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Re: Re: Finding the periphery of a region

  • To: mathgroup at smc.vnet.net
  • Subject: [mg72097] Re: [mg72005] Re: Finding the periphery of a region
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Mon, 11 Dec 2006 04:55:34 -0500 (EST)
  • References: <el8ufm$st3$1@smc.vnet.net> <200612081117.GAA20172@smc.vnet.net> <6DA7D258-EA36-456E-A8F7-B4CBE82001B8@mimuw.edu.pl> <457D0248.7020004@metrohm.ch>

Well, excluding it would of course make things easier. In fact,  
however, in real algebraic geometry one does not usually consider the  
boundary but the closure. So the question is "what is the closure of  
a semi-algebraic set given by a collection of strict inequalities? At  
this point it is natural to expect that the closure should be given   
by relaxing the inequalities but this is not true. A counter example is

A= {{x,y} in R^2 such that x^3-x^2-y^2 >0}
The closure is not {x,y} in R^2 such that x^3-x^2-y^2 >= 0} but {x,y}  
in R^2 such that x^3-x^2-y^2 >0, x>=1}

This can actually be seen from CylindricalDecomposition.
A somewhat curious thing is that eactly the same example is given in  
two large books on this subject, which otherwise deal with quite  
different aspects (one geometry and topology, the other algorithms),  
so perhaps it might be interesting to find a quite different one ;-)

Andrzej


On 11 Dec 2006, at 16:01, Daniel Huber wrote:

> Hi Andrzej,
> thank's a lot for the interesting example. Note that (0,0) is an  
> isolated point. The question is, if an isolated  point belong's to  
> the boundary of an area. I think this is up to how we define  
> "boundary". I would prefere to exclude it. What do you think about  
> this?
> Daniel
>
> Andrzej Kozlowski wrote:
>> *This message was transferred with a trial version of CommuniGate 
>> (tm) Pro*
>> It's well known fact in real algebraic geometry that this does not  
>> work in general. Here is a well known example example (also  
>> included in my response to the OP):
>>
>> x^3 - x^2 - y^2 > 0 && x < 10
>>
>> The boundary is not x^3 - x^2 - y^2 >= 0 && x <= 10
>>
>> This can be seen also on a picture:
>>
>>
>> <<Graphics`InequalityGraphics`
>>
>>
>> InequalityPlot[x^3-x^2-y^2>0&&x<10,{x},{y}]
>>
>> You can see that the point {0,0} which lies on x^3 - x^2 - y^2 ==  
>> 0  is not on the boundary of the region.
>>
>> Andrzej Kozlowski
>>
>>
>> On 8 Dec 2006, at 20:17, dh wrote:
>>
>>>
>>>
>>> Hi,
>>>
>>> try replacing inequalities by equalities. This should work fine  
>>> as long
>>>
>>> as you do not have intersecting regions. E.g.:
>>>
>>> x^2+y^2<100  ==> x^2+y^2=100, obviously a circle
>>>
>>> (5<=x<=25 and -10<=y<=10)  ===>( x==5&&-10<=y<=10) ||
>>>
>>> (x==25&&-10<=y<=10) || (5<=x<=25&&y==-10)  || (5<=x<=25&&y==10),  
>>> four
>>>
>>> line segements.
>>>
>>> Daniel
>>>
>>>
>>>
>>> Bonny Banerjee wrote:
>>>
>>>> I have a region specified by a logical combination of  
>>>> equatlities and
>>>
>>>> inequalities. For example, r(x,y) is a region defined as follows:
>>>
>>>>
>>>
>>>> r(x,y) = x^2+y^2<100 or (5<=x<=25 and -10<=y<=10)
>>>
>>>>
>>>
>>>> How do I obtain the periphery of r(x,y)? I am only interested in  
>>>> finite
>>>
>>>> regions i.e. x or y never extends to infinity.
>>>
>>>>
>>>
>>>> Thanks,
>>>
>>>> Bonny.
>>>
>>>>
>>>
>>>>
>>>
>>>
>>
>>
>>
>
>
> -- 
>
> Daniel Huber
> Metrohm Ltd.
> Oberdorfstr. 68
> CH-9100 Herisau
> Tel. +41 71 353 8585, Fax +41 71 353 8907
> E-Mail:<mailto:dh at metrohm.ch>
> Internet:<http://www.metrohm.ch>
>


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