epsilon$
Originally Posted by Definition 2
$|x-x_0|<\delta \mbox{ implies }x\in \mbox{dom}(f) \mbox{ and }|f(x)-f(x_0)|<\epsilon$
Definition 1 is not the same as Defintion 2.
Furthermore, Defintion 2 is wrong. Defintion 1 is correct.

Now, the reason why I used Definition 2 in my earlier post is not because I made a mistake. I was attempting to demonstrate that it is different from the correct defintion.