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- Newsgroups: sci.math
- Path: sparky!uunet!math.fu-berlin.de!news.tu-chemnitz.de!mb3.tu-chemnitz.de!mapool14.mathematik.tu-chemnitz.de!muelleru
- From: muelleru@Mathematik.TU-Chemnitz.DE (Ulrich Mueller)
- Subject: Re: proof wanted 2
- Message-ID: <muelleru.16@Mathematik.TU-Chemnitz.DE>
- Sender: inews@mb3.tu-chemnitz.de (Internet news)
- Organization: University of Technology Chemnitz, FRG
- References: <1993Jan8.195646.1694@cc.umontreal.ca>
- Date: Mon, 11 Jan 1993 13:59:30 GMT
- Lines: 23
-
- In article <1993Jan8.195646.1694@cc.umontreal.ca> cazelaig@ERE.UMontreal.CA (Cazelais Gilles) writes:
-
-
- > n
- >Is it true that if C is a nonempty closed subset of R and x is a point not
- >in C that there exists a point c in C that is closest in C to x.
- >
- >i.e. such that: |x-c'| >= |x-c| for all c' in C.
- >
- >If it is true I would appreciate if someone could give me a proof
- >of the result.
- >
- > Thanks in advance.
-
- There exists a supremum m of all the distances between x and any other
- element of C.
- For every n hence there is a point x_n from C with d(x_n,x)<1/n.
- {x_n} has infinite elements and it is bordered. Therefore it has an
- "agglomeration point" p ( Haeufungspunkt in German ).
- It's distance to x is m. (To show indirect).
- C is closed, therefore p in C.
-
- Ciao, Ulrich
-