NetNews Usenet Archive 1992 #18/NN_1992_18.iso/spool/sci/math/10514

Text File | 1992-08-22 | 2.6 KB | 54 lines

Newsgroups: sci.math
Path: sparky!uunet!gatech!bloom-beacon!bloom-picayune.mit.edu!athena.mit.edu!zeno
From: [e-mail] (Richard Duffy)
Subject: Re: Is Card(R)=Card(R^2)?
Message-ID: <[e-mail]>
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Organization: Massachvsetts Institvte of Technology
References: <[e-mail]> <[e-mail]> <[e-mail]>
Date: Sun, 23 Aug 1992 01:17:46 GMT
Lines: 41

In article <[e-mail]> 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 point-set 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 nowhere-dense 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 nowhere-dense; 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 sub-ball
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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