Dense vs. Continuous
The poster formerly known as Colleyville Alan
Bruckner. It is well-written, user friendly, and free!
They discuss the fact that the rational numbers are dense and describe the
concept thusly:
"The rational numbers are dense. They make an appearance in every
interval; there are no gaps, no intervals that miss having rational
numbers."
And later, they state
"...every real is as close as we please to a rational..."
Finally, they mention:
"For theoretical reasons this fact is of great importance too. It allows
many arguments to replace a consideration of the set of real numbers with
the smaller set of rationals."
Now intuitively, I understand that the real numbers includes the rational
and the irrational and so it seems to make sense to say in the third quoted
sentence "It allows many arguments to replace a consideration of the set of
real numbers with the smaller set of rationals" (the set of real numbers is
larger than the rationals).
Ok, the rationals are countably infinite but dense and the real numbers are
uncountably infinite. But if, looking only at the rational numbers, we can
always find an infinity of rational numbers between any two given rational
numbers, how can we say that the set is not continuous?
Yes, it is "missing" the irrationals and is a smaller set than the real
numbers, but from the definition of "dense", there does not seem to be any
"holes" in the set. I think that I understand the meaning of "dense", from
the author's description, but perhaps I am not correctly understanding what
the term implies. Any thoughts?
Thanks.
Alan
p.s. if anybody is looking for a free Analysis textbook, they authors have
both an elementary and a graduate level text available for free download
here:
http://classicalrealanalysis.com/download.aspx
Bill
"The poster formerly known as Colleyville Alan" <nos...@nospam.net> wrote in
message news:478ae305$0$5084$4c36...@roadrunner.com...
>I am reading through Elementary Real Analysis by Thompson, Bruckner &
>Bruckner. It is well-written, user friendly, and free!
>
> They discuss the fact that the rational numbers are dense and describe the
> concept thusly:
>
> "The rational numbers are dense. They make an appearance in every
> interval; there are no gaps, no intervals that miss having rational
> numbers."
>
> And later, they state
>
> "...every real is as close as we please to a rational..."
>
> Finally, they mention:
>
> "For theoretical reasons this fact is of great importance too. It allows
> many arguments to replace a consideration of the set of real numbers with
> the smaller set of rationals."
>
>
> Now intuitively, I understand that the real numbers includes the rational
> and the irrational and so it seems to make sense to say in the third
> quoted sentence "It allows many arguments to replace a consideration of
> the set of real numbers with the smaller set of rationals" (the set of
> real numbers is larger than the rationals).
>
> Ok, the rationals are countably infinite but dense and the real numbers
> are uncountably infinite. But if, looking only at the rational numbers,
> we can always find an infinity of rational numbers between any two given
> rational numbers, how can we say that the set is not continuous?
>
What do you mean by "the set is not continuous"? If you mean that the set
of rationals is not a path connected subset of the real numbers--then that
is correct.
The poster formerly known as Colleyville Alan
"Bill" <Bill_...@comcast.net> wrote in message
news:O8GdnVowOvH9eBfa...@comcast.com...
Ok, I was assuming that the rationals were not continuous since they were
"missing" the irrational numbers. The square root of two is between one and
two so that point would be "missing" if one used the rationals rather than
the reals. But perhaps my assumption was wrong; are the rationals a
contiuous set? The definition of "dense" seems to me to imply that they
are, but I assumed that only the reals were continuous and that if one were
looking at, say, the graph of a function it would be plotted along the real
number line (i.e. the standard Cartesian grid). So although we can get
infinitely close to, say, the square root of two, that exact value is not
defined in the set of rational numbers and it would seem that the graph of
the function would have an infinitely small hole at that point and thus not
be continuous.
As I say, I am not certain what dense implies. Does it imply contiuity? Do
we ignore the "missing" irrationals or is that a poor way to think about it?
I only have studied Calculus and so stuff like this is new to me.
The World Wide Wade
You haven't told us what it means for a set to be "continuous". Until
that is given a precise meaning, no one can answer.
victor_me...@yahoo.co.uk
That would be a category mistake: sets cannot be "continuous";
functions
can be continuous.
Victor Meldrew
"I don't believe it!"
The poster formerly known as Colleyville Alan
<victor_me...@yahoo.co.uk> wrote in message
news:934d3352-66f2-4f84...@e4g2000hsg.googlegroups.com...
> On 14 Jan, 04:20, "The poster formerly known as Colleyville Alan"
> <nos...@nospam.net> wrote:
>> how can we say that the set is not continuous?
>
> That would be a category mistake: sets cannot be "continuous";
> functions can be continuous.
Ok. I'm getting there slowly but surely. So we go back to Calculus II and
the definition of a continuous function to see whethere a given function is
continuous.
From Wolfram:
http://mathworld.wolfram.com/ContinuousFunction.html
"More concretely, a function in a single variable is said to be continuous
at point if
1. f(x_o) is defined, so that x_o is in the domain of f.
2. lim x->x_o exists for x in the domain of f.
3. lim x->x_o = f(x_o) "
If I say that x is an element of the rationals rather than the reals, then
the 2nd and 3rd parts would be satisfied for a function like y = f(x) = x.
But the first part, "...f(x_o) being defined so that x_o is in the domain
of f..." does not seem to hold. Yet the function f(x) = x would appear to me
to be continuous if you can get infinitely close to any real even though the
irrationals are not included. Does it make any sense at all to think of the
rationals as a domain for a function or should functions only be considered
in terms of the real number system and thereby eliminating all of this
confusion on my part :)
victor_me...@yahoo.co.uk
<nos...@nospam.net> wrote:
> <victor_meldrew_...@yahoo.co.uk> wrote in message
>
> From Wolfram:http://mathworld.wolfram.com/ContinuousFunction.html
>
> "More concretely, a function in a single variable is said to be continuous
> at point if
Phrases like "function in a single variable" are a bit A-Level :-(
> Does it make any sense at all to think of the
> rationals as a domain for a function
Of course!
> or should functions only be considered
> in terms of the real number system
ABSOLUTELY NOT!!!
A function can have *any* set as domain, and *any* set
as codomain. If one has a function f: A -> B and the sets
A and B have topologies on them then there is a notion
of continuity of the function f.
Dave L. Renfro
>> how can we say that the set is not continuous?
Victor Meldrew wrote:
> That would be a category mistake: sets cannot be "continuous";
> functions can be continuous.
There is the notion of a linearly ordered set (and a linear
order) being continuous, although the more common term for
this property (now at least; older books often used "continuous")
is completeness (the linear order notion, not the the Cauchy
sequence notion).
However, I don't know if the book in question uses "continuous"
in the discussion at hand or if this is the original poster's
injection of the term. (I have the book, and I'm somewhat
familiar with it, but it's at home and I'm not.)
Dave L. Renfro
Dave L. Renfro
> I have the book, and I'm somewhat
> familiar with it, but it's at home and I'm not.
Oh, I fogot. It's freely available in digital form now.
So let me rephrase this as "I'm not sufficiently motivated
in resolving the issue of whether the authors used "continuous"
in the context of describing sets to pursue the matter".
Dave L. Renfro
The poster formerly known as Colleyville Alan
"Dave L. Renfro" <renf...@cmich.edu> wrote in message
news:8ad65a97-db49-465d...@u10g2000prn.googlegroups.com...
To clarify, the authors never mentioned the term "continuous", they
mentioned that the rationals were "dense". I was trying to ascertain what
the difference was between "dense" and "continuous". I was (am)uncertain as
to whether even trying to compare "dense" and "continuous" is a valid
question or if it is an apples-to-oranges type of question.
So, let me try again. If I have a function y = f(x) = x, the graph would be
a 45* line through the origin in a standard Cartesian grid. The graph would
be a continuous line with no breaks since it would be using the reals as the
domain assuming that x is a random variable and that the function holds for
{x|x E R }
But, suppose I defined the x-axis to be the set of rationals, not the set of
reals (assuming it makes sense to do that - please inform me if that is an
idiotic concept). If I were to graph the same function and said the domain
was the set of rationals, would the function be considered a continuous
function?
On the one hand, in the interval between 1.41421 and 1.41422, the point
equal to the square root of 2 would be missing from such a graph, suggesting
to me that it is not continuous. On the other hand, if the set of rationals
is dense and if dense means that you can get infinitely close to the square
root of 2 using rationals, then perhaps the graph is considered to be
continuous. I simply have no idea one way or the other and I have no idea
whether the concept is even sensible. But hopefully, it is now clear that I
am trying to develop an intuitive understanding of what dense means and how
it is or is not related to the concept of continuous, whether through
functions, through sets or through some other thing. If those two concepts
belong in two different arenas, that would be worth knowing as well.
Thanks for your patience.
Alan
Dave Seaman
> "Dave L. Renfro" <renf...@cmich.edu> wrote in message
> news:8ad65a97-db49-465d...@u10g2000prn.googlegroups.com...
>> Dave L. Renfro wrote (in part):
>>> I have the book, and I'm somewhat
>>> familiar with it, but it's at home and I'm not.
>> Oh, I fogot. It's freely available in digital form now.
>> So let me rephrase this as "I'm not sufficiently motivated
>> in resolving the issue of whether the authors used "continuous"
>> in the context of describing sets to pursue the matter".
>> Dave L. Renfro
> To clarify, the authors never mentioned the term "continuous", they
> mentioned that the rationals were "dense". I was trying to ascertain what
> the difference was between "dense" and "continuous". I was (am)uncertain as
> to whether even trying to compare "dense" and "continuous" is a valid
> question or if it is an apples-to-oranges type of question.
We can't hope to determine the difference between two terms unless the
terms in question have previously been defined. We have a definition for
a dense set, but there is no commonly accepted definition for a
continuous set. As previously noted, it makes sense to speak of
*functions* being continuous, but not *sets*.
> So, let me try again. If I have a function y = f(x) = x, the graph would be
> a 45* line through the origin in a standard Cartesian grid. The graph would
> be a continuous line with no breaks since it would be using the reals as the
> domain assuming that x is a random variable and that the function holds for
> {x|x E R }
That's an entirely different and unrelated question. Yes, the identity
function on the reals is a continuous function. We still don't know what
a continuous *set* is.
> But, suppose I defined the x-axis to be the set of rationals, not the set of
> reals (assuming it makes sense to do that - please inform me if that is an
> idiotic concept). If I were to graph the same function and said the domain
> was the set of rationals, would the function be considered a continuous
> function?
More generally, the identity function on any topological space is always
continuous, if we are consistent in our choice of a topology. For
example, we could take the identity function defined on the integers, and
it would be a continuous function, even though there are huge gaps in the
set of integers.
You can find a definition of what it means for a function to be
continuous at
<http://mathworld.wolfram.com/ContinuousFunction.html>. Notice that the
definition doesn't say anything about "gaps". Also, there is nothing on
that page that talks about continuous *sets*, only continuous
*functions*.
> On the one hand, in the interval between 1.41421 and 1.41422, the point
> equal to the square root of 2 would be missing from such a graph, suggesting
> to me that it is not continuous. On the other hand, if the set of rationals
> is dense and if dense means that you can get infinitely close to the square
> root of 2 using rationals, then perhaps the graph is considered to be
> continuous. I simply have no idea one way or the other and I have no idea
> whether the concept is even sensible. But hopefully, it is now clear that I
> am trying to develop an intuitive understanding of what dense means and how
> it is or is not related to the concept of continuous, whether through
> functions, through sets or through some other thing. If those two concepts
> belong in two different arenas, that would be worth knowing as well.
I suspect the concept you are trying to get at is that of a connected
set, rather than a "continuous" set. The reals are a connected set, but
the rationals are not. See
<http://mathworld.wolfram.com/ConnectedSet.html>.
--
Dave Seaman
Oral Arguments in Mumia Abu-Jamal Case heard May 17
U.S. Court of Appeals, Third Circuit
<http://www.abu-jamal-news.com/>
Denis Feldmann
>
> "Dave L. Renfro" <renf...@cmich.edu> wrote in message
> news:8ad65a97-db49-465d...@u10g2000prn.googlegroups.com...
>> Dave L. Renfro wrote (in part):
>>
>>> I have the book, and I'm somewhat
>>> familiar with it, but it's at home and I'm not.
>>
>> Oh, I fogot. It's freely available in digital form now.
>> So let me rephrase this as "I'm not sufficiently motivated
>> in resolving the issue of whether the authors used "continuous"
>> in the context of describing sets to pursue the matter".
>>
>> Dave L. Renfro
>
> To clarify, the authors never mentioned the term "continuous", they
> mentioned that the rationals were "dense". I was trying to ascertain
> what the difference was between "dense" and "continuous".
But what on earth make you believe that the term "continous" has any
meaning in this context ? "Dense" is a very classical term applied to
subsets of a topological space. "Continuous" is only (classiccally)
applied to functions. Your questions seem, to be the least, badly
formulated. For any comparison, you would at least have to define *your*
terms, as you admit yourself that they are not (well, for "continuous")
the author's term
I was
> (am)uncertain as to whether even trying to compare "dense" and
> "continuous" is a valid question or if it is an apples-to-oranges type
> of question.
>
> So, let me try again. If I have a function y = f(x) = x, the graph
> would be a 45* line through the origin in a standard Cartesian grid.
> The graph would be a continuous line
You perhaps mean a connected set, or a path-connected set. "Continuous"
has no clear meaning in this context.
with no breaks since it would be
> using the reals as the domain assuming that x is a random variable
Again, you are using words out of their context. A "random variable" is
not what you mean.
and
> that the function holds for {x|x E R }
>
> But, suppose I defined the x-axis to be the set of rationals, not the
> set of reals (assuming it makes sense to do that - please inform me if
> that is an idiotic concept).
You are now speaking of a Q to Q map
If I were to graph the same function and
> said the domain was the set of rationals, would the function be
> considered a continuous function?
Of course. It is *never* (in math) a matter of opinion, but of
definitions and proofs. For any topological set A, the identity function
x ->x from A to A is continuous.
>
> On the one hand, in the interval between 1.41421 and 1.41422, the point
> equal to the square root of 2 would be missing from such a graph,
> suggesting to me that it is not continuous.
What you mean is that *as a subset of the plane*, the graph is not
"continuous" (actually not connected) ; this is correct.
On the other hand, if the
> set of rationals is dense and if dense means that you can get infinitely
> close to the square root of 2 using rationals, then perhaps the graph is
> considered to be continuous.
No. 1) a set is never said "continuous" (in standard terminology), 2)
nothing is ever 'considered' (well, except for borderline-cases like the
styatus of the empty set, and even then), but proved or disproved.
I simply have no idea one way or the other
> and I have no idea whether the concept is even sensible. But hopefully,
> it is now clear that I am trying to develop an intuitive understanding
> of what dense means
Well, you should work in the opposite direction. Learn the definition,
see if they correspond to your intuition, and eventually develop
ifferent definitions to suit your needs, as the consequences of the
definitions are already fixed and, for instance, you have not even yet
realized that "dense" has no meaning without an external set of
reference (for instance, your graph above is *not* dense in the whole
plane).
The poster formerly known as Colleyville Alan
news:fmhkjj$nmv$1...@netfinity.fr...
> The poster formerly known as Colleyville Alan a écrit :
>>
>> So, let me try again. If I have a function y = f(x) = x, the graph would
>> be a 45* line through the origin in a standard Cartesian grid. The graph
>> would be a continuous line
>
> You perhaps mean a connected set, or a path-connected set. "Continuous"
> has no clear meaning in this context.
I mean that to draw the graph of this function, you never need to lift your
pencil from the paper.
> with no breaks since it would be
>> using the reals as the domain assuming that x is a random variable
>
> Again, you are using words out of their context. A "random variable" is
> not what you mean.
Correct, the word random does not belong there.
> and
>> that the function holds for {x|x E R }
>>
>> But, suppose I defined the x-axis to be the set of rationals, not the set
>> of reals (assuming it makes sense to do that - please inform me if that
>> is an idiotic concept).
>
> You are now speaking of a Q to Q map
Thank you for clarifying.
> If I were to graph the same function and
>> said the domain was the set of rationals, would the function be
>> considered a continuous function?
>
> Of course. It is *never* (in math) a matter of opinion, but of definitions
> and proofs. For any topological set A, the identity function x ->x from A
> to A is continuous.
Yes, I understand that it is a matter of definitions and proofs, when I used
the word "considered" I did not mean to imply that it was a matter of
opinion.
>> On the one hand, in the interval between 1.41421 and 1.41422, the point
>> equal to the square root of 2 would be missing from such a graph,
>> suggesting to me that it is not continuous.
>
> What you mean is that *as a subset of the plane*, the graph is not
> "continuous" (actually not connected) ; this is correct.
I have never studied topology and I do not know what you mean by "a subset
of the plane".
> On the other hand, if the
>> set of rationals is dense and if dense means that you can get infinitely
>> close to the square root of 2 using rationals, then perhaps the graph is
>> considered to be continuous.
>
> No. 1) a set is never said "continuous" (in standard terminology),
One definition of a continuous function that I heard from a math professor
was that it was a function whose graph had no breaks, i.e. if you could draw
such a function without lifting your pencil from the paper, it was
continuous. So, when I askd if the graph is continuous, I mean are there
any gaps in it or is the line unbroken? When I ask if there are any gaps in
it, I mean even at the infinitesimal level.
Does an irrational number like the square root of 2 create a gap in the
graph if the domain is defined to be the set of rationals, or does the fact
that you can get infinitely close to this number using rationals mean that
there is no such gap?
victor_me...@yahoo.co.uk
<nos...@nospam.net> wrote:
> I mean that to draw the graph of this function, you never need to lift your
> pencil from the paper.
Blimey, now maths depends on the properties of pencils! :-)
victor_me...@yahoo.co.uk
wrote:
>
> Well, you should work in the opposite direction. Learn the definition,
> see if they correspond to your intuition, and eventually develop
> ifferent definitions to suit your needs, as the consequences of the
> definitions are already fixed and, for instance, you have not even yet
> realized that "dense" has no meaning without an external set of
> reference (for instance, your graph above is *not* dense in the whole
> plane).
There is another use of the term "dense" which has some relevance
here.
A totally ordered set is dense if betweem any two distinct elements
there is another one (and so infinitely many of them). Under this
definition both Q and R are dense.
The poster formerly known as Colleyville Alan
<victor_me...@yahoo.co.uk> wrote in message
news:15515dc2-1b34-43a8...@k39g2000hsf.googlegroups.com...
> On 15 Jan, 13:09, "The poster formerly known as Colleyville Alan"
> <nos...@nospam.net> wrote:
>
>> I mean that to draw the graph of this function, you never need to lift
>> your
>> pencil from the paper.
>
> Blimey, now maths depends on the properties of pencils! :-)
Hey, the guy who told me that has a Phd in math and I was a Pre-Calculus
student so not exactly in a position to argue!
(plus I couldn't think of anything wrong with the argument).
Dave Seaman
What he told you is reasonable if you are only talking about functions
defined on the reals. It fails for functions from Q -> Q, since such
functions can be continuous even though the graph necessarily contains
gaps.
The poster formerly known as Colleyville Alan
"Dave Seaman" <dse...@no.such.host> wrote in message
news:fmjvk6$7bc$1...@mailhub227.itcs.purdue.edu...
> On Tue, 15 Jan 2008 21:49:25 -0600, The poster formerly known as
> Colleyville Alan wrote:
>
>><victor_me...@yahoo.co.uk> wrote in message
>> news:15515dc2-1b34-43a8...@k39g2000hsf.googlegroups.com...
>>> On 15 Jan, 13:09, "The poster formerly known as Colleyville Alan"
>>> <nos...@nospam.net> wrote:
>>>
>>>> I mean that to draw the graph of this function, you never need to lift
>>>> your
>>>> pencil from the paper.
>>>
>>> Blimey, now maths depends on the properties of pencils! :-)
>
>
>> Hey, the guy who told me that has a Phd in math and I was a Pre-Calculus
>> student so not exactly in a position to argue!
>> (plus I couldn't think of anything wrong with the argument).
>
> What he told you is reasonable if you are only talking about functions
> defined on the reals. It fails for functions from Q -> Q, since such
> functions can be continuous even though the graph necessarily contains
> gaps.
Thanks for the explanation. It seems that all we dealt with in Calculus
were real-valued functions so I have been struggling with functions defined
on other sets of numbers. This helps a lot.
Alan
Dave L. Renfro
> I am reading through Elementary Real Analysis by
> Thompson, Bruckner & Bruckner. It is well-written,
> user friendly, and free!
I'm revisiting this thread because I was thinking
a few days ago (away from internet access) that it
might be a good idea to give you some suggestions
for other books to look at. BBT is a great book,
but given your background (just finished calculus),
there are other books that might be a better fit for
you at this time. Also, given your predisposition
for careful arguments and your habit of self-study
(yes, I've read many of your posts in sci.math), many
slightly lower level books would probably not work very
well for you. The following should fit you very well,
however:
Michael Spivak, "Calculus", 3'rd edition, Publish
or Perish, 1994.
Godfrey Harold Hardy, "A Course of Pure Mathematics",
10'th edition, Cambridge University Press, 1993.
Richard Courant and Herbert Robbins, "What is Mathematics?",
2'nd edition (edited by Ian Stewart), Oxford University
Press, 1993. [Begin with Chapter VI: Functions and
Limits.]
Dave L. Renfro
The poster formerly known as Colleyville Alan
Thanks for the suggestions. It so happens that I have 2 of the 3 mentioned
(Spivak and Courant) and have read them sporadically rather than really made
an attempt to study them. Sounds like that time has come. I've glanced at
the Hardy book at the bookstore and it looked interesting, but I had too
much other stuff to read at that time. I will get to it eventually.