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- Newsgroups: sci.math
- Path: sparky!uunet!paladin.american.edu!gatech!udel!princeton!fine.princeton.edu!tao
- From: tao@fine.princeton.edu (Terry Tao)
- Subject: Re: proof wanted 2
- Message-ID: <1993Jan9.193759.3671@Princeton.EDU>
- Originator: news@nimaster
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- Nntp-Posting-Host: math.princeton.edu
- Organization: Princeton University
- References: <1993Jan8.195646.1694@cc.umontreal.ca> <1ikq9eINNmue@roundup.crhc.uiuc.edu>
- Date: Sat, 9 Jan 1993 19:37:59 GMT
- Lines: 45
-
- In article <1ikq9eINNmue@roundup.crhc.uiuc.edu> hougen@focus.csl.uiuc.edu (Darrell Roy Hougen) writes:
- >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.
- >
- >I'm a little rusty on analysis, but ..., intuitively, the closest
- >point to x in C must be on the boundary of C which is no problem if C
- >is closed.
- >
- >More precisely, if there were no closest point to x in C, that would
- >imply that given any point c in C, one could find a point c'' in C
- >that was closer to x. Therefore, one could construct a sequence of
- >points in C such that each point was closer to x than its predecessor
- >but whose limit point was not in C. But, if C is closed, then it
- >contains all of its limit points, so there exists c' as defined above.
- >
- >Note that c' need not be unique. For c' to be unique, you need
- >convexity of C.
- >
- >Darrell R. Hougen
-
- I think this proof needs a bit of tightening up - after all, I can think of
- some sort of "spiral" sequence of points which get closer and closer to x
- though perhaps staying always a given distance away from x at least, but
- have more than one limit point.
-
- also Darrel's sequence of c's have to satisfy |x - c_i| --> inf_{c \in C}
- |x - c| for the argument to work.
-
- However, the sequence Darrell mentioned is contained inside a closed ball,
- which is a compact set, so there must be at least one accumulation point,
- and this point is in C, as C is closed. Taking limits you should get the
- result.
-
- Is the result true for any metric space? I don't think the space of
- l^2 sequences will work, if you take the set composed of elements which are
- 0 everywhere except at one place where they are 1...
-
- Terry
-
-