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], 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)
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.
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.
To eliminate the confusion, I would never just say "tan(x) is continuous" without qualification.
Rather, I would say "tan(x) is continuous on its domain".
For a real-valued function f(x) whose domain is some subset of the set of real numbers then, in the Precalculus/Calculus context, I don't equate the statement
"f(x) is continuous on its domain"
with the statement
"f(x) is continuous"
unless the domain of f is an interval.
>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.
>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.
>though it need not be "continuous everywhere".
Sorry, that's just too confusing for my taste.
It definitely doesn't match the naive precalculus "no breaks" version.
>In essence, whether the function is "continuous" is an intrinsic > property of the function;
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.
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
(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".
>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.
Moreover, as far as I'm aware, no Precalculus or Elementary Calculus textbook would call a function such as 1/x "a continuous function", without qualifying it with a phrase such as
The function 1/x is continuous on (-oo,0).
The function 1/x is continuous on (0,oo).
The function 1/x is continuous on (-oo,0) \/ (0,oo).
The function 1/x is continuous on its domain.
The function 1/x is continuous at all values of x for which it is defined.
>> By your usage, all of the above functions, including the >> functions 1/x and tan(x) would be regarded as "continuous", >> without qualification. > >Yes. > >> 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.
The function f(x) as you defined it above is continuous everywhere.
Perhaps the example you intended was something like the function g defined by
g(x) = x if x is rational, 0 otherwise
which is continuous only at x = 0.
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.
Certainly the function g as a whole is not a continuous function.
>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.
In an Elementary Calculus course, just don't call a function just plain "continuous" without qualification if it's clear that it "breaks".
>> 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.
In any case, since the graph has no breaks, that version of the Precalculus concept is consistent with the Calculus version.
>(In fact, Spivak brings up this example to note that the >naive notion of "a continuous function is one whose graph can > be drawn without lifting the pencil from the paper" is >"a little too optimistic").
For that function, as I previously noted, actual tracing of the graph in finite time is problematic, but visualizing the trace is easy. In any case, since there are no breaks, the Precalculus Calculus versions of "continuous function" both agree -- the function is continuous.
>> >So the function f(t) = (t^2-9)/(t-3) is continuous; >> >> Very strange. >> >> Can you find a reference to a textbook at the Precalculus >> or Calculus level which has a definition by which a function >> similar to the function f above would be regarded as a >> continuous function? > >Don't know about Precalculus (I would be hard pressed to find >a precalculus textbook that even discusses limits appropriately, >let alone the subtle notions of continuity). It matches the use > in Bourbaki (Elements of Mathematics, General Topology, >Chapter 1, Section 2, part 1) (Yes, I know Bourbaki is not > precalculus or calculus)
But that's exactly my point.
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.
>A mapping f from a topological space X into a topological space
(Irrelevant to our current discussion).
>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?
>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.
I phrased it as "Here's my take on this issue ..."
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.
I've never seen any text at the Elementary Calculus level or below that would call the function 1/x "a continuous function", without qualification.
>Spivak always says "continuous on...";
Which is also my point.
Don't call a function just plain continuous, unless there can be no confusion.
To avoid confusion, _qualify_ it.
What I'm saying is that, at the Elementary Calculus level, if you insist on _not_ qualifying it, then only do it for those functions for which the earlier Precalculus concept would yield the same answer.
>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.
>Anton, Bivens, and Davis only use "continuous on...", except > for one informal statement that they specify and highlight > as being "informal". Stewart talks about "continuous on..." > and "continuous at...", never about just "continuous".
I have no problem with that.
I'm not insisting that one call functions just plain "continuous", without qualification.
But if, at the Elementary Calculus level, you do choose to omit the qualification, then for the function being called just plain continuous, that claim should not be blatantly inconsistent with the Precalculus version.
I suggested a definition that avoids that inconsistency. I'm not insisting on it -- it's just a suggestion.
By your proposed definition
>> f(t) = (t^2 - 9)/(t - 3) >> >> would be called a continuous function.
By mine it would not.
>I have never encountered your definition of "f is continuous" >(without specification of where) that implies the domain is >an interval, either in a precalculus, calculus, or more >advanced level, that I can remember.
As I've indicated, I'm not claiming that the definition I proposed is in any sense an official definition.
As you can see, most authors avoid the issue by taking care (in most cases, or at least in cases where it might cause confusion) to avoid a statement of the form "f is continuous" without qualification, opting instead for a statement of the form "f is continuous on [some set]".
>I would expect, in fact, that most books would consider >sqrt(1-x^2) to be "a continuous funcion", despite the domain >being disconnected.
Presumably you mean the function sqrt(x^2 - 1).
In any case, most books? Really?
In fact, I suspect the opposite is the case.
I suspect there are hardly any textbooks -- recent (last 50 years), in English, that would call the function sqrt(x^2 - 1) "a continuous function" without qualification.
>I've encountered mine, though I free admit that it is not > widespread at that level; most precalculus and calculus > textbooks never refer to a function as just being "continuous".
Exactly. Qualify it -- then there's no issue.
But _you_ were the one who asserted, in this thread, that the function
f(t) = (t^2 - 9)/(t - 3)
could legitimately be called "a continuous function" without qualification. You even justified it with a stated definition of what it meant for a function to be called a continuous function without the need for qualification.
I objected to your version and proposed an alternate definition.
But as you can see, most authors at the Elementary Calculus level completely avoid the issue by always supplying the appropriate qualification whenever omitting it might cause confusion.
Can we at least agree on the following?
At the Elementary Calculus level, The functions
sqrt(x^2 - 1)
should not be called "continuous functions" with some qualification such as "continuous on their domains".