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Open and closed intervals and real numbers 
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#1
Oct1610, 06:48 PM

P: 64

1. The problem statement, all variables and given/known data
Show that: Let S be a subset of the real numbers such that S is bounded above and below and if some x and y are in S with x not equal to y, then all numbers between x and y are in S. then there exist unique numbers a and b in R with a<b such that S is one of the intervals (a,b), [a,b), (a,b], or [a,b]. 2. Relevant equations 3. The attempt at a solution Assume if x and y are elements of S with x not equal to y, then all numbers between x and y are in S and S is bounded above and below. Thus there exists a M, N such that M is greater than or equal to the maximal element of S and N is smaller than the minimal element of S. Also all elements between x and y are inside (M,N). 


#2
Oct1610, 07:06 PM

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hi reb659!
i think i'd start by proving that there must be a greatest lower bound and a least upper bound, and then call them a and b, and carry on from there. 


#3
Oct1610, 07:12 PM

P: 64

Good idea.
Isn't it an axiom that if a nonempty subset of R has an upper bound, then it has a least upper bound/sup(S)? 


#4
Oct1610, 07:20 PM

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Open and closed intervals and real numbers
I don't think it's an axiom, I think it's the Dedekindcompleteness theorem:
A bounded realvalued function has a least upper bound and a greatest lower bound. (See Rolle's theorem in the PF Library ) 


#5
Oct1710, 11:54 PM

P: 64

So far:
Since S is bounded above and below, by Dedekind completeness there exists a supremum of S. Call it b. Again by dedekind completeness we can say there exists an infimum of S. Call it b. By definition of sup and inf, a<b. We are left to show a,b are unique and that S is exactly one of the intervals in the OP. To show this, can't we consider four simple different cases in which a,b are either in S or outside of it? 


#7
Oct1910, 11:45 AM

P: 64

Yay!
How exactly does uniqueness follow though? It seems like its trivial to prove. 


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