Re: Peano's space-filling curve

From: Shmuel (Seymour J.) Metz (spamtrap_at_library.lspace.org.invalid)
Date: 06/08/04 

Date: Tue, 08 Jun 2004 16:11:53 -0300

In <2ij6g0Fmc61aU1@uni-berlin.de>, on 06/07/2004
   at 02:17 PM, "John Morgan" <john.morgan@REMOVECAPSataraxie.fr>
said:

>Here I read that f is continuous and onto. Elsewhere a
>poster tells me that f cannot be all three of '1 to
>1',continuous and onto. So f is not '1 to 1'?

Correct, if you're talking about the same f.

>Re-arrange the following words to make a well-known phrase,
>or saying. "together., Get, act, your,"

Act his together is.

>After reading the post, the response, and the response to the
>response, I think I can see more clearly than ever why I am
>experiencing confusion.

No you can't. The base problem is that you have not been reading what
people actually wrote, and have been responding to things that existed
only in your head. A secondary problem is that you've been conflating
a 1-1 mapping between I and I^2 with an unrelated function attributed
to Peano.

In <2ij6g2Fmc61aU2@uni-berlin.de>, on 06/07/2004
   at 02:42 PM, "John Morgan" <john.morgan@REMOVECAPSataraxie.fr>
said:

>But if you mean (1/3,2/3) you will need to explain to me why.

He did.

>The geometry screams out that it does not.

No. In fact, you don't understand what the Geometry is.

>Indeed every recursion of the Peano construct "creates" further
>points with rational coordinates that will never be crossed.

Why do you believe that matters? None of those functions is the Peano
curve.

>Does the algorithm defining the drawing of the curve

There is none; the curve is the limit of a sequence of functions.

>as one passes from the finite to the infinite?

What does that mean?

>So if there is a bijection from[0,1] to [0,1]^2 why did nobody say
>so at the beginning of all of this.

He did. If David told you that there was a red car in his driveway,
would you ask why he didn't tell you that there was a red thing that
was a car? It is part and parcel of the definition of a bijection that
it has an inverse, which is also a bijection.

>Once more it seems things are "very clear" to you. What Dan Grubb
>said was he hoped he'd cleared "some things up", which is not quite
>the same as "cleared something up".

Correct, but irrelevant. David did not challenge Dan's statement that
he (Dan) believed that he had cleared things up; rather he accepted
Dan's belief and indicated that he (David) thought that belief to be
erroneous. As it was.

>You may be the king pin mathy on this group but that actually seems
>to work to the disadvantage of both of us. You might try leaving
>some of the people whose posts I reply to, to answer for themselves.

This is a news group on Usenet, not private e-mail. Potentially
anybody could respond to any article. In particular, David responding
to your article does not prevent others from also responding, should
they choose. Neither you nor I, nor David, nor anybody else, has the
right to demand that a specific poster respond or refrain from
responding.

>You continue to assume that I can instantly see the truth behind
>these bald statements, despite the fact that I tell you I am finding
>it difficult to do so.

You continually make claims about statements that you admit you don't
understand, instead of asking for clarification.

>> (i) There _is_ a bijection from I^2 onto I. Its inverse
>> _is_a bijection from I onto I^2. Such a function
>> _cannot_ be continuous.
>>
>> (ii) There also _is_ a continuous function mapping I onto
>> I^2. It is not 1-1, hence not a bijection.
>You continue to assume that I can instantly see the truth behind
>these bald statements, despite the fact that I tell you I am finding
>it difficult to do so. I interpret the above as follows: (i) assures
>us that line and square have the same cardinality, while (ii)
>guarantees that the Peano curve, that conforms to (ii), fills the
>square because of (i). If that's wrong, then put me right.

It's wrong because you wrote "because of (i)"; (ii) is true for other,
unrelated, reasons, *NOT* because of (i). As David has already
explained.

>Actually, I have only just registered the fact that
>there are bijections both ways due to being "told"
>previously that there was a surjection from [1,0] to
>[1,0]^2.

By whom? I saw statements that there was a bijections and I saw
statements that there was a continuous surjection; the only statements
that I saw claiming those to be related came from you.

>And before you say "yes that's true" I had
>initially assumed that the fact I was being offered a
>surjection implied that there was no bijection.

You weren't just offered any old surjection, you were offered two very
special surjections. One of them was a bijection and one of them was
continuous. They were not related and were not relevant to the same
context.

Define f,g: Z-> by f(n)=floor(n/2), g(n)=n+1. You will find that f is
a surjection. Does that mean that g is not a bijection?

>Please, all of you, try to get your act together.

PKB. 

-- 
     Shmuel (Seymour J.) Metz, SysProg and JOAT
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