Why are linearly ordered R and R/{0} not isomophic?

  • #1
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Main Question or Discussion Point

i saw a proof that said “in R/{0} , the set [-1,0) has an upper bound ,but no least upper bound. no such set exists in linearly ordered R” ,but i could not understand it.
 

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  • #2
StoneTemplePython
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so ##[-1,0]## is compact and using the identity function has a maximum at zero. If you delete the zero you lose compactness and there is no longer a maximum.
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edit:
I think I misread this. The issue is that the Real Line with zero deleted doesn't have a smallest positive number for the same reason it doesn't have biggest (i.e. smallest magnitude) negative number. Any non-negative number is an upper bound of ##[-1,0)## but the issue is that if you have ##0## removed from the real line, you must use positive numbers and there isn't a smallest positive number hence no tight upper bound on negative numbers.

The point, I suppose is that ##\mathbb R## and ##\mathbb R /{0}## are both open sets but the former is connected while the latter is not
 
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