Re: Peano's space-filling curve
From: hack (hack_at_watson.ibm.com)
Date: 06/29/04
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Date: 29 Jun 2004 20:01:28 GMT
In article <290620041435245659%edgar@math.ohio-state.edu.invalid>,
G. A. Edgar <edgar@math.ohio-state.edu.invalid> wrote:
>>
>> Here is one that is not too complicated -- and it leaves explicit
>> decimal representations out of the picture:
>>
>> A bijection between the unit interval and the unit square.
>> ----------------------------------------------------------
>>
>> Define, for x non-negative rational:
>> g(0) = 0
>> g(x+1) = (g(x)+3)/4
>> g(1/x) = 1 - g(x)
>> Extend to irrationals by limits.
>>
>> Now define f(x,y) of two real variables, based on g as defined above:
>>
>> f(x,y) = (4g(x)+8g(y))/9
>>
>> This function appears to provide a bijection of the closed unit square
>> to the closed unit interval. It is monotonic in each coordinate (but
>> not continuous).
>>
>> Michel.
>
>
>This claim seems quite unlikely to me. If g is discontinuous, then
>"extend to irrationals by limits" needs explanation, too.
>Are you sure you don't have the same problem as the decimal
>construction of countably many exceptions?
There is indeed a similar phenomenon, but it is not a problem here.
The function is continuous at irrationals, and semi-continuous (either
from the right or from the left, but not both) at rationals.
If you need a hint, look at the Continued-Fraction expansion of the
arguments, and how the Partial Quotients determine the digits of the
base-4 representation of the result. The two finite CF expansions of
a rational argument correspond to the digit sequences that describe
the left and right limits, but the definition at rationals picks one
of them. The linear combination of the pieces takes care of the rest.
Michel.
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