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Jan Kubrik...
Posted: Tue May 13, 2008 6:08 am
Guest
Consider R^n with the usual topology and consider an equivalence relation ~ on R^n with classes that consist of 2 and only 2 different elements
(for every x in R^n we have [x]={x,x'}). Is it possible that the quotient space R^n/~ :

i) is Hausdorff,
ii) is indiscrete,
iii) is discrete?
Daniel Lichtblau...
Posted: Tue May 13, 2008 6:20 am
Guest
On May 13, 11:08 am, Jan Kubrik <kubri... at (no spam) gmail.com> wrote:
Quote:
Consider R^n with the usual topology and consider an equivalence relation ~ on R^n with classes that consist of 2 and only 2 different elements
(for every x in R^n we have [x]={x,x'}). Is it possible that the quotient space R^n/~ :

i) is Hausdorff,
ii) is indiscrete,
iii) is discrete?

Are they pairing continuously? That usually factors into whether they
are being indiscreet.

As for the rest, what happened last month with this?

http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist;task=show_msg;msg=0857.0001

Daniel Lichtblau
Wolfram Research
Robert H. Lewis...
Posted: Tue May 13, 2008 7:41 am
Guest
Quote:
Consider R^n with the usual topology and consider an
equivalence relation ~ on R^n with classes that
consist of 2 and only 2 different elements
(for every x in R^n we have [x]={x,x'}). Is it
possible that the quotient space R^n/~ :

i) is Hausdorff,
ii) is indiscrete,
iii) is discrete?


You seem to have posted to the wrong sci.math group. This is for symbolic computation.

Robert H. Lewis
Fordham University
 
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