Definition of a discontinuous function
Andreas Enbacka
Pi/2) ?
The question comes from the fact that discontinuity can be defined in two
ways.
The first definition states that a function f only can be discontinuous at a
point x0, if f(x0) exists. The other definition however states that a
function f also can be said to be discontinuous outside its definition
domain.
Regards,
Andreas Enbacka
e-mail: aen...@geocities.com
The Wizard of Id
are asymptotes at all odd multiples of pi/2 for the tangent
function -- therefore it is discontinuous.
--
The Wizard of Id
Andreas Enbacka wrote in message
<7je4qg$9rf$1...@tron.sci.fi>...
Ronald Bruck
<pig...@vif.com> wrote:
:Yes, tan x is discontinuous at both + and - pi/2 -- there
NO.
tan is a CONTINUOUS function, period. "The other definition..." referred
to by the original poster is in his own imagination; I defy him to cite a
single standard reference wherein that terminology is used or implied.
First, let's deal with the elementary calculus definition of a continuous
function: after defining "continuous at a" to mean "the limit exists at
a, a is in the domain of f, and the limit is f(a)", the books say "a
function f is continuous if it is continuous at each point of its
domain." The tangent satisfies this; pi/2 isn't in the domain, so the
nonexistence of the limit there is irrelevant.
Second, the topological definition. When one has a function f from one
topological space to another, one says the function is CONTINUOUS if the
inverse image of an open set is open. Here we have a function with domain
R \ {odd multiples of pi/2} to the reals, and it has this property.
I'll agree that tan is not continuous on R. That's because it's not
defined on R. I'll agree that it has no continuous EXTENSION to R,
because of the asymptotes. But it is not usual or conventional to call a
function like tan discontinuous.
--Ron Bruck
Ulrich Fahrenberg
>
> Can the function f(x) = tan x be said to be discontinuous (at point x =
> Pi/2) ?
>
> The question comes from the fact that discontinuity can be defined in two
> ways.
> The first definition states that a function f only can be discontinuous at a
> point x0, if f(x0) exists. The other definition however states that a
> function f also can be said to be discontinuous outside its definition
> domain.
>
to the second definition.
I would take the first and thus state, that tan x is continuous IN ALL
POINTS OF ITS DOMAIN. That way you would avoid any misunderstanding.
But you're right, there is some ambiguity with that definition. Take
e.g. the definition of "piecewise continuous". tan x being continuous,
you'd suppose it certainly was piecewise continuous - but it is not! (At
least with the definition of "piecewise continuous" I have in mind -
CORRECT ME IF I'M WRONG)
But of course, the ambiguity only appears when you're not saying what
set your functions are continuous/piecewise continuous on.
-----------------
Ulrich Fahrenberg (u...@math.auc.dk http://www.math.auc.dk/~uli)
"A mathematician is a machine for turning coffee into theorems."
-Paul Erd"os
Farooq
The correct definition of discontinuity is defined when the limit
of the function taken as we approach x0 from the left hand side of
x0 does not equal the limit of the function at x0 as we approach x0
from the right hand side.
So for f(x) = tan x where x0 is pi/2, we approach pi/2 (or 90
degrees) from left. We find the the limit is positively infinite.
Then we take the limit as x approaches pi/2 from the right hand
side, and find that its negative infinity. Since the limits of the two
sides
do not match, there is a discontinuity at f(x) = tan x for x = pi/2.
Andreas Enbacka wrote in message <7je4qg$9rf$1...@tron.sci.fi>...
>Can the function f(x) = tan x be said to be discontinuous (at point x =
>Pi/2) ?
>
>The question comes from the fact that discontinuity can be defined in two
>ways.
>The first definition states that a function f only can be discontinuous at
a
>point x0, if f(x0) exists. The other definition however states that a
>function f also can be said to be discontinuous outside its definition
>domain.
>
Sebastian Moleski
to give some extra understanding:
[1] f(x)=tan x; Domain: all (real) numbers between (-inf, +inf) ==>
discontinuous at n * pi/2
[2] f(x)=tan x; Domain: all (real) numbers between (-inf, +inf) EXCEPT n *
pi/2 ==> continuous
lim(x -> n*pi/2 +) tan x =+INF ==> not a defined real number
lim(x -> n*pi/2 -) tan x =-INF ==> not a defined real number
Cheers,
--
Sebastian
Erik Max Francis
> The correct definition of discontinuity is defined when the limit
> of the function taken as we approach x0 from the left hand side of
> x0 does not equal the limit of the function at x0 as we approach x0
> from the right hand side.
Continuity of some function f(x) at some point x = c means that f[c]
exists and lim{x -> c} f(x) exists and are equal.
tan x is definitely not continuous at x = pi/2. The question was if it
is a "continuous function," which is subtle distinction. If a
continuous function is defined as being continuous everywhere in its
domain, then tan x is a continuous function, since x = pi/2, etc. are
not in its domain and thus it does not have be continuous there to still
be called a continuous function.
My maths books don't seem to go to a great deal of trouble to define
"continuous functions," however; they just concentrate on continuity
itself.
--
Erik Max Francis / email m...@alcyone.com / whois mf303 / icq 16063900
Alcyone Systems / irc maxxon (efnet) / finger m...@members.alcyone.com
San Jose, CA / languages En, Eo / web http://www.alcyone.com/max/
USA / icbm 37 20 07 N 121 53 38 W / &tSftDotIotE
\
/ Procrastination is the thief of time.
/ Edward Young
Ulrich Fahrenberg
>
> [1] f(x)=tan x; Domain: all (real) numbers between (-inf, +inf) ==>
> discontinuous at n * pi/2
You'e wrong about that. The domain, the function tan x "lives" on, can
MAX. be |R\{n*\pi/2 | n \in |Z}. You can let it "live" on a "smaller"
domain, but you cannot extend it.
UNLESS you define (some finite) values in some of the points n * \pi/2.
But then it's YOUR function, NOT tan x.
Take, e.g.,
f(x) = tan x for x != n * \pi/2
17 else
The domain of THIS function is |R.
But you're (in a way) right about the discontinuity. For you canNOT
extend the function tan x to have domain |R and still be continuous.
Brian M. Scott
> The correct definition of discontinuity is defined when the limit
> of the function taken as we approach x0 from the left hand side of
> x0 does not equal the limit of the function at x0 as we approach x0
> from the right hand side.
This is obviously wrong. Consider the function f defined on R as
follows:
f(x) = 0 if x != 0, and f(0) = 1.
The one-sided limits as x approaches 0 are both 0, but f is clearly
discontinuous at x = 0.
Moreover, this 'definition' is limited to R (and other linearly
ordered topological spaces), while continuity is a much more generally
applicable property.
It is pointless to ask whether tan x is discontinuous at pi/2; the
point isn't in its domain, so it isn't even defined there. For
more details see Ron Bruck's post.
Brian M. Scott
Zdislav V. Kovarik
Andreas Enbacka <aen...@geocities.com> wrote:
:Can the function f(x) = tan x be said to be discontinuous (at point x =
:Pi/2) ?
:
:The question comes from the fact that discontinuity can be defined in two
:ways.
:The first definition states that a function f only can be discontinuous
:at a point x0, if f(x0) exists. The other definition however states that
:a function f also can be said to be discontinuous outside its definition
:domain.
There is a lot of disagreement and outright sloppiness in the textbook
circles about it.
Suppose the definitions go (I am starting with the one I do not accept
but it can be found in many textbooks):
f is continuous at c iff all three following conditions are satisfied:
c is in the domain of f,
lim[as x->c] f(x) exists,
lim[as x->c] f(x) = f(c)
(end of definition);
f is discontinuous at c iff it is not continuous at c;
Then tan is discontinuous at pi/2 because f fails the first condition.
Also, according to this definition, a function would be discontinuous at
every isolated point of its domain, which is contrary to the more general
topological definition of continuity. That's why I reject the above
definition.
(An example of a function with an isolated point in its "natural" domain:
f(x) = sqrt(x^4 - x^2), with domain (-inf, -1] union {0} union [1, inf).)
One can amend the definition of continuity of f by adding "or, if c is an
isolated point of the domain of f", but this looks forced.
But: This will not change the conclusion about f=tan and c=pi/2 because c
is not in the domain of f.
To put isolated and non-isolated ("accumulation") points under one roof,
to call f continuous at c, we require that
c be in the domain of f (as above), and
the inverse image of every neighbourhood of f(c) contain a neighbourhood
of c, relative to the domain of f.
(end of definition).
The trick is: c is an isolated point iff the single point set {c} is a
neighbourhood of c. (Look up the missing definitions in a real analysis
book, or wait for the course.)
Even now, tan is not continuous at pi/2.
Some want to distinguish "removable" and "unremovable" discontinuities,
depending on whether the limit exists or not, but that is another topic.
Here, tan has an unremovable discontinuity at pi/2, described as a "doubly
infinite jump".
I have seen other confusing and confused attempts at the definition of
limits and continuity (in textbooks), but I'd rather not clutter my
response with them.
All this is meant to be discussed in the context of real numbers, in their
usual topology, i.e. concept of closeness defined by open bounded
intervals as basic neighbourhoods (no infinities considered).
We can change the playground and the rules to make tan continuous at pi/2,
by attaching the "two-sided infinity" to the target (=codomain) of tan,
and by defining its neighbourhoods, while keeping other neighbourhoods as
they were. But that would be another article. (One-sided infinities will
not help here.)
Cheers, ZVK(Slavek).
Pertti Lounesto
> Suppose the definitions go (I am starting with the one I do not accept
> but it can be found in many textbooks):
>
> f is continuous at c iff all three following conditions are satisfied:
> c is in the domain of f,
> lim[as x->c] f(x) exists,
> lim[as x->c] f(x) = f(c)
> (end of definition);
>
> f is discontinuous at c iff it is not continuous at c;
>
> Then tan is discontinuous at pi/2 because f fails the first condition.
This definition is good for basic calculus courses, while it intesifies
the learning of the concept of limit, taught about the same time.
> Also, according to this definition, a function would be discontinuous at
> every isolated point of its domain, which is contrary to the more general
> topological definition of continuity. That's why I reject the above
> definition.
>
> (An example of a function with an isolated point in its "natural" domain:
> f(x) = sqrt(x^4 - x^2), with domain (-inf, -1] union {0} union [1, inf).)
>
> One can amend the definition of continuity of f by adding "or, if c is an
> isolated point of the domain of f", but this looks forced.
>
> But: This will not change the conclusion about f=tan and c=pi/2 because c
> is not in the domain of f.
>
> To put isolated and non-isolated ("accumulation") points under one roof,
> to call f continuous at c, we require that
> c be in the domain of f (as above), and
> the inverse image of every neighbourhood of f(c) contain a neighbourhood
> of c, relative to the domain of f.
> (end of definition).
>
> The trick is: c is an isolated point iff the single point set {c} is a
> neighbourhood of c. (Look up the missing definitions in a real analysis
> book, or wait for the course.)
>
> Even now, tan is not continuous at pi/2.
This definition is good in a course heading for topology. This latter
definition is conceptual and cannot be checked by a pocket calculator,
to the satisfaction of a lazy student.
Zdislav V. Kovarik
Pertti Lounesto <Pertti....@hut.fi> wrote:
:"Zdislav V. Kovarik" wrote:
:
:> Suppose the definitions go (I am starting with the one I do not accept
:> but it can be found in many textbooks):
:>
:> f is continuous at c iff all three following conditions are satisfied:
:> c is in the domain of f,
:> lim[as x->c] f(x) exists,
:> lim[as x->c] f(x) = f(c)
:> (end of definition);
:>
:> f is discontinuous at c iff it is not continuous at c;
:
:This definition is good for basic calculus courses, while it intesifies
:the learning of the concept of limit, taught about the same time.
(Prof. Lounesto need not be told the facts; I am addressing a wider
audience.)
Some time ago, I was lured into a dispute about "definitions for the
masses vs. definitions for the experts" - that time it was probability.
(The other debater turned out to be trolling, but I digress.)
I am still convinced that holding back a part of the definition which is
so easy to formulate would be cheating:
[...]
:>
:> One can amend the definition of continuity of f by adding "or, if c is
:> an isolated point of the domain of f", but this looks forced.
On second thought, it is not so forced if we draw attention to the fact
that limits are unavailable (or not unique, depending on the definition of
a limit) at isolated points, hence a default clause in in order.
[...]
:> To put isolated and non-isolated ("accumulation") points under one
:> roof, to call f continuous at c, we require that
:> c be in the domain of f (as above), and
:> the inverse image of every neighbourhood of f(c) contain a
:> neighbourhood of c, relative to the domain of f.
:> (end of definition).
[...]
:
:This definition is good in a course heading for topology. This latter
:definition is conceptual and cannot be checked by a pocket calculator,
:to the satisfaction of a lazy student.
:
The sight of students' fingers dancing on the calculator keys is sadly
familiar to me; it leads to generations of students graduating with the
impression that all you have to do to check out a limit is to try it on
the student's single favourite sequence: a student's favourite sequence
convergent to 3 is
{2, 2.9, 2.99, 2.999, ...}
but rarely
{3.1, 3.01, 3.001, ...}
Or vice versa, if the notation uses "3 + Delta x".
The impression (about a single sequence being enough), for less informed
readers, is false.
To wake up the calculator-happy ones, give them the limit of
(sin(x))^(-2) - x^(-2) as x tends to 0,
or variations on this theme.
The calculated values will soon go crazy or settle at 0, depending on the
hardware, and the limit is 1/3.
And there is always the old standby: limit of
a(n) = n^5 * (0.999)^n as n goes to infinity.
Here a(n) increase as long as n < 4997, and a(4997) > 2*10^16 is the
largest. We know, of course, that the limit is 0.
(How do I prove it without the hand-waving L'Hopital style manipulation?
With some foresight, I find the maximum of n^6 * (0.999)^n, estimate it
from above by some M, and then 0 < a(n) < M/n -- Squeeze argument. I
calculated M=2*10^20. And I do not need Calculus for the maximum; just
solve ((n+1)/n)^6 * (0.999) > 1.)
There is a correct sequential formulation of limits and continuity (in
metric spaces), due to Heine, which uses the hypothetical process of
testing _every_ sequence with entries in the domain of f, convergent to c
but different from c. For continuity, one drops the condition "but
different from c".
(And: isolated points are covered by Heine without any trouble.)
The neighborhood definition is easy to illustrate graphically, both by
static and by moving pictures. It is also an opportunity to put the
student "in the driver's seat": instead of sitting back and watching
teeny-weeny cobolds pushing x towards 3 with all their might, but never
reaching it, we answer a challenge:
tell me the tolerance - how close you want f(x) to be to L, the proposed
limit (or to f(c) for continuity) - and I will specify how close x should
be to c. Engineering students would (or should) relate to that.
Regards, ZVK(Slavek).
Jeremy Boden
>Can the function f(x) = tan x be said to be discontinuous (at point x =
>Pi/2) ?
>
>The question comes from the fact that discontinuity can be defined in two
>ways.
>The first definition states that a function f only can be discontinuous at a
>point x0, if f(x0) exists. The other definition however states that a
>function f also can be said to be discontinuous outside its definition
>domain.
>
What is the current definition of tan(x) as it's obviously different
from the version I learned at school!
We were told that tan(x) is defined as sin(x)/cos(x), with the implicit
assumption that the domain of tan(x) is the same as that of the
functions sin, cos.
i.e. The domain of tan(x) is R and so tan(x) has discontinuities.
--
Jeremy Boden mailto:jer...@jboden.demon.co.uk
Ronald Bruck
<jer...@jboden.demon.co.uk> wrote:
:In article <7je4qg$9rf$1...@tron.sci.fi>, Andreas Enbacka
In what follows, I'm going to take "discontinuities" to mean "points of
discontinuity".
The convention for algebraic and trigonometric functions is that when we
write a formula f(x) for such a function, the domain is the set of x for
which f(x) is well-defined. Sometimes we mean over the reals, sometimes
over the complexes.
Thus when one refers to the "function" 1/x -- by which one means the
function x \mapsto 1/x -- one is referring to a function whose domain is
(-\infty,0) union (0,+\infty), i.e. R\{0}, (in an elementary calculus
course where we're working only over the reals).
The tan(x) IS defined as sin(x)/cos(x), but cos(x) is zero for x =
(2n+1)*pi/2. Thus tan(x) is only defined for R \ {(2n+1) pi/2 : n an
integer}. That's its domain, not R. You can't plot tan on R, because it
isn't defined there.
On its domain it is continuous. And that's the definition of continuous
function; a function is continuous iff it is continuous at every point of
its domain. Functions don't "know" anything about points outside of their
domain.
And **I** don't know of any other definition of continuous function. The
continuity of a function is solely a property of the function, another
property of which is its domain. One may define continuity by limits
(ugh!), by limits of sequences (a good way to start thinking about
continuity, at least in metric spaces), or by inverse neighborhoods of
open or closed sets, or presumably still other ways; but all of these
start from the functional formulation f : X \to Y, and there aren't any
other points than X to consider.
This is one of the things I don't like about letting pi/2 be a
discontinuity of tan: with this terminology a continuous function can
have discontinuities!!!
The domain of tan is R\{odd multiples of pi/2}. This function doesn't
know about the rest of R. If we're permitted to assert that it's
discontinuous at pi/2, because that's not in its domain, then we should
also assert that it's not continuous at the color Red, because Red isn't
in its domain either.
OK, this is a whimsically extreme remark. The reals play a special role
here: students can "see" the discontinuity when they look at the graph,
and they think we're weird for calling such a function continuous. But
it's easy to call such points what they are: poles. And to take the
opportunity to introduce students to series, and an explanation of what
tan x looks like in the neighborhood of a pole, by dividing the series for
cos into the series for sin. THAT notion has substance, whereas whether
pi/2 is a point of discontinuity of tan or not is merely a matter of
definition.
--Ron Bruck
Brian M. Scott
> Can I ask a rather basic question here?
> What is the current definition of tan(x) as it's obviously different
> from the version I learned at school!
> We were told that tan(x) is defined as sin(x)/cos(x), with the implicit
> assumption that the domain of tan(x) is the same as that of the
> functions sin, cos.
No, the implied domain would be the intersection of the domain of
sin(x) with {x in dom(cos) : cos(x) != 0}.
> i.e. The domain of tan(x) is R and so tan(x) has discontinuities.
The domain of the tangent function can't be R, since tan(x) isn't
defined on all of R.
Brian M. Scott
Pertti Lounesto
Jeremy's question is legitimate: consider a rope of one meter long
and another rope of the same length with the middle point missing.
Are they both continuous ropes? If mathematicians say "yes", it is
legitimate to ask: do mathematicians serve the real world?
Stan Brown
>What is the current definition of tan(x) as it's obviously different
>from the version I learned at school!
>
>We were told that tan(x) is defined as sin(x)/cos(x), with the implicit
>assumption that the domain of tan(x) is the same as that of the
>functions sin, cos.
>
>i.e. The domain of tan(x) is R and so tan(x) has discontinuities.
Leaving out quibbles about using the definite article (as in "the"
definition), your memory of
tan(x) = sin(x) / cos(x)
is correct.
I think the problem you're having is not with the definition of tan but
with the definition of domain. Let me take a whack at this; doubtless
someone else will put it in more formal language.
The domain of a function is the set of allowable inputs to that function,
or the set of numbers for which the function is defined. Graphically, the
domain corresponds to the set of points on the x axis that are under (or
over) points on the graph of the function -- in other words, the
projection of the function onto the x axis.
You can't assume that tan has the same domain as sin and cos, even though
it's defined in terms of those functions. Why? Because when cos(x) is 0,
tan(x) is undefined. Since cos(x) = 0 does occur, we know there will be
some values in the domain of cos that are not in the domain of tan.
Specifically, as you know, tan(x) is not defined when cos(x) = 0, i.e.
when x = pi/2 + k*pi for any integer k. Therefore those values are not in
the domain of the tan function.
(In the olden days, it was fashionable to say that tan(pi/2) = infinity.
The problem with that is that it has the same form as tan(pi/4) = 1, so
it looks like we're treating "infinity" as a number. But infinity is not
a number, it's shorthand for "increases/decreases forever, beyond any
number you can possibly name". So tan(pi/2) has NO value because the
graph rises beyond any y value you could possibly name.)
--
There's no need to e-mail me a copy of a follow-up; but if you do,
please identify it as such.
Stan Brown, Oak Road Systems, Cleveland, Ohio, USA
http://www.mindspring.com/~brahms/
My reply address is correct as is. The courtesy of providing a correct
reply address is more important to me than time spent deleting spam.
Stan Brown
>Jeremy's question is legitimate: consider a rope of one meter long
>and another rope of the same length with the middle point missing.
>Are they both continuous ropes? If mathematicians say "yes", it is
>legitimate to ask: do mathematicians serve the real world?
It is not reasonable to expect "continuous" to mean the same thing in
math as in real life.
After all, do we complain because there is no grass growing in the
"field" of real numbers?
Randy Poe
Ronald Bruck wrote:
>
> And **I** don't know of any other definition of continuous function. The
> continuity of a function is solely a property of the function, another
> property of which is its domain. One may define continuity by limits
> (ugh!).
The standard epsilon-delta definition is the one I've usually
seen. (Get as close as you like to some value y, say within
delta. Then I can find a neighborhood of size epsilon around
some point x such that f(x) is within delta of y for all x
in the neighborhood.) That's not quite rigorously stated,
but that's the basic idea.
The limit of tan(x) doesn't exist on either side of pi/2.
But it's easy enough to find functions where the limit
exists on both sides of some value x, but is not the
same. These are left-continuous and/or right-continuous
at x, but not "continuous".
Left-continuous, you could define by saying that f(x0)
exists, has some value y0 AND that the limit as
x -> x0 from the left is y0. A carefully phrased definition
like that would let me say tan(x) is not continuous
at pi/2 because tan(pi/2) is not defined (and the limit
does not exist).
Minor semantic note on the topic: People usually define
continuity, not discontinuity. If you asked a mathematician
for a "definition of a discontinuous function", he'd be
likely to say "a function which is not continuous (everywhere)."
> cos into the series for sin. THAT notion has substance, whereas whether
> pi/2 is a point of discontinuity of tan or not is merely a matter of
> definition.
All excellent points. There may or may not be reasons for
fine-tuning your definition of "continuous" in whatever
context you're developing.
Perhaps it is of interest to the more specialized student
to note that sometimes a mathematician may want to either
change or extend a definition of things like "continuous"
for specific purposes.
- Randy
Pertti Lounesto
> [arguments about continuity of the fucntion y = tan(x),
> but discountinuity at x = odd*pi/2]
Ron, you probably want to say that the function y = tan(x)
is continuous, but the graph of y = tan(x) is not connected.
Pertti Lounesto
> Pertti....@hut.fi (Pertti Lounesto) in alt.algebra.help:
> >Jeremy's question is legitimate: consider a rope of one meter long
> >and another rope of the same length with the middle point missing.
> >Are they both continuous ropes? If mathematicians say "yes", it is
> >legitimate to ask: do mathematicians serve the real world?
>
> It is not reasonable to expect "continuous" to mean the same thing in
> math as in real life.
>
> After all, do we complain because there is no grass growing in the
> "field" of real numbers?
The algebraic terms "field", "ring" and "group" were chosen
on purpose to avoid artefact terms or labelling by humans,
and to act as mnemonics. The choices had nothing to do with
everyday fields, rings and groups; as teachers of algebra know,
the chosen mnemonics proved useful. Stan Brown's comparison
shows that he does not know etymology of mathematical terms
about which he chose to argue. The situtation of "continuous"
is different: the term "countinuous" actually describes and
abstracts the property "continuous". The concept "continous"
is also met by students, for the first time, at a level when
the students do not yet have ability to understand the concept
of a mathematical definition: at calculus courses students
still think of definitions as descriptions of reality, not as
agreements of concepts for embarking deductions. Thus, at
that level it is pedagogically more effective to choose
definitions which do not conflict with students' observations
about reality.
J Lee Jaap
"continuity at a point". I have not checked my calculus text to
see if it provides distinct definitions, but I suspect it does.
tan x can be a continuous function and still be discontinuous at some
elements of R.
Also, while I use the label "infinity" or "oo" to mean "increases
without bound", I find it poor usage to say "tan(pi/2) = oo", because
from one side the limit is oo ("increases without bound") and from the
other the limit is -oo ("decreases without bound"). IOW, oo != -oo.
Cheers.
--
J Lee Jaap <Jaa...@POBox.com>
Brian M. Scott
> The algebraic terms "field", "ring" and "group" were chosen
> on purpose to avoid artefact terms or labelling by humans,
> and to act as mnemonics. The choices had nothing to do with
> everyday fields, rings and groups; as teachers of algebra know,
> the chosen mnemonics proved useful.
The terms 'field', 'ring', and 'group' *are* labels chosen by human
beings. The choice of 'group' in particular was obviously related to
its everyday meaning.
> Stan Brown's comparison
> shows that he does not know etymology of mathematical terms
> about which he chose to argue.
The mathematical 'field' was not created ex nihilo; it is in fact
derived from the everyday noun 'field' and therefore shares most of its
etymology with that word.
> The situtation of "continuous"
> is different: the term "countinuous" actually describes and
> abstracts the property "continuous". The concept "continous"
> is also met by students, for the first time, at a level when
> the students do not yet have ability to understand the concept
> of a mathematical definition: at calculus courses students
> still think of definitions as descriptions of reality, not as
> agreements of concepts for embarking deductions. Thus, at
> that level it is pedagogically more effective to choose
> definitions which do not conflict with students' observations
> about reality.
But it is paedagogically very foolish to choose definitions that are in
conflict with those actually used by mathematicians. And it is
paedagogically valuable to have yet another opportunity to point out to
students that mathematical terms have precise definitions and that they
must work with these definitions, not with some vague notion carried
over from everyday experience.
Brian M. Scott
Nathaniel Silver
extended to a larger domain, agrees with untutored
intuitions (as weak as they may be) in the following
respects:
1. No holes
2. When the x values are "close enough,"
the y-values also are close.
3. You can "trace" the curve without lifting
your pencil.
Of course, all this is very rough.
Intuition is not static but developing.
When one "plays" with more sophisticated
examples, one comes to understand
more subtleties of definition, including
functions, defined on the whole real line,
that are nowhere differentiable yet continuous.
If a fertile definition is awkward for the uninitiated
that's unfortunate and also just tough cookies.
Students must understand that what they know
or deem credible is not a criterion for learning.
Rather it is what professionals choose as a right path.
Students need to double, quadruple or 2^n
their efforts when faced with some type of cognitive
dissonance in order to resolve it. Otherwise, one is just
pandering to them.
Jeremy Boden
<loun...@pop.hit.fi> writes
Or perhaps it is also true to say that tan(x) is piecewise continuous.
That it only has a piecewise definition...
Brian M. Scott
<Pertti....@hut.fi> wrote:
>Jeremy's question is legitimate: consider a rope of one meter long
>and another rope of the same length with the middle point missing.
>Are they both continuous ropes?
Not a mathematical question. On the other hand, if you ask about the
functions {<x,0> : x in [0,1]} and {<x,0> : x in [0,1]\{1/2}}, I will
answer that each is continuous on its stated domain. I will not
confuse either with a rope, however.
> If mathematicians say "yes", it is
>legitimate to ask: do mathematicians serve the real world?
Elementary error: besides confusing a technical mathematical question
with a question about the real world, you are assuming that an
everyday word keeps (or at least should keep) exactly its everyday
connotations when used as a precisely defined technical term in
mathematics.
Brian M. Scott
Pertti Lounesto
> Pertti Lounesto wrote:
>
> > The algebraic terms "field", "ring" and "group" were chosen
> > on purpose to avoid artefact terms or labelling by humans,
> > and to act as mnemonics. The choices had nothing to do with
> > everyday fields, rings and groups; as teachers of algebra know,
> > the chosen mnemonics proved useful.
>
> The terms 'field', 'ring', and 'group' *are* labels chosen by human
> beings.
Yes, but they are not labels carrying names of human beings,
like in Newton's law, Euler's formula, Lagrange's theorem.
This point was discussed by mathematicians, who chose the
the labels 'field', 'ring', 'group'.
> The choice of 'group' in particular was obviously related to
> its everyday meaning.
Related yes, in the sense that a sociological group has a leader,
corresponding to the neutral element of a group, but the math
concept of group does not try to abstract or develop a theory
or a model for a sociological groups. In this sense, the concept
of group is in sharp contrast to the concept of continuous,
which abstracts everyday phenomenom of countinuity.
> The mathematical 'field' was not created ex nihilo; it is in fact
> derived from the everyday noun 'field' and therefore shares most
> of its etymology with that word.
While you pretend to know things, can you point out what
'human being' was the one who 'derived' the label 'field'?
In French the concpet of 'field' is labelled 'corps commutative'.
Maybe you, as somebody who knows these things, can tell
us how the the concept of 'non-commutative field' became
labelled by the term 'corps' in French, and who 'derived' it?
> > The concept "continous" is also met by students, for the first time,
> > at a level when the students do not yet have ability to understand
> > the concept of a mathematical definition: at calculus courses
> > students still think of definitions as descriptions of reality, not as
> > agreements of concepts for embarking deductions. Thus, at
> > that level it is pedagogically more effective to choose
> > definitions which do not conflict with students' observations
> > about reality.
>
> But it is pedagogically very foolish to choose definitions that are in
> conflict with those actually used by mathematicians. And it is
> pedagogically valuable to have yet another opportunity to point out to
> students that mathematical terms have precise definitions and that they
> must work with these definitions, not with some vague notion carried
> over from everyday experience.
Have you actually taught mathematics? Calculus I, II? If the
students do not have appropriate cognition to assimilate a new
concept to their existing cognitive structures, then students are
not prepared to learn the concept. Can you give me an example
of an elementary function, discussed in the calculus courses, which
is not continuous, according to your definition of continuity?
As I undertand, according to you tan(x) is continuous, in
addition to x^n, sqrt(x), exp(x), log(x), sin(x) and cos(x). If no
function is 'not countinuous', then 'continuous' is a concept
introduced for no purpose, for the calculus courses.
Nathaniel Silver
>Can you give me an example of an elementary function,
>discussed in the calculus courses, which is not continuous,
y = int(x)
Brian M. Scott
<Pertti....@hut.fi> wrote:
>"Brian M. Scott" wrote:
[...]
>> The choice of 'group' in particular was obviously related to
>> its everyday meaning.
>Related yes, in the sense that a sociological group has a leader,
>corresponding to the neutral element of a group,
I hardly had anything so far-fetched in mind. The relationship is
much simpler: 'group' refers to a collection of objects. Had a
completely arbitrary label really been desired, the mathematical
object could have been called a 'grillip' (complete neologism) or a
'red' (completely unrelated term, not even the right part of speech).
>> The mathematical 'field' was not created ex nihilo; it is in fact
>> derived from the everyday noun 'field' and therefore shares most
>> of its etymology with that word.
>While you pretend to know things, can you point out what
>'human being' was the one who 'derived' the label 'field'?
>In French the concpet of 'field' is labelled 'corps commutative'.
>Maybe you, as somebody who knows these things, can tell
>us how the the concept of 'non-commutative field' became
>labelled by the term 'corps' in French, and who 'derived' it?
I have no idea. It's also irrelevant to the choice of 'field' in
English. Moreover, one of the senses of 'corps' is 'collection', so
it is clearly appropriate in precisely the same way that 'group' is
appropriate.
>> > The concept "continous" is also met by students, for the first time,
>> > at a level when the students do not yet have ability to understand
>> > the concept of a mathematical definition: at calculus courses
>> > students still think of definitions as descriptions of reality, not as
>> > agreements of concepts for embarking deductions. Thus, at
>> > that level it is pedagogically more effective to choose
>> > definitions which do not conflict with students' observations
>> > about reality.
>> But it is pedagogically very foolish to choose definitions that are in
>> conflict with those actually used by mathematicians. And it is
>> pedagogically valuable to have yet another opportunity to point out to
>> students that mathematical terms have precise definitions and that they
>> must work with these definitions, not with some vague notion carried
>> over from everyday experience.
>Have you actually taught mathematics? Calculus I, II?
For over a quarter of a century.
> If the
>students do not have appropriate cognition to assimilate a new
>concept to their existing cognitive structures, then students are
>not prepared to learn the concept. Can you give me an example
>of an elementary function, discussed in the calculus courses, which
>is not continuous, according to your definition of continuity?
>As I undertand, according to you tan(x) is continuous, in
>addition to x^n, sqrt(x), exp(x), log(x), sin(x) and cos(x). If no
>function is 'not countinuous', then 'continuous' is a concept
>introduced for no purpose, for the calculus courses.
Every elementary calculus text has examples of discontinuous
functions. Some of them, like the floor (greatest integer) function
arise naturally. Others are introduced to illustrate the concept of
continuity, e.g., f(x) = 1/x if x != 0, f(0) = 1.
A student should learn that tan is continuous (because it is), and
that the problem with it at, say, pi/2 is that (a) it's undefined
there, and (b) no matter how it's defined, the resulting function
won't be continuous at pi/2. It is technically correct to say that
pi/2 is a point of discontinuity of tan(x), but it's rather like
saying 'Antti can't come to the phone now because he's not awake' when
in fact Antti is dead; it's much better simply to note that the
function isn't defined there.
Brian M. Scott
Pertti Lounesto
> Pertti Lounesto <Pertti....@hut.fi> wrote:
> >"Brian M. Scott" wrote:
>
> [...]
>
> >a sociological group has a leader,
> >corresponding to the neutral element of a group,
>
> I hardly had anything so far-fetched in mind.
That is not far-fetched: to distinguish from sets (of people who do
not have a leader), the label 'group' was chosen to emphasize the
existence of a particulat element, the neurtal element, in a group.
> one of the senses of 'corps' is 'collection', so
> it is clearly appropriate in precisely the same way that 'group' is
> appropriate.
'Corps' means a collection which is not dissoluble, or a group
of people in the same profession. Thus, there is more structure
in a 'corps' than in a 'groupe', which in turn has more structure
than 'ensemble'. This step-wise addition of structure was the
basis of choosing the names 'fields', 'group' and 'sets', which
act as mnemonics for additional mathematical structures.
> Every elementary calculus text has examples of discontinuous
> functions. Some of them, like the floor (greatest integer) function
> arise naturally.
Step function does arise natrually in applications of mathematics,
but it has not appeared in math courses prior to its introduction
in conjunction with illustrations of 'continouity'. But other non-
continuous examples, like y(x) = tan(x), x not=odd*pi/2, y(x) =
0, x = odd*pi/2, are artificial inventions.
> A student should learn that tan is continuous (because it is), and
> that the problem with it at, say, pi/2 is that (a) it's undefined
> there, and (b) no matter how it's defined, the resulting function
> won't be continuous at pi/2. It is technically correct to say that
> pi/2 is a point of discontinuity of tan(x),
Referring to discountinuity might be misleading for students;
better say that the graph of tan(x) is not connected.
But using the other common definition of continuity, left and
right limits exist and equal the value at the point, is not a bad
choice pedagogically: you have the advantage of changing your
definition at more advanced courses, and thus drilling students
in the practice of choosing an appropriate definition for the
particular purpose at hand.
Nathaniel Silver
in elementary calculus are:
Rational functions are continuous where they are defined.
Ditto for trig functions, exponentials, logarithms, and roots,
sums, differences, products, quotients, and compositions, etc.
Now students can practice finding natural domains of
such functions f in order to determine intervals where
f is continuous and leave it at that.
Jeremy Boden
<mat...@worldnet.att.net> writes
at x=0 is 1, which is convenient as this gives a continuous everywhere
function.
Darrell Ryan
>
> In my opinion, the right words to use
> in elementary calculus are:
>
> Rational functions are continuous where they are defined.
> Ditto for trig functions, exponentials, logarithms, and roots,
> sums, differences, products, quotients, and compositions, etc.
Right. And because they are continuous where they are defined, it
makes it easier to understand for cal I folks why you can evaluate
limits of the these functions by direct substitution, (barring an
indeterminate forms of course) since continuity at c means f(c)=lim
x-->c f(x).
--
Regards,
Darrell
Darrell Ryan
>
> The standard limit definition of continuity at a point,
> extended to a larger domain, agrees with untutored
> intuitions (as weak as they may be) in the following
> respects:
>
> 1. No holes
> 2. When the x values are "close enough,"
> the y-values also are close.
> 3. You can "trace" the curve without lifting
> your pencil.
The way it was told to me was "no holes, jumps, or gaps."
>
> Of course, all this is very rough.
> Intuition is not static but developing.
> When one "plays" with more sophisticated
> examples, one comes to understand
> more subtleties of definition, including
> functions, defined on the whole real line,
> that are nowhere differentiable yet continuous.
Agreed.
I have read over this thread and have the opinion that what started as
a simple, legitimate question has turned into a debate of what the
absolute technical definition of "continuous function" should be. I
think what is important here, especially if this is a CAL I question,
is to thouroughly understand the following.
I am not an expert by any means, but the definitions of "continuity" I
have been exposed to talk about only two types: (a)continuity at a
point, and (b)continuity on an interval, where:
1. f(x) is continuous at c if:
lim x-->c f(x) = f(c)
where both the limit and f(c) are defined
2. f(x) is continuous on an open interval (a,b) if it is continous at
every point in (a,b)
3. f(x) is continuous on a closed interval [a,b] if it is continuous
on the open interval (a,b) and
lim x-->a+ f(x) = f(a) and lim x-->b- f(x) = f(b)
(similar definitions can be made to cover continuity on half-open
intervals of the form (a,b] and [a,b) and on infinite intervals.
Although I have not formally seen "continuous function" defined, we
used it informally but very frequently to describe a function that was
continuous on its entire domain (not necessarily on (-oo,oo)). I find
it unreasonable to expect the term "continuous function" to be an
erronious description of functions that are not defined everywhere but
continuous on their entire domain, such as f(x)=tan(x).
If f(x)=tan(x) and you want to talk about f(pi/2), it really doesn't
serve any useful purpose to say that point is part of function that is
not a "continuous function" because you really don't have any
"function" at all at that point. Don't let technical jargon get in
the way of understanding the concept of continuity. The jargon is
important, but don't focus on it too much. Seems to me those that say
tan(x) is not a "continuous function" are focusing on an insignificant
technicality that the term is equivalent to "continuous everywhere."
BTW, I have seen "continuous everywhere" defined as continuous on
(-oo,oo) such as teh sine and cosine functions. But, it's alsa true
that a dog is not a black cat. The fact that it's not a cat at all
does not *technically* make the statement any less true. Of course,
it's really of no use to say a dog is not a black cat, but you can say
that and be technically correct. Likewise, I can see where the
sticklers for rigor may want to say tan(x) is not a "continuous
function," but it serves no useful purpose here since at certain
values the function f(x)=tan(x) is not really a "function" at all,
much less a "continuous function."
IMO, as long as a CAL I student understands the concept of continuity
as defined above then it is very intuitive to realize that:
(a) f(x)=tan(x) is *not* defined on (-oo,oo) therefore it *cannot* be
continuous on (-oo,oo)
(b) f(x) = tan(x) *is* continuous on its entire domain
(c) any quibbles here over the definition of "continuous function" in
all liklihood will not matter that much to a CAL I student as long as
the general concept (and definitions) of continuity at a point and on
an interval are understood. Any ambiguity can be resolved by asking
the instructor just what he/she means when/if they say "continuous
function." Chances are they are going to say a function that is
continuous on its entire domain. Even if he/she says differently, or
the book says differently, at least you will understand *why* they say
that.
--
Regards,
Darrell
Erik Max Francis
> Although I have not formally seen "continuous function" defined, we
> used it informally but very frequently to describe a function that was
> continuous on its entire domain (not necessarily on (-oo,oo)). I find
> it unreasonable to expect the term "continuous function" to be an
> erronious description of functions that are not defined everywhere but
> continuous on their entire domain, such as f(x)=tan(x).
A better example might be f(x) = ln x. It seems even more sensible to
call this a continuous function, even though its domain is not R.
--
Erik Max Francis / email m...@alcyone.com / whois mf303 / icq 16063900
Alcyone Systems / irc maxxon (efnet) / finger m...@members.alcyone.com
San Jose, CA / languages En, Eo / web http://www.alcyone.com/max/
USA / icbm 37 20 07 N 121 53 38 W / &tSftDotIotE
\
/ After a thousand years or so you go native.
/ Camden Benares
Darrell Ryan
>
> In article <7joocl$733$1...@bgtnsc01.worldnet.att.net>, Nathaniel Silver
> <mat...@worldnet.att.net> writes
> >in elementary calculus are:
> >
> >Rational functions are continuous where they are defined.
> >Ditto for trig functions, exponentials, logarithms, and roots,
> >sums, differences, products, quotients, and compositions, etc.
> >
> >such functions f in order to determine intervals where
> >f is continuous and leave it at that.
> >
> What about sin(x)/x -it's a rational function and it's limit
> at x=0 is 1, which is convenient as this gives a continuous everywhere
> function.
Look at sin(x)/x as the product of the trig function sin(x) and the
rational function 1/x. The second is not defined at x=0.
Please do not confuse "continuous everywhere" with a function that is
continuous on it's entire domain, a.k.a. "continuous function."
Continuous everywhere means continuous *everywhere,* as in (-oo,oo)
--
Regards,
Darrell
Brian M. Scott
<Pertti....@hut.fi> wrote:
>"Brian M. Scott" wrote:
>> Pertti Lounesto <Pertti....@hut.fi> wrote:
>> >"Brian M. Scott" wrote:
>> >a sociological group has a leader,
>> >corresponding to the neutral element of a group,
>> I hardly had anything so far-fetched in mind.
>That is not far-fetched: to distinguish from sets (of people who do
>not have a leader), the label 'group' was chosen to emphasize the
>existence of a particulat element, the neurtal element, in a group.
I don't believe it for a moment, if only because in English 'group'
does not connote the existence of a leader.
>> one of the senses of 'corps' is 'collection', so
>> it is clearly appropriate in precisely the same way that 'group' is
>> appropriate.
>'Corps' means a collection which is not dissoluble, or a group
>of people in the same profession. Thus, there is more structure
>in a 'corps' than in a 'groupe', which in turn has more structure
>than 'ensemble'. This step-wise addition of structure was the
>basis of choosing the names 'fields', 'group' and 'sets', which
>act as mnemonics for additional mathematical structures.
If this is in fact the reason for the choices, then you have refuted
your own claim that the choices had nothing to do with the everyday
meanings of the words.
>> Every elementary calculus text has examples of discontinuous
>> functions. Some of them, like the floor (greatest integer) function
>> arise naturally.
>Step function does arise natrually in applications of mathematics,
>but it has not appeared in math courses prior to its introduction
>in conjunction with illustrations of 'continouity'.
This is not generally true: I've seen quite a few pre-calculus texts
that discuss it.
> But other non-
>continuous examples, like y(x) = tan(x), x not=odd*pi/2, y(x) =
>0, x = odd*pi/2, are artificial inventions.
There's nothing wrong with an artificial example if it makes a point,
though in most cases more natural ones could be found. Characteristic
functions of simple sets are quite useful in this regard.
>> A student should learn that tan is continuous (because it is), and
>> that the problem with it at, say, pi/2 is that (a) it's undefined
>> there, and (b) no matter how it's defined, the resulting function
>> won't be continuous at pi/2. It is technically correct to say that
>> pi/2 is a point of discontinuity of tan(x),
>Referring to discountinuity might be misleading for students;
>better say that the graph of tan(x) is not connected.
Better yet to say what I already said in (a) and (b) above.
[...]
Brian M. Scott
Ayan Mahalanobis
is never talked about in this debate or may be I missed it.
Whatever is the reason if we go into the definition of a function and
stick to the point that we are not allowed to talk of any points which
are not in the domain then i guess that the problem is solved.
Cheers
Ayan
Pertti Lounesto
> <Pertti....@hut.fi> wrote:
> >"Brian M. Scott" wrote:
> >> Pertti Lounesto <Pertti....@hut.fi> wrote:
> >> >"Brian M. Scott" wrote:
>
> >> >a sociological group has a leader,
> >> >corresponding to the neutral element of a group,
>
> >> I hardly had anything so far-fetched in mind.
>
> >That is not far-fetched: to distinguish from sets (of people who do
> >not have a leader), the label 'group' was chosen to emphasize the
> >existence of a particulat element, the neurtal element, in a group.
>
> I don't believe it for a moment, if only because in English 'group'
> does not connote the existence of a leader.
In sociology and psychology of groups such connotation is given,
and also in economics of companies under one holding company.
But, it is irrelevant what common English says here, because that
was the way the mathematicians, who chose the words 'ensemble',
groupe', and 'corps' justified their choice as a mnemonics.
> >'Corps' means a collection which is not dissoluble, or a group
> >of people in the same profession. Thus, there is more structure
> >in a 'corps' than in a 'groupe', which in turn has more structure
> >than 'ensemble'. This step-wise addition of structure was the
> >basis of choosing the names 'fields', 'group' and 'sets', which
> >act as mnemonics for additional mathematical structures.
>
> If this is in fact the reason for the choices, then you have refuted
> your own claim that the choices had nothing to do with the everyday
> meanings of the words.
No, and here you are for the second time (after 'labelling by humans')
either misinterpreting or distorting what I wrote. When I mentioned
that the mathemtical terms 'group', 'ring', 'field' had nothing to do with
everyday concpets of 'group', 'ring', 'field', I made it clear that the
terms were chosen as a mnemonics (from their everyday meaning)
and that the mathematical concepts cannot be applied to analyse their
everyday counterparts, in contrast to the concept/term 'continuous'.
I did not extend this discussion from 'continuous' to 'field'; that
extension was made by somebody else; you engaged the debate and
show off poor knowledge of etymology of the term 'field', which
you chose to argue.
> >> A student should learn that tan is continuous (because it is), and
> >> that the problem with it at, say, pi/2 is that (a) it's undefined
> >> there, and (b) no matter how it's defined, the resulting function
> >> won't be continuous at pi/2. It is technically correct to say that
> >> pi/2 is a point of discontinuity of tan(x),
>
> >Referring to discountinuity might be misleading for students;
> >better say that the graph of tan(x) is not connected.
>
> Better yet to say what I already said in (a) and (b) above.
When there are teachers like you, who say that a 'continuous'
function has 'points of discontinuity', no wonder why there are
former students, like Jeremy Boden, who are still confused, after
years of receiving their education, about 'continuity' of tan at its
'points of discontinuity'.
Randy Poe
I have seen applications where +- infinity is allowed
as a value of a function. So the functions map the
reals to R + {+-infinity}. In this case, f(x) = log x
is continuous on the closed interval consisting of
the non-negative reals.
- Randy
Stan Brown
>The way it was told to me was "no holes, jumps, or gaps."
What's the difference between a hole and a gap?
b_m_...@my-deja.com
Pertti Lounesto <Pertti....@hut.fi> wrote:
> "Brian M. Scott" wrote:
> > <Pertti....@hut.fi> wrote:
> > >"Brian M. Scott" wrote:
> > >> Pertti Lounesto <Pertti....@hut.fi> wrote:
> > >> >a sociological group has a leader,
> > >> >corresponding to the neutral element of a group,
> > >> I hardly had anything so far-fetched in mind.
> > >That is not far-fetched: to distinguish from sets (of people who
> > >do
> > >not have a leader), the label 'group' was chosen to emphasize the
> > >existence of a particulat element, the neurtal element, in a group.
> > I don't believe it for a moment, if only because in English 'group'
> > does not connote the existence of a leader.
> In sociology and psychology of groups such connotation is given,
> and also in economics of companies under one holding company.
Technical usage is not common usage. You are talking about
technical usages (and, I might add, ones that may just be
younger than the mathematical term).
> But, it is irrelevant what common English says here, because that
> was the way the mathematicians, who chose the words 'ensemble',
> groupe', and 'corps' justified their choice as a mnemonics.
I believe that the words were chosen with some eye to their common
meanings; I already said as much. I have no idea whether they were
given the elaborate justifications that you imply; it's possible,
but your assertion isn't evidence.
> > >'Corps' means a collection which is not dissoluble, or a group
> > >of people in the same profession. Thus, there is more structure
> > >in a 'corps' than in a 'groupe', which in turn has more structure
> > >than 'ensemble'. This step-wise addition of structure was the
> > >basis of choosing the names 'fields', 'group' and 'sets', which
> > >act as mnemonics for additional mathematical structures.
> > If this is in fact the reason for the choices, then you have refuted
> > your own claim that the choices had nothing to do with the everyday
> > meanings of the words.
> No, and here you are for the second time (after 'labelling by humans')
> either misinterpreting or distorting what I wrote.
Perhaps you should write more clearly.
> When I mentioned
> that the mathemtical terms 'group', 'ring', 'field' had nothing to do
> with everyday concpets of 'group', 'ring', 'field', I made it clear
> that the terms were chosen as a mnemonics (from their everyday
> meaning)
You contradict yourself. Again. If they were chosen as mnemonics
on the basis of their everyday meanings, it is obviously not true
that they had nothing to do with these meanings.
> and that the mathematical concepts cannot be applied to analyse their
> everyday counterparts, in contrast to the concept/term 'continuous'.
> I did not extend this discussion from 'continuous' to 'field'; that
> extension was made by somebody else; you engaged the debate and
> show off poor knowledge of etymology of the term 'field', which
> you chose to argue.
I did not. I pointed out that you apparently don't understand the
meaning of 'etymology'. It is not limited to the reasons possibly
given by someone for choosing the term; it includes the history of
the word that was borrowed into technical use.
> > >> A student should learn that tan is continuous (because it is),
> > >> and
> > >> that the problem with it at, say, pi/2 is that (a) it's undefined
> > >> there, and (b) no matter how it's defined, the resulting function
> > >> won't be continuous at pi/2. It is technically correct to say
> > >> that
> > >> pi/2 is a point of discontinuity of tan(x),
> > >Referring to discountinuity might be misleading for students;
> > >better say that the graph of tan(x) is not connected.
> > Better yet to say what I already said in (a) and (b) above.
> When there are teachers like you, who say that a 'continuous'
> function has 'points of discontinuity',
You're being dishonest. If you go back and check my post before
last -- the one that contains the *rest* of the sentence that begins
'It is technically correct to say ...' -- you'll see that I don't
recommend this terminology at all. I recommend the correct
explanation in (a) and (b) above, which says nothing about the
function having a point of discontinuity. I don't normally say
that 1/x is discontinuous at 0; I say that it's undefined there.
However, if a student asks whether it's correct to say that the
function is discontinuous at 0, I must answer 'yes', because the
function f(x) = 1/x does not satisfy the definition of continuity
at 0. If you don't understand this, you've no business teaching.
> no wonder why there are
> former students, like Jeremy Boden, who are still confused, after
> years of receiving their education, about 'continuity' of tan at its
> 'points of discontinuity'.
Your attitude is better suited to your primary hobby of finding
insignificant errors in published proofs. I will not respond again:
I see no reason to give you more opportunities to 'justify' insults
by selective quotation.
Brian M. Scott
Sent via Deja.com http://www.deja.com/
Share what you know. Learn what you don't.
J Lee Jaap
|>I suppose it would just confuse things if I noted that
|>I have seen applications where +- infinity is allowed
|>as a value of a function. So the functions map the
|>reals to R + {+-infinity}. In this case, f(x) = log x
|>is continuous on the closed interval consisting of
|>the non-negative reals.
In common algebra and calculus, +oo and -oo are not numbers, the fact
that they look as if they are used as numbers notwithstanding. Any time
"infinity" or "infinite" occurs, it's a shorthand for another meaning,
that is, the statement has a different meaning than if a number were
used. In fact, the statement with infinity is often a negation of what
the statement would say if a number were used.
Always check the definitions and the context in which they're given.
Many many careless users of mathematics have gotten fouled up by
ignoring this basic warning. (Remember, Einstein divided by his
"Cosmological Constant", implicitly assuming that it was not zero.)
Virgil
(1) The point in question is a point of the domain of the function.
A function may be discontinuous at such a point because
(a) the limit does not exist at the point, or
(b) the limit exists but does not equal the value of the function
An example of type (a) at x = 0 is the floor function,
f(x) = largest integer <= x
An example of type (b) is g(x) = floor(abs(x))
(2) The point in question is a limit point of the domain but is not in the
domain. If the limit exists, as for h(x) = sin(x)/x at x=0, as in case (b)
above, the point is a "removeable" discontinuity.
--
Virgil
vm...@frii.com
Erik Max Francis
> What's the difference between a hole and a gap?
My guess would be that in this context a "hole" is a removable
discontinuity (discontinuous only at one point). A "jump" is a
non-removable point discontinuity. And a "gap" is a non-point,
non-removable discontinuity (i.e., a finite interval is "missing" from
the domain).
>
> --
> There's no need to e-mail me a copy of a follow-up; but if you do,
> please identify it as such.
>
> Stan Brown, Oak Road Systems, Cleveland, Ohio, USA
> http://www.mindspring.com/~brahms/
> My reply address is correct as is. The courtesy of providing a correct
> reply address is more important to me than time spent deleting spam.
--
Erik Max Francis / email m...@alcyone.com / whois mf303 / icq 16063900
Alcyone Systems / irc maxxon (efnet) / finger m...@members.alcyone.com
San Jose, CA / languages En, Eo / web http://www.alcyone.com/max/
USA / icbm 37 20 07 N 121 53 38 W / &tSftDotIotE
\
/ Be able to be alone. Lose not the advantage of solitude.
/ Sir Thomas Browne
Stan Brown
>(Remember, Einstein divided by his
>"Cosmological Constant", implicitly assuming that it was not zero.)
And it turns out that it's very likely not zero after all.
See the January 1999 /Scientific American/ and the 17 Dec 1998 /Science/.
I was privileged to attend a lecture last month by one of the
researchers.
Darrell Ryan
> No, and here you are for the second time (after 'labelling by humans')
> either misinterpreting or distorting what I wrote.
Ahh, I see Pertti is back with us trolling once again. It would be a
misinterpretation or a distortion of anything Pertti writes if anyone
made any sense out of it or thought he trying to take part in any
constructive debate.
--
Regards,
Darrell
Darrell Ryan
>
> Quoth ry...@edge.net (Darrell Ryan) in alt.algebra.help:
> >The way it was told to me was "no holes, jumps, or gaps."
>
>
> There's no need to e-mail me a copy of a follow-up; but if you do,
> please identify it as such.
>
> Stan Brown, Oak Road Systems, Cleveland, Ohio, USA
> http://www.mindspring.com/~brahms/
> My reply address is correct as is. The courtesy of providing a correct
> reply address is more important to me than time spent deleting spam.
....from the Ryan Dictionary of Mathematical Terms
A hole is a discontinuity consisting of one point where the point
(c,lim x-->c f(x)) would remove the discontinuity. Ya know, a hole!
Gaps are not necessarily one "hole" in size.
--
Regards,
Darrell
Brian M. Scott
> A hole is a discontinuity consisting of one point where the point
> (c,lim x-->c f(x)) would remove the discontinuity. Ya know, a hole!
> Gaps are not necessarily one "hole" in size.
Okay, then what's the difference between a gap and a jump? <g> More
seriously, which of these terms (if any) would you use to describe the
f(x) = sin(1/x) type of discontinuity (define f(0) = 0, say)?
Brian M. Scott
Jeremy Boden
.....
>When there are teachers like you, who say that a 'continuous'
>former students, like Jeremy Boden, who are still confused, after
>years of receiving their education, about 'continuity' of tan at its
>'points of discontinuity'.
Thanks for the citation Pertti.
Pertti Lounesto
> Pertti Lounesto <Pertti....@hut.fi> writes
> .....
> >When there are teachers like you, who say that a 'continuous'
> >function has 'points of discontinuity', no wonder why there are
> >former students, like Jeremy Boden, who are still confused, after
> >years of receiving their education, about 'continuity' of tan at its
> >'points of discontinuity'.
>
> Thanks for the citation Pertti.
You are welcome. I repeat, your questionning was legitimate: poor
teaching can be critisized by bringing forth confusion. Look at the
debating styles of my/our opponents here: they bring in other topics,
labelling of algebraic terms, and show poor competence in the topic
they chose. They distort my writings (admittedly English is not my
mother tongue, but my intension should have been clear to know-
ledgeable people), when I argue that (I rephrase my argument here):
1. the mathematical term 'continuous' was borrowed from its
everyday meaning and can be applied to analyze everyday
concept of 'continuity',
2. the mathematical terms 'group', 'ring', 'field' were borrowed
from their everyday counterparts, but cannot be applied to
analyze the everyday concepts of 'group', 'ring', 'field'.
Brian Scott, who wriggles with non-essentials, does not understand
the foundations of existence of mathematics in the society: math can
be applied for the purpose of benefit of mankind, and that purpose
is evident in the concept/label 'continuous' but not in the labels of
'group', 'ring', 'field', which cannot be applied to everyday concepts
of 'group', 'ring', 'field', although they do have other applications.
Brian Scott declares:
> you've no business teaching.
Well, in my own insitute I have been regularily chosen as one of the
best teachers. And similarly abroad, at universities I have visited.
But maybe you have not attended my classes.
Brain Scott, a looser in this debate, appeals for ad hominem:
> your primary hobby of finding insignificant errors in published proofs.
Scott refers to http://www.math.hut.fi/~lounesto/counterexamples.htm
where I falsify statements published as theorems, by renowned
mathematicians, by counterexamples which satisfy all the assumptions
of the theorem without the conclusion being valid. Scott brings in
his opponent's 'hobbies'; such use of ad hominem shows that Scott
is short of relevant arguments. Scott evaluates my achievements as
'insignificant'. Thus, Scott pretends that he is an expert in my field
of speciality, Clifford algebras, and declares my work 'insignificant'.
To my knowledge, Brian Scott knows nothing about Clifford algebras.
By declaring himself an expert on a research topic, where he knows
nothing, Scott shows megalomanic tendencies, grandious dilusions.
As for evaluating signifigance of my list of counterexamples, some
of the mistake-makers regarded my counterexamples 'significant
enough to publish', but the same mistake-makers declared my list
'insignificant' after it was actually published. The editor of the
journal 'Advances in Applied Clifford Algebras' regarded my list
of counterexamples significant enough to publish in his journal
of 6 (1996), 69-104. He asked opinion of several members of
his editorial board of 10 specialists, 7 of whom I had listed as
mistake-makers. All regarded my list significant enough to
publish. But maybe Brian Scott knows better.
Pertti Lounesto
And what if you define f: R -> R U P(R), by
sin(1/x) for x not= 0
f(x) =
[-1,1] for x = 0?
Brian M. Scott
> And what if you define f: R -> R U P(R), by
>
> sin(1/x) for x not= 0
> f(x) =
> [-1,1] for x = 0?
You will be obliged first to define a topology on
R U P(R) if you wish to talk about continuity of f.
Brian M. Scott
Pertti Lounesto
That is right. If we simplify things, and consider only
mappings f:R->P(R), what interesting topologies could
we use on P(R)?
Pertti Lounesto
> Pertti Lounesto wrote:
>
> > And what if you define f: R -> R U P(R), by
> >
> > sin(1/x) for x not= 0
> > f(x) =
> > [-1,1] for x = 0?
>
> You will be obliged first to define a topology on
> R U P(R) if you wish to talk about continuity of f.
I introduce the usual topology on the xy-plane and
ask: is the graph of my function connected?
Dave Rusin
Pertti Lounesto <Pertti....@hut.fi> wrote:
>> The mathematical 'field' was not created ex nihilo; it is in fact
>> derived from the everyday noun 'field' and therefore shares most
>> of its etymology with that word.
>
>While you pretend to know things, can you point out what
>'human being' was the one who 'derived' the label 'field'?
>In French the concpet of 'field' is labelled 'corps commutative'.
>Maybe you, as somebody who knows these things, can tell
>us how the the concept of 'non-commutative field' became
>labelled by the term 'corps' in French, and who 'derived' it?
There is a rather long discussion of the origin of "field" as a
mathematical term at
http://members.aol.com/jeff570/f.html
(a site which I have found reputable for origins of math terms).
The article begins
FIELD. The term Zahlkörper (body of numbers) is due to Richard
Dedekind (1831-1916) (Kline, page 1146). The term appears in
Stetigkeit und Irrationale Zahlen (Continuity & Irrational Numbers).
...
I would imagine the French term also derives from the German.
Stan Brown
>misinterpretation or a distortion of anything Pertti writes if anyone
>made any sense out of it
I'm a slow learner -- I replied twice to him (her?) before I realized
that.
Maybe s/he's Oh Yeah's alter ego?
Virgil
<Pertti....@hut.fi> wrote:
>And what if you define f: R -> R U P(R), by
>
> sin(1/x) for x not= 0
> f(x) =
> [-1,1] for x = 0?
This is not a real function, since its value at x=0 is not a real number,
but a set of real numbers.
How would you calculate the value of |f(epsilon) - f(0)| in order to
test the function's continuity at x = 0?
--
Virgil
vm...@frii.com
Brian M. Scott
> There is a rather long discussion of the origin of "field" as a
> mathematical term at
> http://members.aol.com/jeff570/f.html
> (a site which I have found reputable for origins of math terms).
Nice site; thanks for the pointer.
Brian M. Scott
Ronald Bruck
<Pertti....@hut.fi> wrote:
:"Brian M. Scott" wrote:
:
:> Pertti Lounesto wrote:
:>
:> > And what if you define f: R -> R U P(R), by
:> >
:> > sin(1/x) for x not= 0
:> > f(x) =
:> > [-1,1] for x = 0?
:>
:> You will be obliged first to define a topology on
:> R U P(R) if you wish to talk about continuity of f.
:
:I introduce the usual topology on the xy-plane and
:ask: is the graph of my function connected?
Of course it is.
But introducing a topology on the plane is NOT the same as introducing a
topology on P(R) (assuming by that you mean the power set of R).
Multivalued functions are perfectly legitimate; they can be described in
terms of classical functions (by using the power set), or by relations.
But continuity becomes a lot more problematic, and impossible until you
describe a topology.
--Ron Bruck
Ronald Bruck
<Pertti....@hut.fi> wrote:
:"Brian M. Scott" wrote:
:
:> Pertti Lounesto wrote:
:>
:> > And what if you define f: R -> R U P(R), by
:> >
:> > sin(1/x) for x not= 0
:> > f(x) =
:> > [-1,1] for x = 0?
:>
:> You will be obliged first to define a topology on
:> R U P(R) if you wish to talk about continuity of f.
:
:That is right. If we simplify things, and consider only
:mappings f:R->P(R), what interesting topologies could
:we use on P(R)?
Ah, missed seeing this before I replied to the earlier post.
On P(R) itself, I dunno of a useful one. Usually one describes a topology
on the closed sets, or the closed bounded sets, or the closed convex sets
etc. For example, the Hausdorff metric on closed bounded sets is defined
by
dist(A,B) = inf { r > 0 : A \subset N_r(B) and B \subset N_r(A)}
where N_r(S) is the collection of all points which are within r of a point
of the set S (in the usual Euclidean metric).
Modifying your function in the obvious way to be
f(x) = {sin(1/x)} for x \ne 0, [-1,1] for x = 0,
(so it's now from R into P(R)), it's still not continuous at 0, because
the Hausdorff distance from f(x) to f(0) is always at least 1.
--Ron Bruck
Brian M. Scott
> In article <37626260...@hut.fi>, Pertti Lounesto
> <Pertti....@hut.fi> wrote:
> :"Brian M. Scott" wrote:
> :> Pertti Lounesto wrote:
> :> > And what if you define f: R -> R U P(R), by
> :> >
> :> > sin(1/x) for x not= 0
> :> > f(x) =
> :> > [-1,1] for x = 0?
> :> You will be obliged first to define a topology on
> :> R U P(R) if you wish to talk about continuity of f.
> :That is right. If we simplify things, and consider only
> :mappings f:R->P(R), what interesting topologies could
> :we use on P(R)?
> Ah, missed seeing this before I replied to the earlier post.
> On P(R) itself, I dunno of a useful one.
If X is a topological space, let A(X) be the collection
of non-empty subsets of X topologized as follows. For
each n in N and each (n+1)-tuple <V(0), ..., V(n)> of
open sets in X, define <<V(0), ..., V(n)>> = {A in A(X):
A is a subset of V(0) U ... U V(n) and A meets each V(i)
in a non-empty set}. The sets <<V(0), ..., V(n)>> form
a base for the Vietoris topology on A(X).
Another topology on A(X) is the Pixley-Roy topology; its
base is the collection of sets [A,V], for A in A(X) and
V open in X, where [A,V] = {S : A is a subset of S and S
is a subset of V}. This is finer than the Vietoris
topology and is most interesting when X is metrizable.
In each case it's often more interesting to look at a
subspace, e.g., the finite, compact, or closed non-empty
subsets of X instead of all of them. For Pertti's
example we could use the compact sets, but it's not
continuous wrt either topology: for each n [-1,1] has a
Vietoris nbhd in which every point is a set of cardinality
at least n.
Brian M. Scott
Pertti Lounesto
Then f is not continuous, but its graph is connected (loosly in
xy-plane).
Define f:R\{odd*pi/2}-> R by f(x) = tan(x).
Then f is continuous, but its graph is not connected.
Thus, it is not appropriate to introduce continuity by handwaving:
a function is continuous if its graph is connected.
Pertti Lounesto
> Quoth ry...@edge.net (Darrell Ryan) in alt.algebra.help:
> >It would be a
> >misinterpretation or a distortion of anything Pertti writes if anyone
> >made any sense out of it
>
> I'm a slow learner -- I replied twice to him (her?) before I realized
> that.
Indeed, you are a slow learner, as I told you and Darrell Ryan on my
posting of April 29, 1998 (reposted below). You two are poor losers:
you have not yet admitted yourself that you lost our debate about
advantages of the use of counterexamples in classroom teaching.
On April 29, 1998 Pertti Lounesto wrote in alt.algebra.help:
> Stan Brown and Darrell Ryan both defended Brown's counterexample,
explaining
> that it is as good as Lounesto's counterexample in classroom teaching.
Then both
> Brown and Ryan admitted that they do not have fresh teaching experience
in the
> classroom. From their postings it is clear that Stan Brown and Darrell
Ryan are not
> familiar with modern learning theories, cognitive psychology, in
particular Piagetian or
> some newer theories of cognition. All relevant terminology of learning
theories is
> missing from their postings. Their lack of adequate cognitive basis
prevents them
> from understanding (= assimilating into their present cognitive
structures) my
> arguments. The discussion has not resulted in any learning in Stan
Brown or Darrell
> Ryan (= cognitive growth did not happen = knowledge structures were not
altered),
> because of the affective problems of Darrell Ryan and Stan Brown (both
focus on
> insults, rather than benefiting my teaching efforts).
Curiously, at that thread on counterexamples, Brian Scott also sided Stan
Brown
and Darrell Ryan. Brown, Ryan and Scott, for short BRS, take advantage of
my
teaching efforts and learn some mathematics and debating styles from my
page
http://www.math.hut.fi/~lounesto/counterexamples.htm
After BRS learn some mathematics, they can publish their results in math
journals, and refer to each other, for instance, so that B refers to R, R
refers to S, and S refers to B. They can hope that no one notices their
method, and have high citations in SCI.
Brian M. Scott
> Quoth ry...@edge.net (Darrell Ryan) in alt.algebra.help:
> >It would be a
> >misinterpretation or a distortion of anything Pertti writes if anyone
> >made any sense out of it
> I'm a slow learner -- I replied twice to him (her?) before I realized
> that.
Him. <Pertti> is a Finnish borrowing of a pet form of some Germanic
name with second element <-bert>, like <A(da)lbert>.
Brian M. Scott
Pertti Lounesto
> In article <375F3A78...@hut.fi>,
> Pertti Lounesto <Pertti....@hut.fi> wrote:
> >> The mathematical 'field' was not created ex nihilo; it is in fact
> >> derived from the everyday noun 'field' and therefore shares most
> >> of its etymology with that word.
> >
> >While you pretend to know things, can you point out what
> >'human being' was the one who 'derived' the label 'field'?
> >In French the concpet of 'field' is labelled 'corps commutative'.
> >Maybe you, as somebody who knows these things, can tell
> >us how the the concept of 'non-commutative field' became
> >labelled by the term 'corps' in French, and who 'derived' it?
>
> There is a rather long discussion of the origin of "field" as a
> mathematical term at
> http://members.aol.com/jeff570/f.html
> (a site which I have found reputable for origins of math terms).
>
> The article begins
> FIELD. The term Zahlkörper (body of numbers) is due to Richard
> Dedekind (1831-1916) (Kline, page 1146). The term appears in
> Stetigkeit und Irrationale Zahlen (Continuity & Irrational Numbers).
> ...
> I would imagine the French term also derives from the German.
The article does not explain why Körper, while translated into French
as corps (which is not commutative, though), was translated into English
as field. If you want to know the definition of a field, take look at
sci.physics thread 'what is the definition of a field?', where it is
written:
On June 13, 1999 Peter Wilkie wrote:
>Fields in physics are extremely closely related to the mathematical
objects
>call fields. Of particular importance in physics are conservative fields,
>which have a mathematical counterpart, also called conservative fields.
I understand that in a conservative field, the total energy is a sum of the
kinetic energy and potential energy. Thus, fields have direct applications
in potential theory. Besides fields, also energy is conserved, due to
symmetry of field equations in time translation. But not all symmetries
of field equations, like rotations in time-like planes, result in
conservation
of quantities (because the Hamiltonian does not commute with the
Lorentz transformations). This leads us to Hamilton, who was the first
one to introduce a non-commutative field, namely the skew-field of
quaternions. But this leads us to other fields, altogether.
Brian M. Scott
> Define f:R->P(R) by f(x) = {sin(1/x)} for x \ne 0, [-1,1] for x = 0.
> Then f is not continuous,
Depends on the topology assigned to P(R). Since you haven't given one,
we don't know.
> but its graph is connected (loosly in
> xy-plane).
The xy-plane is plainly irrelevant, since the graph lives in R x P(R)
(with whatever product topology this has after you decide how to
topologize P(R)). In fact there are functions from R to R that have
connected graphs but are not continuous; you might try to construct one.
> Define f:R\{odd*pi/2}-> R by f(x) = tan(x).
> Then f is continuous, but its graph is not connected.
Hardly surprising: neither is its domain.
> Thus, it is not appropriate to introduce continuity by handwaving:
> a function is continuous if its graph is connected.
Handwaving is probably never appropriate unless it is acknowledged as
such; then it may be an acceptable introduction to something. This
particular bit of handwaving may have some modest value if done properly
-- the graph is connected on a connected chunk of the domain, the image
of an interval is an interval, etc. -- but its value is limited, since
by and large students will be no more comfortable with the explanations
than with the original concept.
Brian M. Scott
Pertti Lounesto
Brian seems to be more cultivated than Stan. There is another borrowing,
from Bartholomeous, this time the first element Bart-. Do you know
what Bartholomeous stands for?
Pertti Lounesto
> For Pertti's example we could use the compact sets, but it's not
> continuous wrt either topology:
At 10:30am Brian Scott had already forgotten his topologies:
> Pertti Lounesto wrote:
>
> > Define f:R->P(R) by f(x) = {sin(1/x)} for x \ne 0, [-1,1] for x = 0.
> > Then f is not continuous,
>
> Depends on the topology assigned to P(R). Since you haven't given one,
> we don't know.
>
> > but its graph is connected (loosly in
> > xy-plane).
>
> The xy-plane is plainly irrelevant, since the graph lives in R x P(R)
> (with whatever product topology this has after you decide how to
> topologize P(R)).
Reread: 'loosly', which should indicate that your comment was already
taken care of. As for irrelevancy of graphing R-> P(R) as a subset of
of RxR, rather than of RxP(R), I can represent each element of P(R)
as a subset of R, and draw the graph, in the original and everyday
meaning of the word 'graph' (of a relation), and if the 'graph' looks
nice, I might get some insight in the function, independent of your
critics. In getting insight, I am not restricted to your concepts.
> students will be no more comfortable with the explanations
> than with the original concept.
Students have access to new concepts only through their earleir
concepts, which they are familiar with. Teaching of the 'original
concept' often requires reformation of existing cognitive structures,
not just adding a new piece of info. Thus, 'original concpet' can
seldom be taught directly.
John R Ramsden
wrote:
Surely use of the word "field" in an algebraic context is based
on its meaning abstracted from "physical expanse" to "area of
activity", for example "She's an expert in her field".
(Not to be confused with the old joke "My Uncle's a farmer, and
he's out standing in his field" :-)
Cheers
John R Ramsden (j...@redmink.demon.co.uk)
Pertti Lounesto
> Surely use of the word "field" in an algebraic context is based
> on its meaning abstracted from "physical expanse" to "area of
> activity", for example "She's an expert in her field".
That is right. The 'field' corresponds to 'corps' in French, and
there is 'corps diplomatique' at the Embassy of France in London,
the licence plates being signalled with CD.
> (Not to be confused with the old joke "My Uncle's a farmer, and
> he's out standing in his field" :-)
Unfortunately, the 'corps' of physicists confuse 'corps' with
'application', that is, 'field' with 'function'.
Darrell Ryan
> More
> seriously, which of these terms (if any) would you use to describe the
> f(x) = sin(1/x) type of discontinuity (define f(0) = 0, say)?
Good question. I'm not sure. I know it takes on all values between
-1 and 1 in an oscillating fashion, oscillating more rapidly as x-->0.
A question for you. Is this function continuous everywhere except
x=0? I know lim x-->0 does not exist so it is definitely not
continuous there, but what about everywhere else? I want to say yes
but am looking for confirmation.
--
Regards,
Darrell
J. Mayer
Darrell Ryan schrieb:
> "Brian M. Scott" wrote:
>
> > More
> > seriously, which of these terms (if any) would you use to describe the
> > f(x) = sin(1/x) type of discontinuity (define f(0) = 0, say)?
>
> A question for you. Is this function continuous everywhere except
> x=0? I know lim x-->0 does not exist so it is definitely not
> continuous there, but what about everywhere else? I want to say yes
> but am looking for confirmation.
Yes, it is continuous everywhere else. The composition of continuous
functions is again continuous and as 1/x is continuous everywhere except for
x=0 and sin is continuous everywhere, you have that f is continuous
everywhere except possibly x=0, where in fact it is discontinuous.
Jan Mayer
Brian M. Scott
> At 6:38am today Brian Scott wrote:
> > For Pertti's example we could use the compact sets, but it's not
> > continuous wrt either topology:
> At 10:30am Brian Scott had already forgotten his topologies:
Oh, I knew *my* topologies. But your post didn't acknowledge or quote
them, and you don't appear to know much about the subject, so I'm
certainly not going to try to guess what topology *you* had in mind --
if any.
> > Pertti Lounesto wrote:
> > > Define f:R->P(R) by f(x) = {sin(1/x)} for x \ne 0, [-1,1] for x = 0.
> > > Then f is not continuous,
> > Depends on the topology assigned to P(R). Since you haven't given one,
> > we don't know.
> > > but its graph is connected (loosly in
> > > xy-plane).
> > The xy-plane is plainly irrelevant, since the graph lives in R x P(R)
> > (with whatever product topology this has after you decide how to
> > topologize P(R)).
> Reread: 'loosly', which should indicate that your comment was already
> taken care of.
Hardly! If you really were talking about the Euclidean
topology on R^2, the graph *is* connected, and 'loosely'
is inappropriate. If you were talking about any other
topology), 'loosely' is meaningless since (a) we don't
know what topology you mean, and (b) you haven't defined
this new concept of 'loosely connected'.
> As for irrelevancy of graphing R-> P(R) as a subset of
> of RxR, rather than of RxP(R), I can represent each element of P(R)
> as a subset of R, and draw the graph, in the original and everyday
> meaning of the word 'graph' (of a relation),
Wrong. The graph of a relation, as that term is generally
used, is a combinatorial object consisting of edges and
vertices. And a picture of the relation (in the sense that
the diagonal is a picture of the identity function) is a
representation of it as a subset of R x P(R). It is
certainly also possible to represent it pictorially as a
subset of R x R, but this is a very different kind of
picture that must be interpreted according to a very
different set of conventions, and it is *not* the ordinary
and everyday meaning of 'graph' *of* *a* *function* *from*
R to P(R).
[...]
> > students will be no more comfortable with the explanations
> > than with the original concept.
> Students have access to new concepts only through their earleir
> concepts, which they are familiar with. Teaching of the 'original
> concept' often requires reformation of existing cognitive structures,
> not just adding a new piece of info. Thus, 'original concpet' can
> seldom be taught directly.
Typical: you snipped enough context to disguise the fact
that your comment isn't relevant. You're wasting our
time again.
Brian M. Scott
Brian M. Scott
> "Brian M. Scott" wrote:
> > More
> > seriously, which of these terms (if any) would you use to describe the
> > f(x) = sin(1/x) type of discontinuity (define f(0) = 0, say)?
> Good question. I'm not sure. I know it takes on all values between
> -1 and 1 in an oscillating fashion, oscillating more rapidly as x-->0.
If students press me to give this type a name, I usually just call it an
oscillatory discontinuity. Then I reassure them by pointing out that
jumps, poles, and removable discontinuities are much more common with
the usual run of functions at that level.
> A question for you. Is this function continuous everywhere except
> x=0?
Yes: except at 0 it's the composition of two continuous functions.
Brian M. Scott
Pertti Lounesto
> > Reread: 'loosly', which should indicate that your comment was already
> > taken care of.
>
> Hardly! If you really were talking about the Euclidean
> topology on R^2, the graph *is* connected, and 'loosely'
> is inappropriate. If you were talking about any other
> topology), 'loosely' is meaningless since (a) we don't
> know what topology you mean, and (b) you haven't defined
> this new concept of 'loosely connected'.
For the third time in 24 hours, you are either misinterpreting or
deliberately distorting what I intend: loosely does not refer to any
of the meanings you label to it, but to replacing RxP(R) by RxR
and visualizating points of RxP(R) as subsets of RxR. You
cannot prevent me from gaining insight from this improper
picture (as you admit, although you do not admit the picture to
be a graph, in the everyday sense on an everyday xy-paper).
Brian, you should have become a lawyer: a judge, a district
attorney or a counsellor. Their job is to find non-intended
interpretations and deliberate distortions of truth. And, at
least in my country, judges are payed for compulsive lying
on everyday paper, called minutes of session.
But, while you have an eye on mathematics, and insist on
orthogoxy, wihtout insight, consider a function f:C->C and
its graph in CxC. The graph you can visualize (or can you)
as a surface in R^4, identified with CxC. The function is
real differentiable, if its graph has a tangent plane. The
condition of complex differentiability can be interpreted as
certain attitude of the tangent plane. Can you describe this
attitude of the tangent plane of the graph, which as you
require 'lives in' CxC (although you have not defined 'living').
Nathaniel Silver
>Thus, it is not appropriate to introduce continuity by
>handwaving: a function is continuous if its graph is
>connected.
Effective teaching does involve some lying.
I invoke Polya, who wrote that if students
cannot discover mathematics, give
them the illusion they do.
The idea being, in modern terms, that the high of
"discovery" will motivate them to redouble their
efforts in order to experience more highs.
In learning, sometimes there is an affective
component, which transcends logical
explanation. When introducing a subtle
concept like continuity, one is forced to
lie at one point or another (no pun intended)
as you mathematicians have demonstrated.
Concocting correct and dry formalisms serve
students as does pouring acid on plant seeds.
Only the very strongest survive. And you risk teaching
only yourselves as you have so ably demonstrated.
Pertti Lounesto
> Effective teaching does involve some lying.
There are two kinds of learning. First, new information
is assimilated into existing cognitive structures. Second,
cognitive structures accomodate themselves into changes
in the environment of the learner. Accomodation proceeds
only slowly and gradually, through several phases. In the
first phases it is often effective to give a simplified model
of reality. When the mental construction is ready and the
scaffoldings are carried away, intermediate steps often seem
oversimplification or cheating.
J Lee Jaap
|>Quoth jaa...@asmobj.larc.nasa.gov (J Lee Jaap) in alt.algebra.help:
|>>(Remember, Einstein divided by his
|>>"Cosmological Constant", implicitly assuming that it was not zero.)
|>
|>And it turns out that it's very likely not zero after all.
|>
|>See the January 1999 /Scientific American/ and the 17 Dec 1998 /Science/.
|>I was privileged to attend a lecture last month by one of the
|>researchers.
But it's still bad math to divide by something that could be zero,
without making a rather big note to the effect.
--
J Lee Jaap <Jaa...@POBox.com>
Sheldon Mandel
Andreas Enbacka wrote:
> Can the function f(x) = tan x be said to be discontinuous (at point x =
> Pi/2) ?
>
yes!. there is no defined value for tan(x) at pi/2
>
> You make a mistake here:
>
> The first definition states that a function f only can be discontinuous at a
> point x0, if f(x0) exists.
It should read : if f(x0) does not exist.
> The other definition however states that a
> function f also can be said to be discontinuous outside its definition
> domain.
>
never heard of that, but it can be true - when discussing continuity, it's
always with the domain of the function in mind - outside of that, the function
isn't even defined, much less continuous!
Shelly