If I ever teach a class on real analysis I be sure to put this question on the exam.
TRUE/FALSE
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A function is continous at if and only if:
.
The actual true definition (as appears in analysis books) is as follows:
A function is continous at iff any sequence in converging to we have that .
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 iff for any there exists a such that for all and imply that .
Now in a Calculus course when talk about continuity they do not talk about this minor discrepancy.
Now limits are only defined on functions whose domain contain an open interval containing except possible at .
So whenever a limit is written it is assumed we are talking about a function defined on an open interval.
Note, the function is continous! But it is not defined on an open interval containing . So writing makes no sense. Consequently, 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.
Not true. Since is being written, it is implicit that we must have for to be defined. Therefore this definition is just shorthand (maybe sloppy shorthand) for your longer definition.
For the domain , the only possible sequence is the sequence of zeroes, so makes sense and is true.So whenever a limit is written it is assumed we are talking about a function defined on an open interval.
Note, the function is continous! But it is not defined on an open interval containing . So writing makes no sense. Consequently, just does not make sense.
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 in the codomain space, 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 the only open sets are and . Thus , which is either or is trivially open for any open 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.
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 being in an open interval
contained in the domain of .
If the domain contains an isolated point then it is vacously continuous
there as every sequence which goes to is constant
so as all the ](eventualy anyway, it can do something else initialy)
RonL
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: 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.Originally Posted by CaptainBlank
You need that term if you want to talk about limits of functions. And hence define continuity as . (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: . Let me do if with the more general (and correct vesion) of .
1)Choose any .
2)Choose .
3)All such that is
4)Then all in #3 satisfy (since )
5)The function is continous. Q.E.D.
Let us use the same function and show that (the limit version fails):
.
Because we need to show,
"For any there exist a so that all are in and .
1)Choose
2)No matter what we work with interval
3)So by #2 needs to be defined on 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 is defined on an open interval.
No, that is incorrect. What you need to show is
For any there exists a so that for all , implies .
Since means , implies is trivially true and is continuous.
But to really understand that, you would need to understand the term "inherited topology." Here is a subspace of , so we can use the topology on defined by the metric to define a topology on
You'll need to study topology to understand what this means.
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 are in and "
is not the same as my statement (which I'm saying is correct)
"for all , implies ."
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.
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 1Definition 1 is not the same as Defintion 2.Originally Posted by Definition 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.