Continuous real function with an "irrational" anomaly
Dustan
I want to know if a function meeting the following description exists
(intuitively, it seems like it should exist, but I can't figure out
how to go about constructing such a function):
Continuous, and mostly closed under the rationals, except for exactly
one rational that yields an irrational value.
In other words:
Does there exist a continuous function F and a rational R such that
for every rational r, F(r) is rational if and only if r != R?
Note the requirement that the function be continuous. So, does the
function exist? If so, can you provide an explicitly constructed
example (and if so, provide). If not, why not? While I would be
interested in a proof, I am more interested in attaining an intuitive
grasp of why such a function should be impossible (so if a proof
should help me get that understanding, then prove away).
I'm sure this probably seems pretty basic; I have to confess that I
am, indeed (*gasp*), a high school student. I know that this newsgroup
is anti-high school (or college freshmen, sophomore, junior... well,
just about anyone without a PhD), so I've posted from Google Groups so
that anyone who really cares enough that a high school student managed
to enter their internet sanctuary will most likely have already
filtered out GG.
All answers appreciated.
José Carlos Santos
> I wasn't sure how exactly to word the title; sorry for the vagueness.
> I want to know if a function meeting the following description exists
> (intuitively, it seems like it should exist, but I can't figure out
> how to go about constructing such a function):
>
> Continuous, and mostly closed under the rationals, except for exactly
> one rational that yields an irrational value.
>
> In other words:
>
> Does there exist a continuous function F and a rational R such that
> for every rational r, F(r) is rational if and only if r != R?
Yes, such a function exists. Let (q_n)_n be a strictly decreasing
sequence of rational functions greater than 0 such that the limit
lim_n q_n exists and it is an irrational number _y_. Now define
f(x) = q_1 if x >= 1. If 1/2 <= x <= 1, then define
f(x) = q_2 + 2(x - 1/2)(q_1 - q_2).
So, f(1/2) = q_2, f(1) = q_1 and _f_ is linear in [1/2,1]. Then, if
1/3 <= x <= 1/2, define
f(x) = q_3 + 6(x - 1/3)(q_2 - q_3).
So, f(1/3) = q_3, f(1/2) = q_2 and _f_ is linear in [1/3,1/2]. And so
on. Define f(0) = y and, when x < 0, define f(x) = f(-x).
This function _f_ is continuous, it maps non-zero rational numbers into
rational numbers, but f(0) = y, which is irrational, by hypothesis.
> Note the requirement that the function be continuous. So, does the
> function exist? If so, can you provide an explicitly constructed
> example (and if so, provide). If not, why not? While I would be
> interested in a proof, I am more interested in attaining an intuitive
> grasp of why such a function should be impossible (so if a proof
> should help me get that understanding, then prove away).
>
> I'm sure this probably seems pretty basic; I have to confess that I
> am, indeed (*gasp*), a high school student. I know that this newsgroup
> is anti-high school (or college freshmen, sophomore, junior... well,
> just about anyone without a PhD),
Really?! I have been posting here for years and I had never noticed it.
> so I've posted from Google Groups so
> that anyone who really cares enough that a high school student managed
> to enter their internet sanctuary will most likely have already
> filtered out GG.
>
> All answers appreciated.
Glad to know that you will appreciate my answer then.
Best regards,
Jose Carlos Santos
G. A. Edgar
F(0) = sqrt(2)
F maps the rationals > 0 bijectively, preserving order, onto the
rationals > sqrt(2)
F maps the rationals < 0 bijectively, preserving order, onto the
rationals < sqrt(2)
Extend F to the reals by continuity
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
fjb...@yahoo.com
wrote:
> Will this work?
>
> F(0) = sqrt(2)
>
> F maps the rationals > 0 bijectively, preserving order, onto the
> rationals > sqrt(2)
Can you explain how to construct such an F?
José Carlos Santos
>> Will this work?
>>
>> F(0) = sqrt(2)
>>
>> F maps the rationals > 0 bijectively, preserving order, onto the
>> rationals > sqrt(2)
>
> Can you explain how to construct such an F?
It seems to me that G. A. Edgar was _asking_ to the OP whether or not
such a function F exists. He certainly did not claim that it does.
Dustan
> Glad to know that you will appreciate my answer then.
Thanks. I totally understood that. Much... erm, appreciated.
:)
Dustan
wrote:
> Will this work?
>
> F(0) = sqrt(2)
>
> F maps the rationals > 0 bijectively, preserving order, onto the
> rationals > sqrt(2)
>
> F maps the rationals < 0 bijectively, preserving order, onto the
> rationals < sqrt(2)
Yes, that'll work, but that doesn't really describe a unique function,
much less an explicit construction of such a function, now does it?
G. A. Edgar
> >
> > F(0) = sqrt(2)
> >
> > F maps the rationals > 0 bijectively, preserving order, onto the
> > rationals > sqrt(2)
> >
> > F maps the rationals < 0 bijectively, preserving order, onto the
> > rationals < sqrt(2)
>
> Yes, that'll work, but that doesn't really describe a unique function,
> much less an explicit construction of such a function, now does it?
quoting:
"I want to know if a function meeting the following description exists"
--
ge...@math.mq.edu.au
> On 24-08-2008 16:30, Dustan wrote:
>
> > Does there exist a continuous function F and a rational R such that
> > for every rational r, F(r) is rational if and only if r != R?
>
> Yes, such a function exists. Let (q_n)_n be a strictly decreasing
> sequence of rational functions greater than 0 such that the limit
> lim_n q_n exists and it is an irrational number _y_.
Rational functions? I think you mean rational numbers.
> Now define
> f(x) = q_1 if x >= 1. If 1/2 <= x <= 1, then define
>
> f(x) = q_2 + 2(x - 1/2)(q_1 - q_2).
>
> So, f(1/2) = q_2, f(1) = q_1 and _f_ is linear in [1/2,1].
If you really mean rational functions, you wouldn't expect
f to be linear in that interval.
--
GM
ge...@math.mq.edu.au
> I know that this newsgroup
> is anti-high school (or college freshmen, sophomore, junior... well,
> just about anyone without a PhD),
What garbage. Many of us are anti-moron, even if the moron has a PhD
(there is more than one PhD holder who has been given a hard time
here),
and we are pro-mathematics, even if the person wanting to talk math
is
in high school. We ask for civility, and return it.
--
GM
Dustan
Eh?
Sorry, but my definition of civility is slightly more restrained than
freely criticizing people for things that they simply don't know
about. See William Elliot & Michael Press:
http://groups.google.com/group/sci.math/browse_frm/thread/b2d34ef7ebba4c59/
Yes, I hold a grudge.
Angus Rodgers
Perfectly understandable! I was wondering if you had evidence, perhaps
from discussions outside sci.math, of an "anti-high school" bias. Does
sci.math have some kind of nasty reputation somewhere else, or are your
feelings entirely the result of your own experiences here? I can't say
I've seen any such bias myself, although I'm well aware that people can
be aggressive here. FWIW, I don't have a PhD; I have bruises of my own
from sci.math (but they've healed); I think you have conducted yourself
very well, in the threads that I've seen; and I hope that you settle in,
and that you find sci.math valuable (as I do). You already seem capable
of making valuable contributions yourself. The clarity of your writing
puts most of us to shame. Just carry on as you are, and you'll be fine.
P.S. I think perhaps people took offence at your use of the word "nerd"
in the Subject line of the thread you referred to. That, too, would be
perfectly understandable! (No doubt those involved will explain whether
this was the case.) That word is often used in an unjustly contemptuous
way, expressive (to my mind at least) of a dumbed-down culture. Perhaps
you were being defensive, having been wrongly looked down on as a "nerd"
yourself? If so, fear not! There is need for that here. If you settle
in, you'll soon learn what liberties you can take; also, why people here
are quite often defensive. ("Defensive? Defensive like Genghis Khan!" -
John Cleese.)
--
Angus Rodgers
(twirlip@ eats spam; reply to angusrod@)
Contains mild peril
Angus Rodgers
In a previous thread, this post by Dave Renfro was mentioned:
<http://groups.google.ca/group/sci.math/msg/c46a3913f6a3189>
The paper by Barth & Schneider [5] shows that such a function
can be constructed.
(Usual caveats: I haven't actually seen the paper in question;
and it's long past my bedtime, so I may be talking ********.)
Dustan
wrote:
> > > Will this work?
>
> > > F(0) = sqrt(2)
>
> > > F maps the rationals > 0 bijectively, preserving order, onto the
> > > rationals > sqrt(2)
>
> > > F maps the rationals < 0 bijectively, preserving order, onto the
> > > rationals < sqrt(2)
>
> > Yes, that'll work, but that doesn't really describe a unique function,
> > much less an explicit construction of such a function, now does it?
>
> quoting:
> "I want to know if a function meeting the following description exists"
quoting myself:
"If so, can you provide an explicitly constructed example (and if so,
provide)."
(but don't worry about it; José took care of it for you. ;)
Dustan
> On Sun, 24 Aug 2008 16:48:23 -0700 (PDT), Dustan
>
>
>
> <DustanGro...@gmail.com> wrote:
> >On Aug 24, 5:45 pm, ge...@math.mq.edu.au wrote:
> >> On Aug 25, 1:30 am, Dustan <DustanGro...@gmail.com> wrote:
>
> >> > I know that this newsgroup
> >> > is anti-high school (or college freshmen, sophomore, junior... well,
> >> > just about anyone without a PhD),
>
> >> What garbage. Many of us are anti-moron, even if the moron has a PhD
> >> (there is more than one PhD holder who has been given a hard time
> >> here),
> >> and we are pro-mathematics, even if the person wanting to talk math
> >> is
> >> in high school. We ask for civility, and return it.
>
> >Eh?
>
> >Sorry, but my definition of civility is slightly more restrained than
> >freely criticizing people for things that they simply don't know
> >about. See William Elliot & Michael Press:
>
> >Yes, I hold a grudge.
>
> Perfectly understandable! I was wondering if you had evidence, perhaps
> from discussions outside sci.math, of an "anti-high school" bias. Does
> sci.math have some kind of nasty reputation somewhere else, or are your
> feelings entirely the result of your own experiences here?
Nah, I don't think sci.math has _any_ reputation outside of sci.math
itself. I could, of course, be wrong, but that's my experience.
> I can't say
> I've seen any such bias myself, although I'm well aware that people can
> be aggressive here. FWIW, I don't have a PhD; I have bruises of my own
> from sci.math (but they've healed); I think you have conducted yourself
> very well, in the threads that I've seen; and I hope that you settle in,
> and that you find sci.math valuable (as I do). You already seem capable
> of making valuable contributions yourself. The clarity of your writing
> puts most of us to shame. Just carry on as you are, and you'll be fine.
I've actually settled outside of Usenet lately; this is my first post
(well, thread) in quite a while (look at the date of the thread I
linked to). But of course I'll pop in every now and then with a
question or perhaps even notice an interesting thread to join.
> P.S. I think perhaps people took offence at your use of the word "nerd"
> in the Subject line of the thread you referred to. That, too, would be
> perfectly understandable! (No doubt those involved will explain whether
> this was the case.) That word is often used in an unjustly contemptuous
> way, expressive (to my mind at least) of a dumbed-down culture. Perhaps
> you were being defensive, having been wrongly looked down on as a "nerd"
> yourself? If so, fear not! There is need for that here. If you settle
> in, you'll soon learn what liberties you can take; also, why people here
> are quite often defensive. ("Defensive? Defensive like Genghis Khan!" -
> John Cleese.)
I didn't really think about that. I would hope that most people into
math enough to join a Usenet group devoted entirely to it would be
perfectly comfortable being labeled a nerd. I've personally gotten way
past the stage of thinking of it as a derogatory term and more as
simply who I am. I've seen self-declared nerds all over the place,
both on the internet and in real life.
It's certainly nice to hear (well, read) from someone who understands.
Thanks for sharing your thoughts.
Angus Rodgers
D'oh! Cantor (1895). I'm going to bed! (And I'll try not to have
nightmares picturing what such a function actually "looks like".)
ge...@math.mq.edu.au
>
> Yes, I hold a grudge.
So your opinion that this newsgroup is hostile to anyone without a
PhD
is based on ... two posts in one thread? Two posts, out of the
hundreds
of thousands - two posts?? And, at that, two posts that you may have
misinterpreted?
Look again at the post from Michael Press: he wrote, "You can discount
my
abilities as much as you like without risk of being shown to have
underestimated them."
Well, it's a strange thing to have posted, but on my reading he's
saying
that he doesn't have any mathematical ability. The only one he's
criticizing
there is himself!
He also wrote, "You seriously underestimate the mathematical
abilities
of the folks with whom you argue." While I wouldn't have written
that,
it doesn't strike me as anything to carry a grudge about for six
months.
Meanwhile, you did get help from David Cantrell (which you
acknowledged)
and from several other posters in the thread - why don't you take
that
as evidence that sci.math is pro-highschool?
I don't know whether you have a chip on your shoulder, or were just
having
a bad day, or read a lot of stuff into two posts that wasn't really
in
them, but I'd say you've failed to make any kind of case against
sci.math.
And by the way, Michael corrected your "free reign" to "free rein,"
and
you went and wrote "free reign" again! If you're not going to bother
to
take on board what people write, why do you bother to post?
--
GM
José Carlos Santos
>>> Does there exist a continuous function F and a rational R such that
>>> for every rational r, F(r) is rational if and only if r != R?
>> Yes, such a function exists. Let (q_n)_n be a strictly decreasing
>> sequence of rational functions greater than 0 such that the limit
>> lim_n q_n exists and it is an irrational number _y_.
>
> Rational functions? I think you mean rational numbers.
Sure. Thanks for the correction.
>> Now define
>> f(x) = q_1 if x >= 1. If 1/2 <= x <= 1, then define
>>
>> f(x) = q_2 + 2(x - 1/2)(q_1 - q_2).
>>
>> So, f(1/2) = q_2, f(1) = q_1 and _f_ is linear in [1/2,1].
>
> If you really mean rational functions, you wouldn't expect
> f to be linear in that interval.
Of course. Besides the fact that the notation would then be incoherent.
I should have written then, for instance,
f(x) = q_2(x) + 2(x - 1/2)(q_1(x) - q_2(x)).
Angus Rodgers
>D'oh! Cantor (1895).
In more detail:
Georg Cantor, "Contributions to the theory of transfinite numbers",
first article, Mathematische Annalen 46 (1895), pp. 481-512. English
translation in Dover paperback (1955). The proof is in section 9,
"The Ordinal Type eta of the Aggregate R of all Rational Numbers
which are Greater than 0 and Smaller than 1, in their Natural Order
of Precedence" (pp. 504-506 of the original).
Strangely enough, this result doesn't seem to be mentioned in Halmos,
/Naive Set Theory/, although it's probably in most other introductory
books on set theory. For instance Moschovakis, /Notes on Set Theory/
(2nd ed.), p. 208:
"Theorem (Cantor). Any two countable, linear, dense in themselves
orderings (A, <=_A) and (B, <=_B) without minimum or maximum element
are similar, i.e., there exists an order-preserving correspondence
f: A -> B."
I must admit I'd never thought about the surprising consequences of
this! I think I saw a proof of it (essentially Cantor's own) in
Kamke's book /Theory of Sets/ (Dover reprint) when I was at school
... an infinity of time ago ... I still can't quite believe that a
function of the type described by Prof. Edgar exists. "I see it,
but I don't believe it." (No, that's not Victor Meldrew speaking!
It's Cantor himself, writing to Dedekind in another context.)
Dustan
> On Aug 25, 9:48 am, Dustan <DustanGro...@gmail.com> wrote:
>
>
>
> > On Aug 24, 5:45 pm, ge...@math.mq.edu.au wrote:
>
> > > On Aug 25, 1:30 am, Dustan <DustanGro...@gmail.com> wrote:
>
> > > > I know that this newsgroup
> > > > is anti-high school (or college freshmen, sophomore, junior... well,
> > > > just about anyone without a PhD),
>
> > > What garbage. Many of us are anti-moron, even if the moron has a PhD
> > > (there is more than one PhD holder who has been given a hard time
> > > here), and we are pro-mathematics, even if the person wanting to talk math
> > > is in high school. We ask for civility, and return it.
>
> > Eh?
>
> > Sorry, but my definition of civility is slightly more restrained than
> > freely criticizing people for things that they simply don't know
> > about. See William Elliot & Michael Press:http://groups.google.com/group/sci.math/browse_frm/thread/b2d34ef7ebb...
>
> > Yes, I hold a grudge.
>
> So your opinion that this newsgroup is hostile to anyone without a
> PhD
> is based on ... two posts in one thread? Two posts, out of the
> hundreds
> of thousands - two posts?? And, at that, two posts that you may have
> misinterpreted?
No, that's just the only posts I have record of. I've seen many
others.
> Look again at the post from Michael Press: he wrote, "You can discount
> my
> abilities as much as you like without risk of being shown to have
> underestimated them."
>
> Well, it's a strange thing to have posted, but on my reading he's
> saying
> that he doesn't have any mathematical ability. The only one he's
> criticizing
> there is himself!
He said that I (note: me, myself and I, not himself) had
underestimated his abilities; I hadn't estimated shit.
> He also wrote, "You seriously underestimate the mathematical
> abilities
> of the folks with whom you argue." While I wouldn't have written
> that,
> it doesn't strike me as anything to carry a grudge about for six
> months.
Again... Oh, what the hell; you don't understand.
> Meanwhile, you did get help from David Cantrell (which you
> acknowledged)
> and from several other posters in the thread - why don't you take
> that
> as evidence that sci.math is pro-highschool?
>
> I don't know whether you have a chip on your shoulder, or were just
> having
> a bad day, or read a lot of stuff into two posts that wasn't really
> in
> them, but I'd say you've failed to make any kind of case against
> sci.math.
Well, maybe I'll start following this newsgroup to accumulate more
evidence. Or maybe I honestly don't care enough to do so, since no
one's going to listen anyway.
> And by the way, Michael corrected your "free reign" to "free rein,"
> and
> you went and wrote "free reign" again! If you're not going to bother
> to
> take on board what people write, why do you bother to post?
Yeah, I failed to realize that he was making a correction the first
time, having focused so much on the latter part of his post. I'm
human; sorry to disappoint.
Dustan
> On 24-08-2008 23:41, ge...@math.mq.edu.au wrote:
>
> >>> Does there exist a continuous function F and a rational R such that
> >>> for every rational r, F(r) is rational if and only if r != R?
> >> Yes, such a function exists. Let (q_n)_n be a strictly decreasing
> >> sequence of rational functions greater than 0 such that the limit
> >> lim_n q_n exists and it is an irrational number _y_.
>
> > Rational functions? I think you mean rational numbers.
>
> Sure. Thanks for the correction.
Whoops! I missed the mistake. I suppose I was just scanning past that
paragraph because I instinctively knew that he was describing a Cauchy
sequence with an irrational limit.
José Carlos Santos
>> D'oh! Cantor (1895).
>
> In more detail:
>
> Georg Cantor, "Contributions to the theory of transfinite numbers",
> first article, Mathematische Annalen 46 (1895), pp. 481-512. English
> translation in Dover paperback (1955). The proof is in section 9,
> "The Ordinal Type eta of the Aggregate R of all Rational Numbers
> which are Greater than 0 and Smaller than 1, in their Natural Order
> of Precedence" (pp. 504-506 of the original).
Thanks. I like to be aware of references like this one.
> Strangely enough, this result doesn't seem to be mentioned in Halmos,
> /Naive Set Theory/, although it's probably in most other introductory
> books on set theory. For instance Moschovakis, /Notes on Set Theory/
> (2nd ed.), p. 208:
>
> "Theorem (Cantor). Any two countable, linear, dense in themselves
> orderings (A, <=_A) and (B, <=_B) without minimum or maximum element
> are similar, i.e., there exists an order-preserving correspondence
> f: A -> B."
>
> I must admit I'd never thought about the surprising consequences of
> this! I think I saw a proof of it (essentially Cantor's own) in
> Kamke's book /Theory of Sets/ (Dover reprint) when I was at school
> ... an infinity of time ago ... I still can't quite believe that a
> function of the type described by Prof. Edgar exists. "I see it,
> but I don't believe it." (No, that's not Victor Meldrew speaking!
> It's Cantor himself, writing to Dedekind in another context.)
Perhaps that I am missing something, but it looks that the construction
that I described to the OP can be used to build a function like the one
described by G. A. Edgar. Consider that construction (I mean, without
the typo which consisted in writing "rational functions" instead of
"rational numbers"), except that the sequence (q_n)_n should satisfy the
extra requirement that lim_n q_n = sqrt(2). Finally, for x >= 1, instead
of defining f(x) = q_1, define f(x) = x - 1 + q_1. This function _f_
(restricted to the rationals greater than sqrt(2)) will be then an
order-preserving bijection from the set {q in Q : q > 0} onto the set
{q in Q : q > sqrt(2)}. Of course, this will not prove Cantor's theorem,
but it allows you to see how such a function can look like.
Herman Rubin
>> F(0) = sqrt(2)
This is a standard procedure. Given any two countable
ordered sets with no gaps, and no first and last elements,
one can construct an order preserving bijection between
them.
Enumerate both sets by the integers. Then assign the
first elements of each to each other. Alternately
assign the next available element of each to the next
available element in the appropriate open interval.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
Herman Rubin
>> F(0) = sqrt(2)
Of course it does not describe a unique function; why
should it? But it does describe how to construct such
a function.
Angus Rodgers
<Dustan...@gmail.com> wrote:
>Well, maybe I'll start following this newsgroup to accumulate more
>evidence. Or maybe I honestly don't care enough to do so, since no
>one's going to listen anyway.
It's part of a bigger picture. I've dipped into and out of sci.math
over a long period, and (as in Usenet generally) there is s surprising
amount of aggression here that sometimes seems to come out of nowhere.
Some of it is explicable as arising out of a long-running war between
"cranks" and professional mathematicians. There are indeed cranks,
who can indeed be a nuisance, and worse (e.g. one has made complaints
to college authorities to try to get teachers fired from their jobs),
but it's also very easy for innocents to be caught in the crossfire.
More than that I can't say, as I don't have a very good grasp of what
is going on. But your experiences are not as isolated as you seem to
believe.
You mentioned that you have "settled outside of Usenet lately".
What other useful forums have you found? How do they compare?
Angus Rodgers
That does indeed look comparatively sane.
Perhaps if I make an effort I will also be able to visualise Cantor's
construction, without having nightmares - I only glanced at the proof
last night, and intend to have another look, after coffee transfusion.
David C. Ullrich
I doubt that. It's very easy to see that an Edgar function exists.
Trying to write the proof down here might be tedious, but
if the two of us were standing at a blackboard in five minutes
you'd see exactly how the construction went and agree that it
was clear that it worked.
>Best regards,
>
>Jose Carlos Santos
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
Dustan
> You mentioned that you have "settled outside of Usenet lately".
> What other useful forums have you found? How do they compare?
I haven't found any other math forums, if that's what you're looking
for, with the exception of the well known math forum (which I've never
used, so I know nothing about it):
http://www.mathforum.org/
I can't say I've found any other really useful forums for that matter;
I fell out of Usenet because other place I had found was extremely
busy and quickly filled my precious hour a day (weekday, that is) that
I devoted to the internet (no, I didn't fall out of Usenet because of
this newsgroup). But it's just a junky time-waster, with (very)
occasional "useful" references to the news. If you really want to
know, email me and I'll send a link back.
Angus Rodgers
No, that's OK, I'm "settled" here now. (It's taken me about a year.)
I was just curious to know if there were any comparable forums, as
I've never really looked for any.
As far as I know, the Math Forum is just a Web interface to sci.math
... hmm, there seems to be a bit more to it than that, e.g. Historia-
Matematica ... although again it seems to be only an interface to an
existing Internet facility (in this case a mailing list).
I don't think I'm quite ready for Historia-Matematica yet, although
my interest in the history of mathematics is growing. It looks as
though it should be the next thing for me to join (as a supplement
to sci.math rather than a replacement).
I wonder if there are a lot of specialist mathematical mailing lists
I don't know about? (I know there's at least one on category theory,
but I'm staying well clear of that for the time being!)
Dave L. Renfro
> "Theorem (Cantor). Any two countable, linear, dense
> in themselves orderings (A, <=_A) and (B, <=_B) without
> minimum or maximum element are similar, i.e., there
> exists an order-preserving correspondence f: A -> B."
>
> I must admit I'd never thought about the surprising
> consequences of this!
Back in Fall 1987 I used Cantor's result to prove that
Q^2 union P^2
is pathwise connected in R^2 (the plane), where Q is the
set (subspace of R, actually) of rational numbers and P is
the set of irrational numbers. For more about this, and how
someone besides me wound up publishing it, see the first post
below. Incidentally, I misread the problem being considered
by the original poster in that thread, thinking it was the
problem I had solved in 1987, but it was something different.
However, I think the rest of the post is fine, except for
one tangential comment I made in order to work in a Star Wars
pun, which is taken care of in the 16 May 2002 post. The last
post below includes an update where I found a simpler proof
of the pathwise connectedness of Q^2 union P^2 in a recent
topology text. [In all, you can find 3 proofs in these posts
that this space is pathwise connected, and at least one proof
independent of these 3 proofs that this space is connected.]
"Topological Sandpaper" (15 May 2002)
http://groups.google.com/group/sci.math/msg/25850ce28ab37b19
"Topological Sandpaper" (16 May 2002)
http://groups.google.com/group/sci.math/msg/4882cf8bae5f3352
" R^2\Q^2 connected?" (19 October 2005)
http://groups.google.com/group/sci.math/msg/eb8c0d1057d06e73
Dave L. Renfro
G. A. Edgar
<twi...@bigfoot.com> wrote:
> I wonder if there are a lot of specialist mathematical mailing lists
> I don't know about? (I know there's at least one on category theory,
> but I'm staying well clear of that for the time being!)
A list of some is kept by the American Mathematical Society
http://www.ams.org/mathweb/mi-listserv.html
A N Niel
> And by the way, Michael corrected your "free reign" to "free rein,"
> and
> you went and wrote "free reign" again!
Write a paragraph in which "free reign", "free rein", and "free rain"
are all used correctly.
Angus Rodgers
Free rain ended with the privatisation of the water supply during the
free reign of Margaret Thatcher, who gave free rein to market forces.
José Carlos Santos
>>>> Will this work?
>>>>
>>>> F(0) = sqrt(2)
>>>>
>>>> F maps the rationals > 0 bijectively, preserving order, onto the
>>>> rationals > sqrt(2)
>>> Can you explain how to construct such an F?
>> It seems to me that G. A. Edgar was _asking_ to the OP whether or not
>> such a function F exists. He certainly did not claim that it does.
>
> I doubt that. It's very easy to see that an Edgar function exists.
You are right. I didn't even try and, on the other hand, I was not aware
of Cantor's theorem.
> Trying to write the proof down here might be tedious, but
> if the two of us were standing at a blackboard in five minutes
> you'd see exactly how the construction went and agree that it
> was clear that it worked.
No need for that. Herman Rubin's post contains a perfectly clear proof
of Cantor's theorem. And one of my posts posted today also contains a
proof in the specific context of finding such a bijection between two
sets of the form { r in Q: r > a } for some real number _a_. But my
construction is rather tedious indeed.
Angus Rodgers
Hey, I still like it! The world is big enough for both constructions.
David C. Ullrich
<jcsa...@fc.up.pt> wrote:
>On 25-08-2008 13:07, David C. Ullrich wrote:
>
>>>>> Will this work?
>>>>>
>>>>> F(0) = sqrt(2)
>>>>>
>>>>> F maps the rationals > 0 bijectively, preserving order, onto the
>>>>> rationals > sqrt(2)
>>>> Can you explain how to construct such an F?
>>> It seems to me that G. A. Edgar was _asking_ to the OP whether or not
>>> such a function F exists. He certainly did not claim that it does.
>>
>> I doubt that. It's very easy to see that an Edgar function exists.
>
>You are right. I didn't even try and, on the other hand, I was not aware
>of Cantor's theorem.
>
>> Trying to write the proof down here might be tedious, but
>> if the two of us were standing at a blackboard in five minutes
>> you'd see exactly how the construction went and agree that it
>> was clear that it worked.
>
>No need for that. Herman Rubin's post contains a perfectly clear proof
>of Cantor's theorem.
Looks to me like a perfectly clear construction of an order-preserving
bijection. We want a continuous order-preserving bijection.
Wait. Never mind what I was about to say - the continuity
follows for free, since the rationals are dense in R and
any disconinuity of a monotone function is a jump discontinuity.
>And one of my posts posted today also contains a
>proof in the specific context of finding such a bijection between two
>sets of the form { r in Q: r > a } for some real number _a_. But my
>construction is rather tedious indeed.
>
>Best regards,
>
>Jose Carlos Santos
David C. Ullrich