Re: Peano's space-filling curve
From: Michael Stemper (mstemper_at_siemens-emis.com)
Date: 06/09/04
- Next message: Eric Hocking: "Re: My paranomal data for Summoning"
- Previous message: conesetter: "Analysis and the Substitution of Vieta."
- In reply to: John Morgan: "Re: Peano's space-filling curve"
- Next in thread: John Morgan: "Re: Peano's space-filling curve"
- Reply: John Morgan: "Re: Peano's space-filling curve"
- Messages sorted by: [ date ] [ thread ]
Date: Wed, 9 Jun 2004 12:49:01 -0500
In article <2ie15fFmb6g6U3@uni-berlin.de>, John Morgan writes:
>Thanks for your very welcome and comprehensible post.
>Between yourself and Michael F. Stemper, who replied
>independantly to the same post, you seem to have arrived at
>an excellent explanation for the layperson of the
>relationship between cardinality and functions.
Thanks for the kind words. One misunderstanding that I should correct,
however, is that I *am* a layperson. I do software project management
for a living.
Most of what I know about set theory came from a single book: _Axiomatic
Set Theory_, by Patrick Suppes. Amazon has copies for sale for as low as
$7.65 (plus S&H), according to:
<http://www.amazon.com/exec/obidos/tg/detail/-/0486616304/102-5199121-5241725?v=glance>
> Perhaps the
>two of you should concoct a synthesis from your posts for
>the online encyclopedia - en.wikipedia.org
Actually, Wikipedia already has an entry:
<http://en.wikipedia.org/wiki/Cardinality>
>> In other words, it is impossible to find a function
>f:[0,1]->[0,1]^2
>> which is one-to-one, onto *and* continuous.
>
>Is this true for all functions? If not, for what class of
>function(s) has it been proved?
When a mathematician says "impossible to find a <x> that <y>", it does
indeed mean that it's true for any <x> that it doesn't <y>. So, in the
above, saying that it's impossible to find a function that has all of
the listed properties says that, for any function at all, it's true
that the function doesn't have *all* of those properties.
>> Yes. A bijection f:A->B always has an inverse function
>f^-1:B->A.
>> It turns out that f^-1 is a bijection also.
>
>Once again I must be missing the crucial point, somehow! In
>the case of the sets [0,1]^2 and [0,1], I understand that
>there is a bijection f: from the square to the line.
>This must have an inverse f^-1:, a bijection, and I
>wonder why this inverse won't map [0,1] to [1,0]^2.
It will. But, it won't be continuous.
-- Michael F. Stemper #include <Standard_Disclaimer> Visualize whirled peas!
- Next message: Eric Hocking: "Re: My paranomal data for Summoning"
- Previous message: conesetter: "Analysis and the Substitution of Vieta."
- In reply to: John Morgan: "Re: Peano's space-filling curve"
- Next in thread: John Morgan: "Re: Peano's space-filling curve"
- Reply: John Morgan: "Re: Peano's space-filling curve"
- Messages sorted by: [ date ] [ thread ]
Relevant Pages
|