Re: Perfect set proof

From: Ron Sperber (ronsperber_at_optonline.net)
Date: 10/29/04 

Date: Fri, 29 Oct 2004 10:12:05 GMT

Agapito Martinez wrote:
> israel@math.ubc.ca (Robert Israel) wrote in message news:<clravt$ppm$1@nntp.itservices.ubc.ca>...
>
>>In article <d2b49620.0410280738.5bdb304b@posting.google.com>,
>>Agapito Martinez <agapito6314@aol.com> wrote:
>>
>>
>>>>"Agapito Martinez" <agapito6314@aol.com> wrote in message
>>>>news:d2b49620.0410260726.ddccdf2@posting.google.com...
>>>>
>>>>>P is a perfect subset of R^n. A perfect set (Rudin) is one all of
>>>>>whose points are limit points, and which contains all its limit
>>>>>points.
>>
>>
>>
>>>>>There exists an (open) neighborhood V in R^n so that P intersect V is
>>>>>not empty. If W is a subset of V and also a neighborhood in R^n, how
>>>>>does one show that P intersect W is not empty?
>>
>>
>>
>>>Of course, I asked the wrong question. The question is how does one
>>>show that there exists a subset of V, name it W, which is also a
>>>neighborhood in R^n, such that P intersect W is not empty. Thanks.
>>
>>What about W = V?
>>
>>Maybe this was also the wrong question.
>>
>
> Indeed. I should have saID W, proper subset of V. Thanks.
This still seems pretty trivial. Let x be some element of V.
if x is not an element of P, then its clear V-{x} is a neighborhood that
intersects P.

If x is an element of P, then x is a limit point of P. and V is a
neighborhood of x. Thus by definition there is a point y with y not
equal to x, y in P intersect V. Thus again V-{x} is a neighborhood that
intersects P.

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