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Topic: Is (t^2-9)/(t-3) defined at t=3?
Replies: 166   Last Post: Oct 30, 2013 9:41 AM

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 magidin@math.berkeley.edu Posts: 11,749 Registered: 12/4/04
Re: Is (t^2-9)/(t-3) defined at t=3?
Posted: Oct 9, 2013 1:29 AM

On Tuesday, October 8, 2013 9:07:54 PM UTC-5, quasi wrote:
> Arturo Magidin wrote:
>

> >quasi wrote:
>
> >>
>
> >> In the Precalculus/Calculus context, here is my take ...
>
> >>
>
> >> A function f is continuous on an interval if it is continuous
>
> >> at all points of that interval (with one-sided continuity
>
> >> acceptable if the interval has an endpoint).
>
> >
>
> >Yes, I agree with that.
>
> >
>
> >> A function f is "continuous" if the domain of f is an interval
>
> >> and f is continuous on that interval.
>
> >
>
> >But not with that. That would mean that, for example,
>
> >
>
> >f(x) = sqrt(1-x^2)
>
> >
>
> >is a fortiori not continuous, since its domain is (-oo,-1] \/ [1,oo), which is not an interval.
>
>
>
> The domain of the function sqrt(1-x^2) _is_ an interval, namely
>
> the interval [-1,1],

Drats; that's what I get for typing and rushing. I meant sqrt(x^2-1).

> hence, since its domain is an interval

>
> and since it's continuous on that interval, it qualifies
>
> (both by my version and yours) as a continuous function.
>
>
>
> Note that this is consistent with the Precalculus version
>
> (no breaks, no holes, no gaps, no vertical asymptotes), since
>
> the graph of the function sqrt(1-x^2) is a semicircle.
>
>
>
> Perhaps the example that you intended is the function
>
>
>
> f(x) = sqrt(x^2-1)

Yes.

> The domain of f is (-oo,1] \/ [1,oo).
>
>
>
> Certainly f is continuous on its domain, but since the domain
>
> is not an interval, by my version, f would not be called a
>
> continuous function without qualification.

The advantage of the definition I gave is that it is perfectly "upwards compatible": it agrees with the definition from topology when interpreted in the situation at hand: a function is continuous if, for every point in the domain x, and every neighborhood W of the image f(x), there is a neighborhood V of x such that f(V) is contained in W; and we say that a function f is "continuous on A" (for some subset A of the space of discourse X) if the restriction of f to A is defined on all of A and is continuous.

> That's consistent with the Precalculus concept since the domain
>
> has "a gap" (and hence the graph "breaks" into two parts).
>
>
>

> >I consider tan(x) to be a continuous function, because it is
>
> >continuous wherever it is defined. This does *not* mean that
>
> >tan(x) is continuous at each number.
>
>
>
> As I see it, that would just creates confusion for what is an
>
> already difficult concept for students.

And that is why most good calculus books do NOT say "is continuous" without qualification. Your proposed definition, on the other hand, will only postpone the confusion to later in their lives, when they study the more general concept.

> To eliminate the confusion, I would never just say
>
> "tan(x) is continuous" without qualification.

And I do not say that in Calculus (though even the better books sometimes commit small "crimes" against this provision by saying things like "composition of continuous functions is continuous" rather than "composition of a functions that are continuous everywhere is continuous everywhere").

> Rather, I would say "tan(x) is continuous on its domain".

In calculus, I say "tan(x) is continuous *at each point* of its domain" (that is, I expand the macro "tan(x) is continuous" which I have in my head but don't use in the class, precisely to prevent that confusion).

[...]

> >So, for example, under the usage I mention, a sum,
>
> >difference, product, and quotient of continuous functions is
>
> >necessarily continuous, whatever the domains may be;
>
>
>
> No, I think that's an oversimplification which needs to be
>
> qualified.
>
>
>
> I would say it this way:
>
>
>
> If functions f,g are continuous on an interval then the
>
> functions f+g, f-g, f*g are continuous on that interval.
>
>
>
> If functions f,g are continuous on an interval and g is
>
> nonzero at all points of that interval the function
>
> f/g is continuous on that interval.
>
>
>
> Qualifying that way fully avoids the issue.

Which is worthwhile in a calculus class, but no necessary later on.

All you are saying (and I'm not disagreeing) is that we should not use the concept of a function being "continuous" without qualifications in a precalculus or calculus setting. Again: I don't disagree. And the only reason I brought up the distinction between "continuous everywhere" and "continuous" above was because **someone else** decided to use the statement "f is continuous" as a synonym for "f is continuous at all real numbers" (which is not the case under either my usage, or yours).

> >In particular, a rational function is "continuous",
>
>
>
> In the version I'm familiar with, a rational function would
>
> not be regarded as "continuous" unless the denominator has
>
> no real zeros.

I understand your usage; I simply find it problematic because it clashes with the more general notion of continuity, which is supposed to *subsume* the notion for real functions of real variable. The simplest solution, of course, is to avoid the use of "continuous" without qualification in the setting of beginning calculus students. Spivak does that, for instance.

> In a Topology or Analysis context perhaps, but in a Precalculus
>
> or Elementary Calculus context, one should not call a function
>
> just plain "continuous" without qualification, unless it's
>
> at least consistent with the "no breaks" precalculus version.

I disagree with your "unless". The "no breaks" percalculus version is pretty bad to begin with, because it just leads to trouble when students continue to try to apply it as if it were a proper formal definition. The "no breaks" version should be discarded **as soon as possible and never mentioned again**. The actual solution is your prior statement: "not call a function just plain continuous without qualification", no unlesses.

> To achieve that consistency, the price is not so outrageous.
>
> One simply has to qualify the statement that a function is
>
> continuous by indicating, for a given discussion, on what set
>
> that continuity holds, _unless_, as I've previously suggested,
>
> the function is such that

No; adding that "unless" *does* make it a price too high: it means that you have one *formal* notion of continuity for calculus that is **inconsistent** with the *formal* notion of continuity from analysis and topology, even though the former is supposed to be a special case of the latter. That's just no good.

There's a very big difference between contradicting a naive notion that is not good enough in any case, and contradicting a formal notion.

The real solution is to not talk about plain "continuous" in calculus.

>
>
>
> (1) The domain of the function is an interval.
>
>
>
> (2) The function is continuous on its domain.
>
>
>
> Then there's no confusion -- just call it "continuous".

No confusion today. Lots of problems tomorrow. I'm not willing to sacrifice tomorrow for today, when there is no need to sacrifice anything.

> >whether a function is "continuous on X" is a *contextual*
>
> >property of the function in relation to X.
>
> >
>
> >This is signaled by the fact that "continuous" refers only to
>
> >the function, whereas "continuous on X" specifies both the
>
> >function and the context.
>
>
>
> My argument is that, at the Elementary Calculus level, any
>
> new definition of "continuous function" should at least
>
> maintain consistency with the naive Precalculus "no breaks"
>
> concept.

And I never said this was a definition to be introduced at "Elementary Calculus level", a level at which I would object just as strongly to the introduction of **your** proposed definition of 'continuous'.

And as I object to the fetish of maintaining consistency with the **deeply flawed** precalculus notion of "no breaks". That notion should be exiled beyond a naive intuition early and often.

> >> I've never seen that usage.
>
> >>
>
> >> Worse, it jars with the naive, Precalculus concept of
>
> >> continuity, typically expressed as:
>
> >>
>
> >> "A function is continuous if the graph has no breaks."
>
> >
>
> >Ah, but that naive notion jars with the idea that you can
>
> >have functions that are continuous at exactly one point
>
> >and nowhere else, as well (e.g., xsin(1/x) for x=/=0,
>
> >and 0 for x=0), and yet such functions exist.
>
>
>
> Not really.
>
>
>
> The function f(x) as you defined it above is continuous
>
> everywhere.

Again, hoisted by lack of time.

Take the Dirichlet function D(x) with value 1 at rationals and 0 at irrationals. Then take f(x) = xD(x) to get a function that is continuous at one and only one point.
> But at the Precalculus level, continuity at a point is not
>
> discussed -- just continuity of the function as a whole.
>
> Discussion of continuity at a point is left for Calculus.

Quasi, I do know what happens at Precalculus level. I've taught such classes. And I did not advocate introducing a formal definition in Calculus or in Precalculus. Nor was I the person who conflated "continuous everywhere" with "continuous". Nor am I advocating hammering "continuous" into students in Calculus.

On the other hand, I'm also not granting your proposed definition of "continuous", either at the calculus or precalculus level, and certainly not at a higher one.

> >Jarring with naive notions is not in and of itself a bad thing,
>
> >given that naive notions are so often wrong.
>
>
>
> But in this case, it doesn't have to be wrong.

Now, sorry, but that's nonsense. The "naive notion" of continuity as "no breaks" **is** wrong. It's naive, incomplete, and does not get you anywhere. Hiding the fact that the naive notion does not accurately capture continuity does not make it "right", it just makes us complicit in a lie.

> >> or equivalently,
>
> >>
>
> >> "One can trace the graph of the function without lifting
>
> >> the pen from the paper."
>
> >
>
> >See above. The function f(x) = xsin(1/x) for x=/=0 and f(0)=0
>
> >is undeniably continuous at 0; but that idea simply cannot be
>
> >reconciled with the naive notion.
>
>
>
> I'll grant that actually "tracing the graph" is problematic,
>
> but it's not hard to "visualize" the trace.

Which just accomplishes the student lying to him or herself with the complicity of the professor who tells him its doable.

Look: I don't object not telling a beginning student the whole truth; I **do** object telling him lies.

> From the start of the thread, the entire discussion has the
>
> presumption, both implied and stated (I've stated it many
>
> times) that the context is at a level no higher than
>
> Elementary Calculus.

And the OP used the expression "f is continuous" without qualification to refer to a function that does not satisfy **either** your (in my opinion flawed) definition, nor my (in your opinion needlessly confusing) definition. The **only reason** I brought it up was to point out to him that he was being careless yet again, and thus saying incorrect things because of the carelessness.

> >Won't help you because it is in Spanish, but this is also the
>
> >convention used in Arizmendi, Carrillo, and Lara's "Calculo
>
> > Diferencial e Integral" (Editorial Trillas), which was the
>
> >textbook used when I took the course as an undergraduate.
>
> >
>
> >While I don't find other books on my shelf with the explicit
>
> >usage I give,
>
>
>
> Then why be so sure of it?

Because I have one that does, and none that contradict it; and I have none whatsoever that agree with yours, and yours contradicts the more advanced notion.

> >I also find no books on my shelf with the usage *you* give on
>
> > "continuous" to mean "defined on an interval and continuous on
>
> > that interval".
>
>
>
> I didn't claim that I was stating an official definition.

Could'a fooled me, given your insistence in this post...

> But my main point is that it makes no sense, at the
>
> Elementary Calculus level to call a function just plain
>
> "continuous", without qualification, if the function clearly
>
> is _not_ continuous based on Precalculus concepts.

First, I will note, yet again, that I was **not** the one who started calling functions "continuous" without qualification.

Second, I will say that I find your fetishism with the deeply flawed and misguided "precalculus concepts" to be a bad thing per se.

"Not confusing the students" is **not** a good excuse for *lying* to them. It is perfectly possible to keep things simple and nonconfusing without having to rely on lies. And the "precalculus concepts" of continuity are pretty much down and out lies.

> >Hughes-Hallet, Gleason, and McCallum suddenly use "continuous"
>
> >on page 57 of their Single Variable 5th Edition without every
>
> >having said what it means to be "continuous" (they have only
>
> > talked about "continuous at x=c" and
>
> >"continuous on an interval");
>
>
>
> A lack of care by the author. Such terminology needs to be
>
> defined before being used.

You think? Nice of you to point it out to me, given that I said so myself later on in the post...

Are you actually talking to me, or to some strawman you've erected?

> But if, at the Elementary Calculus level, you do choose to
>
> omit the qualification,

Which I do not, nor did I ever propose to do. So, what exactly are you arguing against, then?

> then for the function being called
>
> just plain continuous, that claim should not be blatantly
>
> inconsistent with the Precalculus version.

First, it would not be a "claim", it would be a DEFINITION. Second, I never advocated to use this definition in a calculus course, nor did I imply I do; and thirdly, I have absolutely no problem whatsoever with being inconsistent with lies that my students were told by others before me. The Precalculus version of "continuity" that has been shoehorned unnecessarily into textbooks is pure, simple, unadulterated crap, and I see absolutely no reason to offer it any respect whatsoever. I do not care how in love with them you are, they are still crap and they should be afforded no respect; the sooner the students get the correct definition and abandon the precalculus crap, the better. And no, that is not an advocation for introducing the definition from topology.

> Exactly. Qualify it -- then there's no issue.

I'm quite glad you've told me what to do in my classes. It's not as if I've taught calculus dozens of times before and known what to do. Wish I had come here earlier for my lesson on how to teach; thank you for taking the time to let me know how to do it

--
Arturo Magidin

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