Topological Sandpaper
William Elliot
Let E = { (x,y), (y,x) in RxR | x algebraic, y transcendental }
Let S and E have the induced subspace topology of standard RxR.
S is sandpaper and E is emory cloth. Who first invented these?
Claim: S,E are connected, not path connected.
Conjecture: sandpaper homeomorphic emory cloth.
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Jyrki Lahtonen
William Elliot wrote:
>
> Let S = { (x,y), (y,x) in RxR | x rational, y irrational }
> Let E = { (x,y), (y,x) in RxR | x algebraic, y transcendental }
> Let S and E have the induced subspace topology of standard RxR.
>
> S is sandpaper and E is emory cloth. Who first invented these?
>
> Claim: S,E are connected, not path connected.
> Conjecture: sandpaper homeomorphic emory cloth.
If I read your definitions correctly, then the claim seems to
be false: The line x=y doesn't intersect either of your sets,
so it gives a partition of both S and E into two disjoint open
subsets (one with points x<y, the other with x>y)
Or did you mean to identify there mirror image points?
Cheers,
Jyrki Lahtonen
Phil Carmody
Terve Jyrki,
I read
S = { (x,y), (y,x) in RxR | x rational, y irrational }
as
S = Union (
{ (x,y) in RxR | x rational, y irrational },
{ (y,x) in RxR | x rational, y irrational }
)
However, the subject matter is beyond me, so I look forward to some interesting posts in this thread.
Phil
Phil Carmody
> I read
... with my elbows.
Ooops.
Phil
William Elliot
> Jyrki Lahtonen wrote:
> > William Elliot wrote:
> > >
> > > Let S = { (x,y), (y,x) in RxR | x rational, y irrational }
> > > Let E = { (x,y), (y,x) in RxR | x algebraic, y transcendental }
> > > Let S and E have the induced subspace topology of standard RxR.
> > >
> > > S is sandpaper and E is emory cloth. Who first invented these?
> > >
> > > Claim: S,E are connected, not path connected.
> > > Conjecture: sandpaper homeomorphic emory cloth.
> >
> > If I read your definitions correctly, then the claim seems to
> > be false: The line x=y doesn't intersect either of your sets,
> > so it gives a partition of both S and E into two disjoint open
> > subsets (one with points x<y, the other with x>y)
> >
> > Or did you mean to identify there mirror image points?
>
> Terve Jyrki,
> I read
> S = { (x,y), (y,x) in RxR | x rational, y irrational }
> as
> S = Union (
> { (x,y) in RxR | x rational, y irrational },
> { (y,x) in RxR | x rational, y irrational }
> )
>
William Elliot
William Elliot wrote:
> Let S = { (x,y), (y,x) in RxR | x rational, y irrational }
> Let E = { (x,y), (y,x) in RxR | x algebraic, y transcendental }
> Let S and E have the induced subspace topology of standard RxR.
> > S is sandpaper and E is emory cloth. Who first invented these?
> > Claim: S,E are connected, not path connected.
> Conjecture: sandpaper homeomorphic emory cloth.
_ be false: The line x=y doesn't intersect either of your sets,
_ so it gives a partition of both S and E into two disjoint open
_ subsets (one with points x<y, the other with x>y)
Well gee, wasn't that somewhat abrasive? -)
Hm, any line thru (0,0) with rational slope also totally gone.
So S,E are more than disconnected, likely they're shredded.
Let L_m,a = { (x,y) | y = mx + a } for rational a,m
Then all such lines L_m,a, are totally missing.
Thus for all x /= y, can I not make a partition that
puts x on one side and y one the other?
Well then S,E would be completely disconnected.
But can I not shred S,E even more.?
If I make parallelograms with four L_m,a, the space inside is clopen
and as I can make them as small as I wish, then have I not a base of
closed and open sets? Then S,E are zero dimensional.
_ Or did you mean to identify there mirror image points?
Huh?
-- reruns
"Does every open, bounded, connected set, having infinitely many
elements, in a metric domain, have to have a closed, connected
subset, also having infinitely many elements?"
A connected, completely Hausdorff space (which includes metric spaces)
with more than one point has uncountably many points.
Metric spaces are perfectly normal, ie every
open set is union countable # of closed sets.
Put all that together and I've yet still to show the uncountable closed
set is connected. Did anybody in that jumbled thread actually find a
proof of the above likely conjecture? I've some doubt that bounded is
actually needed.
----
Chan-Ho Suh
"William Elliot" <ma...@xx.com> wrote in message
news:20020515051114...@agora.rdrop.com...
> A connected, completely Hausdorff space (which includes metric spaces)
> with more than one point has uncountably many points.
I don't think this is true.
Roy's Lattice Space is connected, completely Hausdorff and has countably
infinite many points.
Description of this space:
Take S_1, S_2, .... to be a countable collection of dense subsets of the
rational which are pairwise disjoint.
Then put S = { (x, i) | x in S_i } and w an ideal point. The resulting
space is called Roy's Lattice space; of course, we must now define the
topology.
Each point (x, 2n) has a basic nbd corresponding to an open interval
containing x in the rationals, by adding a second coordinate 2n to the
points of this interval. Each point (x, 2n-1) has a basic nbd corresponding
to an open interval containing x in the rationals, but for each point in
that interval we now have three corresponding points by adding as second
coordinate, 2n-1, 2n-1, 2n. A basic nbd of w, call it W_n, consists of w
and all points in S with last coordinate at least 2n.
Roy's space is completely Hausdorff since no two points have the same first
coordinate (and if one of the points is w, we can easily check it is
completely separated from any other point). Connectedness is easily seen if
one considers a clopen set U containing w: such a set must contain a basis
nbd W_n, and then since U is closed, {(x, 2n-1)| x in S_2n-1}is contained in
U. Similarly since U is open, {(x, 2n-2) | x in S_2n-2} is contained in U.
Continuing on this way, we get that the whole space is contained in U.
Dave L. Renfro
[sci.math Tue, 14 May 2002 23:36:25 -0700]
http://mathforum.org/epigone/sci.math/voltwenstee
wrote
> Let S = { (x,y), (y,x) in RxR | x rational, y irrational }
> Let E = { (x,y), (y,x) in RxR | x algebraic, y transcendental }
> Let S and E have the induced subspace topology of standard RxR.
>
> S is sandpaper and E is emory cloth. Who first invented these?
>
> Claim: S,E are connected, not path connected.
> Conjecture: sandpaper homeomorphic emory cloth.
Ah, a chance to set the record straight.
First, both S and E are pathwise connected. I even mentioned
this in a sci.math post a while back, wondering if doing so would
generate some interest in trying to find a proof (it didn't, at
least as far as I know of) --->>>
What is the point of rigor?" [Oct. 23, 2000]
http://mathforum.org/epigone/sci.math/phukheldpre/t4qmiv...@forum.mathforum.com
Now to my story . . .
In a topology course taught by Peter Nyikos at the University of
South Carolina during Fall 1987 the following problem was given
for extra credit: "In the subspace topology inherited from R^2,
is (Q x Q) U (P x P) connected?" [Q denotes the set of rational
numbers and P denotes the set of irrational numbers.]
For some reason I couldn't figure out how to give a direct proof
that it was connected (a straightforward proof that I came up with
later is given below), but I managed to prove a stronger result,
namely that it is pathwise connected. My proof made use of
Cantor's theorem that any two unbounded countable dense linear
orderings are order-isomorphic. What got me thinking this way was
I could see that polygonal curves wouldn't work and, at the
same time, I had just happened to come across the following
interesting comment in Felix Hausdorff's "Set Theory" book
(3'rd edition, translated from the German by John R. Aumann,
et al, Chelsea, 1937/1978):
(from p. 60) "The set of all rational numbers > a is of type
eta. Thus there exists, for instance, a function
s = f(r) that preserves order, i.e., increases
monotonically with r, which assigns a rational
number s > a to every rational number r > 0,
and vice versa. Clearly this can be extended to
a continuous monotonically increasing function
y = f(x) that maps the half-line x > 0 onto the
half-line y > a and assigns to every rational x
a rational y, and to every irrational x an
irrational y. For rational a this is trivial, of
course, as y = x + a induces such a correspondence."
Nyikos was impressed with the result and he also didn't think
that it was known. I was thinking of writing it up for publication
but other events got in the way --- I didn't properly prepare for
my Ph.D. exams for the second time and found myself effectively
"on the street" the next semester, Spring 1988.
See the following posts (first three) if you're interested in what
I did the next semester, or what I did the following academic year
(4'th URL below). I also taught an intermediate algebra course
in a prison, did night data entry work for Allstate, taught
high school remedial math (i.e. arithmetic) during the summer
in a very impoverished school district (not the same district I
taught in Fall 1988-89), and some other things to survive.
http://mathforum.org/epigone/math-teach/herpixpor
http://mathforum.org/epigone/math-teach/jixmalva
http://mathforum.org/epigone/math-teach/kunsnoncrer
http://mathforum.org/epigone/math-teach/shuntwolsner/pg9q1y...@forum.mathforum.com
Fall 1989 found me back in school again, this time at North Carolina
State University. [I could write a book about the previous year,
and one day I might --- perhaps when I'm old with no money and no
other means to support myself . . .]
Anyway, I'd never been to a math conference before, and so when
I learned that the Southeastern Section of the MAA was having
it's yearly meeting the following April at Davidson College
(less than 30 miles from where my parents live), I decided to go
and give a talk on this topology result.
I couldn't find any on-line details for the program of this meeting
(I do have a print copy of the program, though), but for the record
here are the details --->>>
** "Q^2 U P^2 is arcwise connected" by Dave Renfro (student)
**
** 11:40 - 11:55 A.M. on Saturday April 7, 1990 in Section H
** "Topology", which was held in room 310 of the Chambers Building.
** [There was a separate student section, but I suppose given
** the nature of my talk and the fact that there was a section
** on topology, I was put in this section.]
**
** 69'th Annual Meeting of the Southeastern Section, MAA, April 6-7,
** 1990 at Davidson College (Davidson, NC)
I asked my Ph.D. advisor whether he thought the result was
worthy of publication. He said he'd think about it for a day or
two and see if he could come up with a more straightforward proof.
The next day he said there's a fairly easy way to do it, so
it's probably not publishable. I agreed when he gave the argument,
and so it became one of those things-for-your-drawer.
[[ Here's the easy way. Given two points in the space,
it's easy to zig-zag your way in a piece-wise linear
way from one point to the other. There's a limit
involved at each end, because you want to get within
the (1/n)-ball about each point for each n = 1, 2, ...,
but this isn't going to affect continuity. ]]
Note this continues to hold for D^2 U (R^2 - D^2), where D is
any countable dense subset of R. [Anyone get my Star Wars joke?]
And so it remained for several years, while I went into other
areas of study (nowhere differentiable continuous functions and
sigma-porous sets). Then, sometime during 1995 (Fall, I think it
may have been), during one of my numerous lengthy, tedious, and
expensive trips to a research library (for the first 6 years or
so after my Ph.D. I was fairly distant from any decent library),
I came across the Zentralblatt Math review of a paper that made
my head spin --->>>
Guanshen Ren, "A note on QxQ U IxI", Questions and Answers in
General Topology 10 (1992), 157-158. [MR 1 180 475; Zbl 786.54037]
[[ (Zbl review) "Let Q and I denote the sets of rational and
irrational numbers, respectively. Consider (Q x Q) U (I x I)
to be the subspace of R x R with the usual topology. Then
this subspace is connected. In this note is shown that
this subspace is also arcwise connected." ]]
It just so happens that Ren was a fellow student of mine in that
very same Fall 1987 topology class. In fact, he was even one of
my two office mates. Worse yet, when I managed to get a copy of
this paper, the proof was very similar to mine, even down to some of
my writing mannerisms. I ultimately got into touch with him about
this paper (1996 or 1997, I think), and he seemed a bit vague and
hesitant about the situation. Here's what I think happened, although
this is my guess and not anything Ren told me. I believe Nyikos had
handed out photocopies of my proof. Several years later Ren must
have come across it and forgot where it was from. He might have
suspected it was one of those "Nyikos starred theorems" we so
often got in class. [These were typically minor (but every now
and then not so minor) improvements of theorem statements in the
text we used, Willard's "General Topology".] Ren may have asked
Nyikos if he could write up the result for a short paper and
Nyikos said that would be fine, forgetting where it came from.
[For what it's worth, at the end of Ren's paper it says that it
was received on July 22, 1991.]
Now for some math.
THEOREM 1: Q^2 U P^2 is connected.
PROOF: If not, then there is a disconnection G U H. Choose g in G
and h in H. Because G and H are open and Q^2 is dense,
there exists r in (G intersect Q^2) and there exists
s in (H intersect Q^2). Let L be the line containing
r and s. Then L is connected. However,
(L intersect G) U (L intersect H) gives a disconnection
of L, and therefore we have a contradiction.
THEOREM 2: Q^2 U P^2 is pathwise connected.
[[ "Arcwise connected" is when the paths can be chosen to be
homeomorphisms from [0,1]. It is equivalent to pathwise
connectedness for T_2 spaces. ]]
PROOF (hard way): We'll be making repeated use of the following fact:
If {A_i: i in I} is a collection of pathwise connected sets
whose (complete) intersection is not empty, then the union
of this collection is pathwise connected. For nonzero r,s
in Q, let L(r,s) be the line containing (0,0) and (r,s).
Then each L(r,s) belongs to our space. Because each L(r,s)
is pathwise connected and contains (0,0), their union, call
it A, is pathwise connected. Next, for each nonzero r in Q,
let L(0,r) be the line containing (1,1) and (0,r), and let
L(r,0) be the line containing (1,1) and (r,0). Then for each
nonzero r in Q, both A U L(0,r) and A U L(r,0) are pathwise
connected. Since {A U L(0,r): r in Q, r \=\ 0} has nonempty
intersection, the union of this collection, call it B, is
pathwise connected. Similarly, the union, call it C, of
{A U L(r,0): r in Q, r \=\ 0} is pathwise connected. Finally,
D = B U C is pathwise connected.
Clearly, Q^2 is a subset of D is a subset of (Q^2 U P^2).
Unfortunately, we don't have D = (Q^2 U P^2), since a point
such as (sqrt(2), sqrt(3)) doesn't belong to D. To take care
of these other points we abandon the use of line and try to
find, for each (x,y) in P^2, a continuous curve containing
(x,y) that lies in (Q^2 U P^2).
Given (x,y) in P^2 we have by Cantor's theorem that
"(x, infinity) intersect Q" is order-isomorphic to
"(y, infinity) intersect Q". Hence, there exists a
strictly increasing function f_{(x,y)} from
"(x, infinity) intersect Q" onto "(y, infinity) intersect Q".
This can be extended to a continuous function g_{(x,y)}
from [x, infinity) to [y, infinity) such that x maps to y.
Note that the graph of g_{(x,y)} lies in (Q^2 U P^2).
Call this graph C(x,y). For each x,y in P, let D(x,y)
be D U C(x,y). Then each D(x,y) is pathwise connected.
Finally, the union of all the D(x,y)'s is pathwise connected,
and this union equals (Q^2 U P^2).
Incidentally, a standard textbook problem is to show that
R^2 - Q^2 is connected. I'm not sure many people realize that
this is due to G. Cantor. I posted some historical information
about this at
"Is this set connected ?" [April 16, 2001]
http://mathforum.org/epigone/sci.math/blexspehwhin/q4du0d...@forum.mathforum.com
Dave L. Renfro
David C. Ullrich
wrote:
>William Elliot <ma...@xx.com>
>[sci.math Tue, 14 May 2002 23:36:25 -0700]
>http://mathforum.org/epigone/sci.math/voltwenstee
>
>wrote
>
>> Let S = { (x,y), (y,x) in RxR | x rational, y irrational }
>> Let E = { (x,y), (y,x) in RxR | x algebraic, y transcendental }
>> Let S and E have the induced subspace topology of standard RxR.
>>
>> S is sandpaper and E is emory cloth. Who first invented these?
>>
>> Claim: S,E are connected, not path connected.
>> Conjecture: sandpaper homeomorphic emory cloth.
>
>Ah, a chance to set the record straight.
>
>First, both S and E are pathwise connected.
??? People have already posted very simple explanations
of why they're not even connected. S is contained in A union
B, where A = {(x,y) : x > y} and B = {(x,y) : x < y}...
>I even mentioned
>this in a sci.math post a while back, wondering if doing so would
>generate some interest in trying to find a proof (it didn't, at
>least as far as I know of) --->>>
>
>What is the point of rigor?" [Oct. 23, 2000]
>http://mathforum.org/epigone/sci.math/phukheldpre/t4qmiv...@forum.mathforum.com
>
>Now to my story . . .
>
>In a topology course taught by Peter Nyikos at the University of
>South Carolina during Fall 1987 the following problem was given
>for extra credit: "In the subspace topology inherited from R^2,
>is (Q x Q) U (P x P) connected?" [Q denotes the set of rational
>numbers and P denotes the set of irrational numbers.]
(Q x Q) U (P x P) is not S. Not that you said explicitly that
it was.
The rest of your post is very interesting as always, but I
don't see that it applies here.
David C. Ullrich
David C. Ullrich
wrote:
>William Elliot <ma...@xx.com>
Actually I stopped reading here - wanted to see if I could
figure it out. No doubt polygonal paths don't work, but
piecwise-polygonal paths do. After I realized this I saw
a big hint in the quote from Hausorff below:
If L is a line which passes through a point of QxQ and
has rational slope that L lies in QxQ U PxP. Given any
p in R^2 it's clear that there exists a sequence
0 = p_0, p_1, ... which converges to p, and which is
such that the slope of the line through p_n and p_{n+1}
is always rational. So the curve consisting of the segment
from p_0 to p_1, etc, is a path in QxQ U PxP from 0 to p.
>and, at the
>same time, I had just happened to come across the following
>interesting comment in Felix Hausdorff's "Set Theory" book
>(3'rd edition, translated from the German by John R. Aumann,
>et al, Chelsea, 1937/1978):
>
>(from p. 60) "The set of all rational numbers > a is of type
> eta. Thus there exists, for instance, a function
> s = f(r) that preserves order, i.e., increases
> monotonically with r, which assigns a rational
> number s > a to every rational number r > 0,
> and vice versa. Clearly this can be extended to
> a continuous monotonically increasing function
> y = f(x) that maps the half-line x > 0 onto the
> half-line y > a and assigns to every rational x
> a rational y, and to every irrational x an
> irrational y. For rational a this is trivial, of
> course, as y = x + a induces such a correspondence."
>
[...]
>
>** "Q^2 U P^2 is arcwise connected" by Dave Renfro (student)
>**
[...]
>
> [[ Here's the easy way. Given two points in the space,
> it's easy to zig-zag your way in a piece-wise linear
> way from one point to the other. There's a limit
> involved at each end, because you want to get within
> the (1/n)-ball about each point for each n = 1, 2, ...,
> but this isn't going to affect continuity. ]]
Oh, so you already knew that. (I take it the above is supposed
to be just a hint - not just any zigzag path works... wait,
I know, I bet you never figured out how the details work.
Yeah, that's it - I should write a paper on this.)
David C. Ullrich
David C. Ullrich
wrote:
[...]
>
> [[ Here's the easy way. Given two points in the space,
> it's easy to zig-zag your way in a piece-wise linear
> way from one point to the other. There's a limit
> involved at each end, because you want to get within
> the (1/n)-ball about each point for each n = 1, 2, ...,
> but this isn't going to affect continuity. ]]
Of course when I said "rational slope" I should have said
"non-zero rational slope".
>Note this continues to hold for D^2 U (R^2 - D^2), where D is
>any countable dense subset of R. [Anyone get my Star Wars joke?]
Too bad the joke doesn't work as well when you fix the typo
(D^2 U (R^2 - D^2) is convex, after all...)
Presumably you meant D^2 U (R - D)^2. I don't see how the argument
works if all we know is that D is a countable dense set.
David C. Ullrich
William Elliot
William Elliot <ma...@xx.com> wrote
> Let S = { (x,y), (y,x) in RxR | x rational, y irrational }
> Let E = { (x,y), (y,x) in RxR | x algebraic, y transcendental }
> Let S and E have the induced subspace topology of standard RxR.
>
_ In a topology course taught by Peter Nyikos at the University of
_ South Carolina during Fall 1987 the following problem was given
_ for extra credit: "In the subspace topology inherited from R^2,
_ is (Q x Q) U (P x P) connected?" [Q denotes the set of rational
_ numbers and P denotes the set of irrational numbers.]
That's not my S which is QxP \/ PxQ.
_ I managed to prove a stronger result,
_ namely that it is pathwise connected.
Let's see. For rational m,a, m /= 0
L_m,a = { (x,y) | y = mx + a } subset QxQ \/ PxP
So can't I go zig-zagging around to get from one place to another?
Hm, your space appears the complement of my space that got shredded
by the very same bunch of L_m,a lines.
_ Here's the easy way. Given two points in the space,
_ it's easy to zig-zag your way in a piece-wise linear
_ way from one point to the other. There's a limit
_ involved at each end, because you want to get within
_ the (1/n)-ball about each point for each n = 1, 2, ...,
_ but this isn't going to affect continuity. ]]
Oh, I get it. You've got to home in, for example, on an irrational when
starting from a rational.
----
William Elliot
> A connected, completely Hausdorff space (which includes metric spaces)
> with more than one point has uncountably many points.
completely Hausdorff: distinct points functionally separated
connected multi-point completly Hausdorff S ==> S uncountable
Assume premise. Thus some x,y in S with x /= y
some continuous f:S -> [0,1] with f(x) = 0, f(y) = 1
f(S) connected as S connected; f(S) subset [0,1]
If some r in [0,1] not in f(S), then f(S) disconnected at r
Thus f(S) = [0,1] and |S| >= |[0,1]| = c
How do you define completely Hausdorff?
_ Roy's Lattice Space is connected, completely Hausdorff
_ and has countably infinite many points.
_ Description of this space:
_ Take S_1, S_2, .... to be a countable collection of
_ dense subsets of the rational which are pairwise disjoint.
That's not an easy construction. Anyway, in following I see
no need of dense, only disjoint. How's dense used?
_ Then put S = { (x,i) | x in S_i } and w an ideal point.
What's this ideal point w? Another point other than the (x,i)?
_ The resulting space is called Roy's Lattice space;
_ of course, we must now define the topology.
What lattice?
_ Each point (x,2n) has a basic nbd corresponding to an open interval
_ containing x in the rationals, by adding a second coordinate 2n to
_ the points of this interval.
The local base for (x,2n) is { (U/\S_2n)x{2n} | x in U, U open in Q }
_ Each point (x,2n-1) has a basic nbd corresponding to an open interval
_ containing x in the rationals, but for each point in that interval we
_ now have three corresponding points by adding as second coordinate,
_ 2n-1, 2n-1, 2n.
Something looks wrong with that last line.
The local base for (x,2n) is (guessing) (\/ union)
{ (U/\S_2n-1)x{2n-1} \/ (U/\S_2n)x{2n} | x in U, U open in Q }
_ A basic nbd of w, call it W_n, consists of w
_ and all points in S with last coordinate at least 2n.
Let W_n = {w} \/ { (x,k) in S | k >= 2n }
The local base for w is { W_n | n in N }.
It's necessary these local bases be intersection closed. Are they?
No they weren't, but then I changed my guess to make them so.
_ Roy's space is completely Hausdorff since no two points have the same
_ first coordinate (and if one of the points is w, we can easily check
_ it is completely separated from any other point).
Hm. Ok, it's Hausdorff, as distinct points separated by open sets.
You've yet to convince me distinct points are functionally separated.
_ Connectedness is easily seen if one considers a clopen set U
_ containing w: such a set must contain a basis nbd W_n, and then since
_ U is closed, {(x, 2n 1)| x in S_2n-1} is contained in U. Similarly
_ since U is open, {(x,2n-1) | x in S_2n-1} is contained in U.
_ Continuing on this way, we get that the whole space is contained in
_ U.
Ok, have I puzzled this correctly? Given any point a = (x,2n-1)
any open set containing a intersects W_n. Hence a in U. So some
open set containing a is subset U and hence U contains W_(n-1)
By infinite descent we come to U = W_0. Oh boy, is Roy connected.
Why did Roy ever invent such a Roy Rogers space?
Dave L. Renfro
[sci.math Wed, 15 May 2002 20:20:38 GMT
and Wed, 15 May 2002 23:55:41 GMT]
http://mathforum.org/epigone/sci.math/voltwenstee
wrote (in parts, respectively)
> ??? People have already posted very simple explanations
> of why they're not even connected. S is contained in A union
> B, where A = {(x,y) : x > y} and B = {(x,y) : x < y}...
and
>> Note this continues to hold for D^2 U (R^2 - D^2), where D is
>> any countable dense subset of R. [Anyone get my Star Wars joke?]
>
> Too bad the joke doesn't work as well when you fix the typo
> (D^2 U (R^2 - D^2) is convex, after all...)
>
> Presumably you meant D^2 U (R - D)^2. I don't see how the argument
> works if all we know is that D is a countable dense set.
Well, this isn't the first time I've misread a post and answered
V when W as asked, nor do I suspect it'll be the last time, but
this has got to rank among the longest delusional posts I've ever
made. I would have said (now) that at least the math part should
be correct (if not approprate to to William Elliot's original post),
but I can't even say that. Darn it, I wanted that Star Wars
joke to work so much that I lost track of what the math was saying.
But hey, if we take off that initial union with D^2, it does
work: R^2 - D^2 is pathwise connected (for any countable D, dense
or otherwise). [But the question is, does R^2 - D^2 stay connected
throughout the movie? I guess he has to, or at least he can't
become totally disconnected, since the present epsiode takes place
before the Hans Solo era.]
Maybe I should do some more James Hunter spoofs -- I seem to do
better with those.
Dave L. Renfro
James Hunter
"Dave L. Renfro" wrote:
> David C. Ullrich <ull...@math.okstate.edu>
> [sci.math Wed, 15 May 2002 20:20:38 GMT
> and Wed, 15 May 2002 23:55:41 GMT]
> http://mathforum.org/epigone/sci.math/voltwenstee
>
> wrote (in parts, respectively)
> Well, this isn't the first time I've misread a post and answered
> V when W as asked, nor do I suspect it'll be the last time, but
> this has got to rank among the longest delusional posts I've ever
> made. I would have said (now) that at least the math part should
> be correct (if not approprate to to William Elliot's original post),
> but I can't even say that. Darn it, I wanted that Star Wars
> joke to work so much that I lost track of what the math was saying.
> But hey, if we take off that initial union with D^2, it does
> work: R^2 - D^2 is pathwise connected (for any countable D, dense
> or otherwise). [But the question is, does R^2 - D^2 stay connected
> throughout the movie? I guess he has to, or at least he can't
> become totally disconnected, since the present epsiode takes place
> before the Hans Solo era.]
>
> Maybe I should do some more James Hunter spoofs -- I seem to do
> better with those.
It is physically *impossible* for a mathematician to do a
James Hunter proof, since he has to constantly
remind them that they are quite obviously
confusing their worthless Academia pearls
of "wisdom" with *INTELLIGENCE*.
We are making some progress, since at least with
Hans Solo we don't have to listen to your
faggot Plato stories for more than an hour in
an otherwise successful career.
>
>
> Dave L. Renfro
Chan-Ho Suh
> cs...@math.ucla.edu
> > A connected, completely Hausdorff space (which includes metric spaces)
> > with more than one point has uncountably many points.
> _ I don't think this is true.
> completely Hausdorff: distinct points functionally separated
> connected multi-point completly Hausdorff S ==> S uncountable
> Assume premise. Thus some x,y in S with x /= y
> some continuous f:S -> [0,1] with f(x) = 0, f(y) = 1
> f(S) connected as S connected; f(S) subset [0,1]
> If some r in [0,1] not in f(S), then f(S) disconnected at r
> Thus f(S) = [0,1] and |S| >= |[0,1]| = c
> How do you define completely Hausdorff?
>
As far as I can tell, it's common to call the T_2.5 axiom "completely
Hausdorff", meaning that every two distinct points have open nbds whose
closures are disjoint. But ok, I see what you mean with your definition
though.
> _ Roy's Lattice Space is connected, completely Hausdorff
> _ and has countably infinite many points.
> _ Description of this space:
>
> _ Take S_1, S_2, .... to be a countable collection of
> _ dense subsets of the rational which are pairwise disjoint.
> That's not an easy construction. Anyway, in following I see
> no need of dense, only disjoint. How's dense used?
>
You need it for the connectedness. See below.
> _ Then put S = { (x,i) | x in S_i } and w an ideal point.
> What's this ideal point w? Another point other than the (x,i)?
>
Yes.
> _ The resulting space is called Roy's Lattice space;
> _ of course, we must now define the topology.
> What lattice?
>
I think the name is more descriptive than technical. The idea is to think
of S_i x{ i} as a "line".
> _ Each point (x,2n) has a basic nbd corresponding to an open interval
> _ containing x in the rationals, by adding a second coordinate 2n to
> _ the points of this interval.
> The local base for (x,2n) is { (U/\S_2n)x{2n} | x in U, U open in Q }
>
Yes.
> _ Each point (x,2n-1) has a basic nbd corresponding to an open interval
> _ containing x in the rationals, but for each point in that interval we
> _ now have three corresponding points by adding as second coordinate,
> _ 2n-1, 2n-1, 2n.
> Something looks wrong with that last line.
> The local base for (x,2n) is (guessing) (\/ union)
> { (U/\S_2n-1)x{2n-1} \/ (U/\S_2n)x{2n} | x in U, U open in Q }
>
Aaah! Sorry about that, the line I originally wrote should say "...2n-2,
2n-1, 2n."
So in your notation we would have:
{ (U/\S_2n-2)x{2n-2}\/ (U/\S_2n-1)x{2n-1}\/ (U/\S_2n)x{2n}| x in U, U open
in Q}
> _ A basic nbd of w, call it W_n, consists of w
> _ and all points in S with last coordinate at least 2n.
> Let W_n = {w} \/ { (x,k) in S | k >= 2n }
> The local base for w is { W_n | n in N }.
>
Yes.
> _ Roy's space is completely Hausdorff since no two points have the same
> _ first coordinate (and if one of the points is w, we can easily check
> _ it is completely separated from any other point).
> Hm. Ok, it's Hausdorff, as distinct points separated by open sets.
> You've yet to convince me distinct points are functionally separated.
>
> _ Connectedness is easily seen if one considers a clopen set U
> _ containing w: such a set must contain a basis nbd W_n, and then since
> _ U is closed, {(x, 2n 1)| x in S_2n-1} is contained in U. Similarly
> _ since U is open, {(x,2n-1) | x in S_2n-1} is contained in U.
> _ Continuing on this way, we get that the whole space is contained in
> _ U.
> Ok, have I puzzled this correctly? Given any point a = (x,2n-1)
> any open set containing a intersects W_n. Hence a in U. So some
> open set containing a is subset U and hence U contains W_(n-1)
>
Wait a minute, you are misquoting me. But I suppose my initial typo above
confused you.
I originally wrote:
"Connectedness is easily seen if one considers a clopen set U
containing w: such a set must contain a basis nbd W_n, and then
since U is closed, {(x, 2n-1)| x in S_2n-1}is contained in U. Similarly
since U is open, {(x, 2n-2) | x in S_2n-2} is contained in U.
Continuing on this way, we get that the whole space is contained in U."
What you just wrote shows that U contains the 2n-1 "line", but now you need
to use that and the fact that U is open to show the 2n-2 "line" is contained
in U. Note that you need denseness of the S_i sets to do these two things.
> By infinite descent we come to U = W_0. Oh boy, is Roy connected.
> Why did Roy ever invent such a Roy Rogers space?
>
Why do point set topologists invent such things in general? <shrug>
David C. Ullrich
wrote:
>David C. Ullrich <ull...@math.okstate.edu>
>[sci.math Wed, 15 May 2002 20:20:38 GMT
> and Wed, 15 May 2002 23:55:41 GMT]
>http://mathforum.org/epigone/sci.math/voltwenstee
>
>wrote (in parts, respectively)
>
>> ??? People have already posted very simple explanations
>> of why they're not even connected. S is contained in A union
>> B, where A = {(x,y) : x > y} and B = {(x,y) : x < y}...
>
>and
>
>>> Note this continues to hold for D^2 U (R^2 - D^2), where D is
>>> any countable dense subset of R. [Anyone get my Star Wars joke?]
>>
>> Too bad the joke doesn't work as well when you fix the typo
>> (D^2 U (R^2 - D^2) is convex, after all...)
>>
>> Presumably you meant D^2 U (R - D)^2. I don't see how the argument
>> works if all we know is that D is a countable dense set.
>
>Well, this isn't the first time I've misread a post and answered
>V when W as asked, nor do I suspect it'll be the last time, but
>this has got to rank among the longest delusional posts I've ever
>made.
Could be. When I said I didn't see how the argument showed that
D^2 U (R - D)^2 was connected for any countable dense D I was
referring to the piecewise-polygon argument. Seems to me the
argument using the fact that any two dense countable orders
does work, doesn't it? If so that's good because it means the
argument has a reason to exist.
We only need to show we can connect any two points of D^2 by
a path - then a piecewise-one-of-those shows we can connect
any two points of D^2 U (R - D)^2. (At least if the path
joining any two points of D^2 doesn't wander far from the
points; if it's, say, contained in the rectangle with the
two points as corners that's enough.)
Lemme revise that, like I added the "non-zero slope" the
other day: We need only show that if p = (x,y) and q = (s,t)
are points of D^2 such that x < s and y < t then we
can connect p and q by a path in D^2 U (R - D)^2 that
doesn't leave the rectangle with p and q as corners.
But the bit about countable orders being isomorphic says
that there is an increasing function f mapping D intersect
[x,s] 1-1 onto D intersect [y,t]. This function must be
continuous, since the only alternative is a jump
discontinuity, which would mean the range would miss some
points of D intersect [y,t]. So it extends to a continuous
mapping f from [x,s] onto [y,t], and that extension
must map points not in D to points not in D, since f
is strictly increasing and all the points of D intersect
[y,t] are already hit by points of D. So the graph of
f is the path we want.
Pretty keen proof, Dave<wink>.
>I would have said (now) that at least the math part should
>be correct (if not approprate to to William Elliot's original post),
>but I can't even say that. Darn it, I wanted that Star Wars
>joke to work so much that I lost track of what the math was saying.
>But hey, if we take off that initial union with D^2, it does
>work: R^2 - D^2 is pathwise connected (for any countable D, dense
>or otherwise). [But the question is, does R^2 - D^2 stay connected
>throughout the movie? I guess he has to, or at least he can't
>become totally disconnected, since the present epsiode takes place
>before the Hans Solo era.]
>
>Maybe I should do some more James Hunter spoofs -- I seem to do
>better with those.
My vote is you should do both.
>Dave L. Renfro
David C. Ullrich
James Hunter
"David C. Ullrich" wrote:
Your vote really should be that mathematicians actually
learn something about *proof* theory rather than their
standard bullshit unlearning "theory" of quantum tarts.
And many along the way the morons will acually
pick up a few hints about *probability* theory.
David C. Ullrich
wrote:
>
>
>"David C. Ullrich" wrote:
>
>> On 16 May 2002 08:36:23 -0700, renf...@cmich.edu (Dave L. Renfro)
>> wrote:
>>
>> >
>> >Maybe I should do some more James Hunter spoofs -- I seem to do
>> >better with those.
>>
>> My vote is you should do both.
>
> Your vote really should be that mathematicians actually
> learn something about *proof* theory rather than their
> standard bullshit unlearning "theory" of quantum tarts.
>
> And many along the way the morons will acually
> pick up a few hints about *probability* theory.
Excellent. Except you left out the CAPS - doesn't look
realistic. But excellent except for that.
>> >Dave L. Renfro
>>
>> David C. Ullrich
>
David C. Ullrich
James Hunter
"David C. Ullrich" wrote:
> On 16 May 2002 08:46:23 -0700, renf...@cmich.edu (Dave L. Renfro)
> wrote:
> >
> >
> > Your vote really should be that mathematicians actually
> > learn something about *proof* theory rather than their
> > standard bullshit unlearning "theory" of quantum tarts.
> >
> > And many along the way the morons will acually
> > pick up a few hints about *probability* theory.
>
> Excellent. Except you left out the CAPS - doesn't look
> realistic. But excellent except for that.
I didn't leave out the caps, since CAPS only
concern mathematicians and their fellow moron
economist's (see physicist) laffy curves.
David C. Ullrich
<James....@Jhuapl.edu> wrote:
>
>
>"David C. Ullrich" wrote:
>
>> On 16 May 2002 08:46:23 -0700, renf...@cmich.edu (Dave L. Renfro)
>> wrote:
>> >
>> >
>> > Your vote really should be that mathematicians actually
>> > learn something about *proof* theory rather than their
>> > standard bullshit unlearning "theory" of quantum tarts.
>> >
>> > And many along the way the morons will acually
>> > pick up a few hints about *probability* theory.
>>
>> Excellent. Except you left out the CAPS - doesn't look
>> realistic. But excellent except for that.
No, what I actually wrote was this:
<quote>
On 16 May 2002 08:46:23 -0700, renf...@cmich.edu (Dave L. Renfro)
wrote:
>
>
>"David C. Ullrich" wrote:
>
>> On 16 May 2002 08:36:23 -0700, renf...@cmich.edu (Dave L. Renfro)
>> wrote:
>>
>> >
>> >Maybe I should do some more James Hunter spoofs -- I seem to do
>> >better with those.
>>
>> My vote is you should do both.
>
> Your vote really should be that mathematicians actually
> learn something about *proof* theory rather than their
> standard bullshit unlearning "theory" of quantum tarts.
>
> And many along the way the morons will acually
> pick up a few hints about *probability* theory.
Excellent. Except you left out the CAPS - doesn't look
realistic. But excellent except for that.
>> >Dave L. Renfro
>>
>> David C. Ullrich
>
</quote>
Let's be careful with those attributions.
James Hunter
"David C. Ullrich" wrote:
We *absolutely* have to be careful..
Since "science" morons invented HTML.
Stands for: Hiccup Transient Mushy Linguistics.
William Elliot
> > with more than one point has uncountably many points.
> _ I don't think this is true.
> completely Hausdorff: distinct points functionally separated
_ "completely Hausdorff", meaning that every two distinct points have
_ open nbds whose closures are disjoint.
Yup, that's 2.5. I've also heard it called Urysohn.
> _ Roy's Lattice Space is connected, completely Hausdorff
> _ and has countably infinite many points.
> _ Description of this space:
>
> _ Take S_1, S_2, .... to be a countable collection of
> _ dense subsets of the rational which are pairwise disjoint.
When not thinking subgroups, it's easy. How about
S_n = { k/p^m | k in Z, m in N, k,p coprime, p = n-th prime } ?
> That's not an easy construction. Anyway, in following I see
> no need of dense, only disjoint. How's dense used?
_ You need it for the connectedness. See below.
All I use is disjoint for Hausdorff.
> _ Then put S = { (x,i) | x in S_i } and w an ideal point.
> What's this ideal point w? Another point other than the (x,i)?
w acts like a one-point connectification of S.
> _ The resulting space is called Roy's Lattice space;
> _ of course, we must now define the topology.
> What lattice?
_ I think the name is more descriptive than technical.
_ The idea is to think of S_i x{ i} as a "line".
Let L_n = S_n x {n}
> _ Each point (x,2n) has a basic nbd corresponding to an open interval
> _ containing x in the rationals, by adding a second coordinate 2n to
> _ the points of this interval.
Local base for (x,2n) is { UxN /\ L_2n | x in U, U open in Q }
> _ Each point (x,2n-1) has a basic nbd corresponding to an open interval
> _ containing x in the rationals, but for each point in that interval we
> _ now have three corresponding points by adding as second coordinate,
> _ 2n-2, 2n-1, 2n.
Local base for (x,2n-1) is
{ UxN /\ (L_2n-2 \/ L_2n-1 \/ L_2n) | x in U, U open in Q }
> _ A basic nbd of w, call it W_n, consists of w
> _ and all points in S with last coordinate at least 2n.
> The local base for w is { W_n | n in N }.
where W_n = {w} \/ { L_k | k >= 2n }. Ok this is a valid topology
as the collection of all local base sets is intersection closed.
> _ Roy's space is completely Hausdorff since no two points have the same
> _ first coordinate (and if one of the points is w, we can easily check
> _ it is completely separated from any other point).
> Hm. Ok, it's Hausdorff, as distinct points separated by open sets.
Hausdorff uses disjointness of the S_k. It's T_2.5 because Q is and
cl W_n = {w} \/ { L_k | k >= 2n-1 } = L_2n-1 \/ W_n
_ Connectedness is easily seen if one considers a clopen set U
_ containing w: such a set must contain a basis nbd W_n, and then
_ since U is closed, { (x, 2n-1) | x in S_2n-1 } is contained in U.
cl W_n = L_2n-1 \/ W_n. Neither disjoiness nor denseness of S_k used.
As every open set containing (x,2n-1), contains (x,2n) in W_n,
(x,2n-1) in cl W_n
_ Similarly since U is open, {(x, 2n-2) | x in S_2n-2}
_ is contained in U. Continuing on this way, we get that the whole
_ space is contained in U.
If (x,2n-1) is in open U, then so is (x,2n-2) in U because any open set
containing (x,2n-1) contains (x,2n-2),
_ Note that you need denseness of the S_i sets to do these two things.
I do? Except for Hausdorff, I haven't even used disjointness.
> By infinite descent we come to U = W_0. Oh boy, is Roy connected.
> Why did Roy ever invent such a Roy Rogers space?
_ Why do point set topologists invent such things in general? <shrug>
Enjoy my new post, "real weird space".
Chan-Ho Suh
Think very carefully. If (x, 2n-1) is an element of Roy's space, then (x,
2n) cannot be. Remember that no two distinct points have the same first
coordinate! Do you see why denseness is needed now?
> _ Similarly since U is open, {(x, 2n-2) | x in S_2n-2}
> _ is contained in U. Continuing on this way, we get that the whole
> _ space is contained in U.
> If (x,2n-1) is in open U, then so is (x,2n-2) in U because any open set
> containing (x,2n-1) contains (x,2n-2),
>
> _ Note that you need denseness of the S_i sets to do these two things.
> I do? Except for Hausdorff, I haven't even used disjointness.
>
> > By infinite descent we come to U = W_0. Oh boy, is Roy connected.
> > Why did Roy ever invent such a Roy Rogers space?
> _ Why do point set topologists invent such things in general? <shrug>
> Enjoy my new post, "real weird space".
>
I'll look for it.
William Elliot
_ dense subsets of the rational which are pairwise disjoint.
_ Then put S = { (x,i) | x in S_i } and w an ideal point.
_ The resulting space is called Roy's Lattice space;
Let L_n = S_n x {n}, W_n = {w} \/ { L_k | k >= 2n }
_ Each point (x,2n) has a basic nbd corresponding to an open interval
_ containing x in the rationals, by adding a second coordinate 2n to
_ the points of this interval.
Local base for (x,2n) is { UxN /\ L_2n | x in U, U open in Q }
_ Each point (x,2n-1) has a basic nbd corresponding to an open interval
_ containing x in the rationals, but for each point in that interval we
_ now have three corresponding points by adding as second coordinate,
_ 2n-2, 2n-1, 2n.
Local base for (x,2n-1) is
{ UxN /\ (L_2n-2 \/ L_2n-1 \/ L_2n) | x in U, U open in Q }
_ A basic nbd of w, call it W_n, consists of w
_ and all points in S with last coordinate at least 2n.
The local base for w is { W_n | n in N }
_ Roy's space is completely Hausdorff since no two points have the same
_ first coordinate (and if one of the points is w, we can easily check
_ it is completely separated from any other point).
It's T_2.5 because Q is and cl W_n subset L_2n-1 \/ W_n
> _ Connectedness is easily seen if one considers a clopen set U
> _ containing w: such a set must contain a basis nbd W_n, and then
> _ since U is closed, { (x, 2n-1) | x in S_2n-1 } is contained in U.
_ Think very carefully. If (x,2n-1) is an element of Roy's space, then
_ (x, 2n) cannot be. Remember that no two distinct points have the
_ same first coordinate! Do you see why denseness is needed now?
If W is open set about (x,2n-1): some Q-open V contains x
some y in S_2n /\ V as S_2n is dense; W intersects W_n
Hence (x,2n-1) in cl W_n subset U
> _ Similarly since U is open, {(x, 2n-2) | x in S_2n-2}
> _ is contained in U. Continuing on this way, we get that the whole
> _ space is contained in U.
If (x,2n-1) in U: some Q-open V with x in V
VxN /\ (L_2n-2 \/ L_2n-1 \/ L_2n) subset U
some z in S_2n-2 /\ V as S_2n-2 is dense; (z,2n-2) in U
for all a in L_2n-1, some x_a in L_2n-2 /\ U
So I get lots of points from L_2n-2 in U
How do you get all of the points of L_2n-2 into U?