Re: Peano's space-filling curve

From: David C. Ullrich (ullrich_at_math.okstate.edu)
Date: 06/06/04 

Date: Sun, 06 Jun 2004 06:53:32 -0500

On Sat, 5 Jun 2004 19:40:36 +0000 (UTC), Dave Seaman
<dseaman@no.such.host> wrote:

>On Sat, 05 Jun 2004 13:53:18 -0500, David C Ullrich wrote:
>> On Sat, 5 Jun 2004 13:50:34 +0200, "John Morgan"
>><john.morgan@REMOVECAPSataraxie.fr> wrote:
>
>>>I have a problem applying "onto" for the Peano construct. If
>>>the line passes abitrarily close to a point, so that all
>>>neighbourhoods of that point contain point(s) through which
>>>the line passes, does that make it "onto".
>
>> No. To be onto it has to actually pass through every point.
>> Which it does.
>
>Actually, since f is continuous, f([0,1]) is compact and therefore
>closed. Hence, the fact that f([0,1]) is dense in [0,1]^2 is a
>sufficient condition for f to be onto.

We all understand that. Well, most of us do - in a context where
one of us is having so much trouble understanding simple definitions
(as evidenced by _repeated_ confusion over how I and I^2 can have
the same cardinality even though space-filling curves are not
bijections) it seems like pointing this out is going to cause
more confusion than claification.

>But that's a special case. In general, saying a set is dense in another
>is not sufficient to establish equality.

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

David C. Ullrich

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