Re: Peano's space-filling curve

From: David C. Ullrich (ullrich_at_math.okstate.edu)
Date: 07/08/04 

Date: Thu, 08 Jul 2004 13:58:57 -0500

On 8 Jul 2004 14:48:25 GMT, grubb@lola.math.niu.edu (Daniel Grubb)
wrote:

>So now, assume f:A->B is a one-to-one correspondence and so is
>f':A'->B'.

Under the circumstances it probably would be better not
to corrupt the notation. In the problem we were given
A ~ A' and B ~ B', not A ~ B and A' ~ B'. So it's
f : A -> A' and g : B -> B', and then F(x,y) = (f(x), g(y)).

(I actually think the literally correct F((x,y)) would be
better than the F(x,y) that a person would typically write,
although it's maybe not clear which is going to cause
less confusion.)

>>>[...]So how about F(x,y)=(f(x),f'(y))?

Well, my opinion is that he would have eventually got
this himself (because once we get straight exactly
what we know and exactly what we're trying to do
it's more or less the only thing to try), and if
he had got it himself he would have attained a
deeper understanding.

Otoh you may be right (or rather what I imagine your
motivation in giving it away may be right) - if this
is our first time with this sort of abstraction we
need more than the hints I was giving.

************************

David C. Ullrich

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