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Re: looking for example of closed set that is *not* complete in a metric space
Posted:
Feb 2, 2013 12:10 AM
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On Saturday, February 2, 2013 12:52:55 AM UTC+8, peps...@gmail.com wrote: > ...To say that a space is "closed" > (as in your statement "closed -> complete") doesn't really mean anything. > To make progress replace "closed -> complete" by something more > formal and rigorous and precise.
This is certainly good advice and many apologies for my sloppy original posting. Is the following any better?...
Let (X,d) be a metric space. Let T be the topology induced by d and (X,T) be the resulting topological space. Let Y be a subset of X. Then (Y,d) is complete ==> Y is closed in (X,d). Alternatively, (Y,d) is complete ==> Y is closed in (X,T).
But what about the converse? That is, is this true? Y is closed in (X,d) ?==>? (Y,d) is complete
One might guess that it is not true. So would someone happen to know of a counterexample in which the set Y is closed in (X,d), but yet (Y,d) is *not* complete?
References: 1. Kubrusly(2011) Theorem 3.40 page 129: books.google.com.tw/books?vid=ISBN0817649980&pg=PA129
2. Haaser(1991) 6.10 Proposition page 75: books.google.com.tw/books?vid=ISBN0486665097&pg=PA75
Many thanks in advance, Dan
On Saturday, February 2, 2013 12:52:55 AM UTC+8, peps...@gmail.com wrote: > On Friday, February 1, 2013 4:37:40 PM UTC, Daniel J. Greenhoe wrote: > > > Let (Y,d) be a subspace of a metric space (X,d). > > > > > > > > > > > > If (Y,d) is complete, then Y is closed with respect to d. That is, > > > > > > > > > > > > complete==>closed. > > > > > > > > > > > > Alternatively, if (Y,d) is complete, then Y contains all its limit > > > > > > points. > > > > > > > > > > > > Would anyone happen to know of a counterexample for the converse? That > > > > > > is, does someone know of any example that demonstrates that > > > > > > closed --> complete > > > > > > is *not* true? I don't know for sure that it is not true, but I might > > > > > > guess that it is not true. > > > > > > > > > > > > Many thanks in advance, > > > > > > Dan > > > > You need to understand that "closed" and "open" don't characterize topologies. > > Rather "X is open in Y" describes a relationship between X and Y. > > To say that a space is complete or compact or Hausdorff makes a statement about a topological space. To say that a space is "closed" (as in your statement "closed -> complete") doesn't really mean anything. > > To make progress replace "closed -> complete" by something more formal and rigorous and precise. > > > > Paul Epstein
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