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- Path: sparky!uunet!gatech!destroyer!cs.ubc.ca!unixg.ubc.ca!liuli
- From: liuli@unixg.ubc.ca (Li Liu)
- Newsgroups: sci.math
- Subject: Re: proof wanted 2
- Date: 10 Jan 1993 10:53:31 GMT
- Organization: University of British Columbia, Vancouver, B.C., Canada
- Lines: 32
- Message-ID: <1iov7bINNobh@skeena.ucs.ubc.ca>
- References: <1993Jan8.195646.1694@cc.umontreal.ca> <1993Jan8.211809.21338@galois.mit.edu>
- NNTP-Posting-Host: unixg.ubc.ca
-
- >>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.
-
- This is not true for general metric spaces.
-
- In a previous followup, I made
- a stupid mistake. My example was
- {1/n | n=1,2,...} = "The metric space" =
- C. And x=0. Notice that x is not in the metric space!
-
-
- Here is a better example:
-
- Let X be the set of all functions on [0,1]. For any two functions f and
- g on [0,1], the distance d(f,g) is defined as
-
- d(f,g)= sup{ |f(r)-g(r)|, where r is in [0,1] }.
-
- Then <X,d> is a metric space.
-
- Define:
-
- f_n( 1/n) = 0.5 + 1/n
- f_n( r ) = 0 for all other r in [0,1].
-
- Then C={ f_n |n=1,2,3, ...} is a closed set. You can verify that there is no
- converging sequence in this set because the distance between every pair
- of functions is larger than 0.5 .
-
- Take a point f_0 = 0. There is no closet point in C to f_0.
-