From: zeno@athena.mit.edu (Richard Duffy)
Newsgroups: sci.math
Subject: Re: Is Card(R)=Card(R^2)?
Date: 23 Aug 92 01:17:46 GMT
In article <16g754INNsmk@function.mps.ohiostate.edu> Gerald Edgar writes:
>>How about the unit line minus the end points, and the interior of
>>the unit square?
>
>A continuous bijection from the open interval onto the open square
>is also not possible. The reason for it is a bit harder, this time.
>But still not beyond a first course in pointset topology.
>
I'm curious to know whether the method Prof. Edgar has in mind is simpler or
just radically different than the following:
Given a continuous bijection f: I > I x I where I is an open interval.
I is a countable union of closed subintervals K_n , hence I x I is the
countable union of the f(K_n). But I x I is open in R x R, hence has the
Baire property that it is not a countable union of nowheredense subsets.
[If you want to avoid the detail of proving that, just use I = R in the
first place, via a homeomorphism, and the fact that R x R is a complete
metric space.]
So some particular f(K_n) is not nowheredense; being also closed (since
it is compact, as the continuous image of the compact set K_n ), f(K_n)
therefore contains an open ball B, which in turn contains a closed subball
C . Now the inverse image f^{1}(C) is a closed subset of K_n and
thus compact, so the restriction of f to this set is a homeomorphism
onto C , i.e. f^{1} restricted to C is continuous. But then f^{1}(C)
must be a connected subset of K_n , so it's a closed interval J. (We've
essentially reduced the problem to the earlier one). Now just remove three
points from C  you still have a connected set, but its image under
f^{1}, which is J \ {three points} since f is bijective, can't
possibly be connected. [Note that a closed interval with *two* points
removed could still be connected.] So we've contradicted f out of
existence.

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From: wft@math.canterbury.ac.nz (Bill Taylor)
Newsgroups: sci.math
Subject: [0,1] homeomorphic to most of [0,1]^2
Date: 23 Aug 92 01:51:21 GMT
There was a flurry of concern in sci.math a week or two ago, as to whether
or not there was a bicontinuous bijection (i.e. a homeomorphism) between
[0,1] and [0,1]^2 . Several posters wrote in with their conviction that
such *did* exist, probably confusing vaguely remembered classic examples such
as Cantor's noncontinuous bijection, and Peano's continuous many1 mapping.
There were some excellent replies, ranging from the esoteric, to pointing
out the obvious such as that [0,1] could be disconnected by removing a
single point, whereas [0,1]^2 couldn't, so they could not possibly be
homeomorphic. Hopefully they have killed off this potential piece of
mathematical folklore.
During the course of the discussion, I mentioned in passing that......
> there *is* a homeomorphism from the unit line to "most of" the unit square.
> i.e. so that the image in the unit square has measure arbitrarily close to 1 .
An "almosthomeomorphism", one might say, from [0,1] to [0,1]^2 .
I have been asked (by email) to substantiate this claim. Being unable to find a
reference for it, I have reconstructed and asciiedup a standard example as it
was shown to me long ago. I thought I would post it here, in case there should
be some others who would like to see such an example.
Though it sounds amazing at first, the existence of an almosthomeomorphism
will come as no surprise to topologists, who know that there is almost no
connection between topology and measure theory. For instance there are
dense sets of measure zero (the rationals), and nowheredense sets of
measure 1e , in the unit interval, for arbitrarily small positive e .
These last could be called "thick Cantor sets". A thick Cantor set is
constructed by removing central intervals from [0,1] and its subsequent
subintervals, NOT all of length 1/3 of their intervals, as in the standard
Cantor set, but of proportional lengths e, (e^2)/2, (e^3)/4 etc, so that
the measure of the remnant is > 1ee^2e^3... > 12e . The intervals removed
can be open, leaving a closed Cantorlike set; or closed, leaving the
"quasiinterior" (i.e. no internal endpoints) of one; having the same measure.
The almosthomeomorphism I display, will have as its range, a subset of [0,1]^2
consisting (mostly) of the crossproduct with itself, of one of these
quasiinteriorofthickCantorsets.
So this range will have measure very close to 1.
Our first step is to create the set [0,1]^2 from which countably many
crossshaped sections have been removed, but crosses of decreasingly small
measure. I believe this remnant set is sometimes called "Sierpinski dust".
~~~~~~~~~~~~~~~
It is totally disconnected, nowhere dense, but of very large measure.
I illustrate it here....

           
 +  +   +  +   +  +   +  + 
           
============ ============ ============ ============
           
 +  +   +  +   +  +   +  + 
           
' ` ' `
   
. . . .
           
 +  +   +  +   +  +   +  + 
           
============ ============ ============ ============
           
 +  +   +  +   +  +   +  + 
           
 
 
 area e removed in this central cross 
 
 
           
 +  +   +  +   +  +   +  + 
           
============ ============ ============ ============
           
 +  +   +  +   +  +   +  + 
           
' ` ' `
   area (e^2)/4 removed here 
. . . .
           
 +  +   +  +   +  +   +  + 
           
============ ============ ============ =(e^3)/16==removed
           
 +  +   +  +   +  +   +  etc. 
           

We will draw a continuous nonselfintersecting image of [0,1] in the unit square,
that covers this Sierpinski dust.
First join the 4 large subsquares with straight line segments as shown...

   
   ***** * 
 ******   * * 
 * *   * * 
 * *   * * 
 * *   * * 
 * *   * * 
 * ************ * 
 *   * 
 *   * 
 *   * 
 *   * 
 *   * 
 *   * 
 *   * 
* *
 1  
  
 0 B 
* *
 *   * 
 *   * 
 *   * 
 *   * 
 *   * 
 *   * 
 *  A * 
 * ******************* * 
 * *   ***** 
 *****   
   
   
   
   
   

The dotted lines show where further connections still have to be made.
"0" and "1" label the ends of our imageset. Note that the three solid lines
form a fixed part of the final curve; they will not be overwritten later
(unlike the Peano example). Points "A" & "B" appear again below.
To define the curve inside each quarter, join up as shown for the magnified bottom
right quarter... (again, the dotted sections to be filled later) 


B 
*
    
    
    
  /  
 /  
 ************ ********** 
 *  / * 
 *  /  * 
 *  /  * 
 *    * 
 *    * 
*  *
 \   
A *\ \/  
 \  
* *
 *   * 
 *   * 
 *   * 
 *   * 
 *   * 
 **************************** 
   
   
   
   
   

The other three quarters are to have internal joins in a similar way.
Then make internal joins in the 16 size(1/16) subsquares in a similar way
to this; and so on in reducing recursive fashion; countably many times.
The resulting curve is continuous and nonselfintersecting (i.e. 11),
though both of these facts need a little proof. The curve is made up of
countably many straight line sections, of total (areal) measure zero,
and an uncountable number of "connecting" limit points, the Sierpinski dust,
of measure close to 1.
Q.E.D.

Bill Taylor wft@math.canterbury.ac.nz

Free will  the result of chaotic amplification of quantum events in the brain.
Galaxies  the result of chaotic amplification of quantum events in the big bang.

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