R^2\Q^2 connected?
SusanP
at least one irrational coordinate).
Is S connected?
My thoughts: I know that R is connected,
but that every subset of R, except the
singleton set and interval, is disconnected...
quasi
>Let S= R^2\Q^2. (Points (x,y)in S have
>at least one irrational coordinate).
>
>Is S connected?
>
Yes, S is connected.
Hint:
It's a 2 line proof!
quasi
Lee Rudolph
This is one of the best hints I've seen in a long, long time.
Lee Rudolph
Dave L. Renfro
>>> Let S= R^2\Q^2. (Points (x,y)in S have
>>> at least one irrational coordinate).
>>>
>>> Is S connected?
quasi wrote:
>> Yes, S is connected.
>>
>> Hint:
>>
>> It's a 2 line proof!
Lee Rudolph wrote:
> This is one of the best hints I've seen in a long, long time.
I agree. That was a great hint!
Incidentally, this result is due to Cantor (1882),
who actually proved this for any countable set being
removed and not just Q^2. I discussed this a little in
the following post (the reference to Cantor should be
"Math. Annalen 20", however, and not "Math. Annalen 19"):
http://groups.google.com/group/sci.math/msg/49d13d079238d5b6
The paper that Cantor proved this result in can also
be found on the internet, at
http://dz-srv1.sub.uni-goettingen.de/sub/digbib/loader?ht=VIEW&did=D35881
If the above URL doesn't work, use
http://dz-srv1.sub.uni-goettingen.de/cache/toc/D35867.html
and choose the paper by Cantor that begins on p. 113.
Dave L. Renfro
SusanP
be found on the internet, at
http://dz-srv1.sub.uni-goettingen.de/sub/digbib/loader?ht=VIEW&did=D35881
If the above URL doesn't work, use
http://dz-srv1.sub.uni-goettingen.de/cache/toc/D35867.html
and choose the paper by Cantor that begins on p. 113.
Dave L. Renfro
The paper is not in English. Can you suggest somewhere
else to find a translation? Or somewhere else to see
his proof of this fact that I am asking?
Valeriu Anisiu
The two lines can be replaced by a picture showing that a
polygonal path acb exists for any a,b in S.
The similar set T = Q^2 union (R\Q)^2 is also connected
(a 4 line proof).
How many lines are neeeded for the (dis)proof of the path-connectedness?
V. Anisiu
Peter L. Montgomery
If all lines are horizontal or vertical, we need three lines
to get from, (0, y1) to (0, y2), where y1, y2 are irrational.
Is there a simple argument that two lines always suffice?
Hopefully the argument works if Q is replaced by another subset of R.
--
The 2nd John Adams president lost to the winner of the Battle of New Orleans.
The 2nd George Bush president lost the Battle of New Orleans.
pmon...@cwi.nl Microsoft Research and CWI Home: Bellevue, WA
quasi
<Peter-Lawren...@cwi.nl> wrote:
>In article <dj3pe0$i10$1...@panix2.panix.com> lrud...@panix.com
>(Lee Rudolph) writes:
>>quasi <qu...@null.set> writes:
>>
>>>On Tue, 18 Oct 2005 16:31:43 EDT, SusanP <sus...@hotmail.com> wrote:
>>>
>>>>Let S= R^2\Q^2. (Points (x,y)in S have
>>>>at least one irrational coordinate).
>>>>
>>>>Is S connected?
>>>>
>>>
>>>Yes, S is connected.
>>>
>>>Hint:
>>>
>>>It's a 2 line proof!
>>
>>This is one of the best hints I've seen in a long, long time.
>>
>>Lee Rudolph
>
> If all lines are horizontal or vertical, we need three lines
>to get from, (0, y1) to (0, y2), where y1, y2 are irrational.
>Is there a simple argument that two lines always suffice?
>Hopefully the argument works if Q is replaced by another subset of R.
Yes.
Remove any countable set Sfrom R^2.
Claim R^2-S is path connected.
Take any 2 distinct points p,q in R^2-S.
There are uncountably many line through p, only countably many of
which meet S. Ignoring those, there are still uncountably many lines
through p remaining, so choose one at random -- call it L1.
If L1 actually goes through q (what amazing luck) then L1 is already a
path from p to q.
If not, consider the lines through q.
There are uncountably many lines through q which do not meet S, so
choose one at random -- call it L2. If L2 happens to be parallel to L1
(what incredible bad luck), simply replace L2 with another one,
calling the new one L2 (forget the old one).
Then L1 intersects L2, and since neither line meets S, the union of
the 2 lines gives a path from p to q.
Since any 2 points of R^2-S can be connected by a such a path, R^2-S
is path connected, hence is connected.
quasi
Ross A. Finlayson
You say that there exist straight lines in the 2-D plane for each at
least one of the coordinates is always irrational.
There are vertical or horizontal lines with that property, those with
irrational intercepts, one of the coordinates is always irrational.
There are lines with rational slope with that property, those with
irrational x-intercepts. If a line is not horizontal it intercepts the
x-axis, here necessarily at an irrational point. The sum of a rational
and irrational is irrational, so at each rational point in the domain
there is one irrational coordinate. So they would have rational x
coordinates and irrational y coordinates, or irrational x coordinates
and rational y coordinates.
The sum of two irrationals can be rational, it takes the sum of
infinitely many rationals to be irrational.
If a line has irrational slope, then it may intersect the x-axis at a
rational point or at an irrational point where for no rational x is
y=mx+b, the equation of the non-vertical/asymptotic line, rational. If
m is irrational, and x as one coordinate of the pair rational, then the
sum of of mx+b can not also be rational. For example, the line with
intercept sqrt(2) can not have slope -sqrt(2), because at rational x
coordinate 1, the y coordinate is rational 0. So, neither the product
nor sum of the slope and intercept could be rational.
If the irrationals are P, then is not R^2\P^2 connected?
Say, there exist straight lines in the 2-D plane for each at least one
of the coordinates is always rational.
For vertical and horizontal lines, the intercept is rational. So, for
two points with two rational components, or for one with two rational
and another with rational and irrational coordinates, there exist the
intersecting lines. For two points each with each of a rational and an
irrational coordinate, they can't both have the x coordinate or y
coordinate irrational.
So, any two points with rational coordinates are connect-able, as are
any two points in R^2\P^2 where one of the points has both coordinates
rational.
In the neighborhood of a point with a rational and irrational
coordinate, as the reals are connected in any neighborhood, the
irrational coordinate of the pair is connected to a rational
coordinate.
So, R^2\P^2 is connected. How do other sets dense in the reals
maintain connectedness in the disjoint of their Cartesian product?
Where the sets are dense in the reals, everywhere discontinuous,
etcetera, NCD sets, it is those properties of those sets that determine
the connectedness characteristic of the Boolean removal of their
products from the plane.
Ross
--
"Re-Vitali-ize measure theory. Well-order the reals."
http://groups.google.com/group/sci.math/browse_thread/thread/37605b6d7271c54a/
Robert Low
This has my vote for sci.math post of the week.
Jyrki Lahtonen
> Hint:
>
> It's a 2 line proof!
>
> quasi
ROTFL! A great job, man.
Cheers,
Jyrki
David C. Ullrich
Although it's already been said several times
it's worth saying again: this is truly excellent.
>quasi
************************
David C. Ullrich
David C. Ullrich
<r...@tiki-lounge.com> wrote:
>That is somewhat out of my area, but I don't see that.
Well of course you don't! As you can see from the detailed
version quasi posted in reply to a question, the proof
depends on the fact that R is uncountable.
************************
David C. Ullrich
Dave L. Renfro
> The similar set T = Q^2 union (R\Q)^2 is also connected
> (a 4 line proof).
> How many lines are neeeded for the (dis)proof of the
> path-connectedness?
This example has some personal relevance to me, which
I wrote about back in May 2002. Incidentally, I misread
what the original poster actually said, so the results
I discussed do not directly apply to the post I replied to.
http://groups.google.com/group/sci.math/msg/25850ce28ab37b19
In this post I gave a straightforward proof that the set
is connected, an "infinite line" proof of path-connectedness,
and a second proof of path-connectedness using the fact
that any two countable, unbounded, and dense linear
orderings are order isomorphic.
A few years ago I came across another proof that this
set is path-connected, one that is essentially a simpler
version of the "infinite line" proof. This other proof
pre-dates the publication I cited in my post above,
although not the conference talk where I gave the proof.
I suspect the path-connectedness of T has been in the
folklore for many years, but the text below is the earliest
publication of this fact that I'm aware of.
The proof is on p. 187 (Example 2 in Chapter 6.3) of
Allan J. Sieradski, "An Introduction to Topology and
Homotopy", PWS-KENT Publishing Company, 1992, xiv + 479 pages.
Here's the proof that Sieradski gives, modulo some
notational changes. Recall that we're dealing with
T = Q^2 union (R\Q)^2.
"T contains each non-vertical and non-horizontal line
with rational slope through a point of Q^2. Thus,
the origin O is joined in T to (x,y) in Q^2 by
a path consisting of one or two line segments
(below left). Also, the origin O is joined in T
to each point (x,y) in (R\Q)^2 by a path consisting
of infinitely many line segments in T joining
consecutive entries in a sequence
O = (0,0), (r_1, s_1), (r_2, s_2), ..., (r_n, s_n), ...,
where {r_n} and {s_n} are strictly increasing
sequences of rational numbers with limits x and y,
respectively (below right) [here the idea is the same
as I described in my post, so I'll omit trying to
give the diagram]."
(0,y) *
/
/
/
*
\ *
\ / \
____________\/___\____________
/O (x,0)
/
(x,y) *
Dave L. Renfro
Valeriu Anisiu
BTW, the set
Qx(R\Q) union (R\Q)xQ
is disconnected.
V. Anisiu
Dave L. Renfro
> BTW, the set
> Qx(R\Q) union (R\Q)xQ
> is disconnected.
Yes, and I think I even mentioned it in my talk.
I recall giving answers to all the variations I
could think of along these lines (Q, R, and R-Q
coordinate possibilities in R^2 using basic
set operations), but I don't have the write-up
where I can get to it now. By the way, if this
is in connection with another thing I said
incorrectly in that post, I should point out
that David C. Ullrich called my attention to
it, which lead to this follow-up post:
http://groups.google.com/group/sci.math/msg/4882cf8bae5f3352
There are at least two puns in the previous paragraph,
by the way, and there might even be one in this sentence.
Dave L. Renfro
Ross A. Finlayson
Well, actually I just didn't get it the first couple times I saw it.
You're right, though, because I say infinite sets are equivalent.
R^2\P^2 is connected, right?
That's a three line proof, except one of them is very short, too small
to write in the margin.
Then, in the signature, I refer to another thread where irrationals
have enough of the same properties in terms of being dense in the reals
that for either X=Q or X=P that the result is so.
That is pretty funny about "quasi's" "two line" proof, quasi's good.
At first I was thinking that there was a two sentence proof, and
sitting there wondering what it was. Then, Low noted the item, and I
still didn't get it. Then Lahtonen said something and I went duh.
"No, pun intended."
There are proofs of these things without resort to cardinality. So,
you can remove products of "uncountable" sets from the unit square and
it's still connected. Proof obviously doesn't depend on the reals'
"uncountability."
Borel vs. combinatorics, anyone?
Ross
Keith Ramsay
quasi wrote:
|Remove any countable set Sfrom R^2.
|
|Claim R^2-S is path connected.
|
|Take any 2 distinct points p,q in R^2-S.
|
|There are uncountably many line through p, only countably many of
|which meet S. Ignoring those, there are still uncountably many lines
|through p remaining, so choose one at random -- call it L1.
|
|If L1 actually goes through q (what amazing luck) then L1 is already a
|path from p to q.
|
|If not, consider the lines through q.
|
|There are uncountably many lines through q which do not meet S, so
|choose one at random -- call it L2. If L2 happens to be parallel to L1
|(what incredible bad luck), simply replace L2 with another one,
|calling the new one L2 (forget the old one).
|
|Then L1 intersects L2, and since neither line meets S, the union of
|the 2 lines gives a path from p to q.
There are uncountably many (a continuum of) disjoint circular
arcs connecting the two points.
Keith Ramsay
quasi
But that's circular reasoning!
David C. Ullrich
Yeah, but saying it's a "one-arc proof" is not nearly
as much fun as saying it's a "two-line proof"!
Let's get with the program, eh?
>Keith Ramsay
************************
David C. Ullrich
ma...@mimosa.csv.warwick.ac.uk
But the point is that it's a circular argument.
Derek Holt.
Lee Rudolph
Ronald Bruck
Keith Ramsay <kra...@aol.com> wrote:
So, with your solution, I can get from one to the other in my Ferrarri?
:-)
[I remember having this problem on a topology exam, and spending a LONG
time on it--never quite seeing the idea. The rest of the exam was
trivial. If you haven't seen this idea before, it's tricky.]
--Ron Bruck