Question about Bolzano-Weierstrass Theorem Proof

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This is the proof of Serge Lang in Undergraduate Analysis. I can't quite understand what he meant in his proof. I read different sources about the theorem but Lang's proof is quite odd. Any help?
BTW. theorem 1.1 just states that Every bounded and monotonic sequence is convergent.
 

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  • #2
lavinia
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What don't you understand about the proof?
 
  • #3
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I don't understand the proof of the Bolzano-Weierstrass Theorem according to Serge Lang...
 
  • #4
lavinia
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Yes but what about it don't you understand?
 
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How can Cn be increasing? How can we be sure that the Xn will less than Xn+1???
 
  • #6
lavinia
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How can Cn be increasing? How can we be sure that the Xn will less than Xn+1???
##C_{n}## is the greatest lower bound of the sequence of ##x_{n}##'s except for the first ##n-1## of them. If you remove some more of the ##x_{n} ##'s then the greatest lower bound can not be less than ##C_{n}##.
 
  • #7
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Yes but how can we be sure that Cn+1 will not be less than Cn if ever those Xn's oscillate in a very random manner? For example, if Xn+1 is less than Xn, then Cn+1 is the GLB of the set Xn+1's, and Cn is the GLB of the Xn's but this implies Cn+1 is less than Cn.
 
  • #8
lavinia
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Yes but how can we be sure that Cn+1 will not be less than Cn if ever those Xn's oscillate in a very random manner? For example, if Xn+1 is less than Xn, then Cn+1 is the GLB of the set Xn+1's, and Cn is the GLB of the Xn's but this implies Cn+1 is less than Cn.
Because they are greatest lower bounds. It doesn't matter if the X's oscillate. ##C_{n}## is lower than all of them except the first ##n-1##. ##C_{n+1}## is lower than all of them except one less so that one removed might be very low.
 
  • #9
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Oh! Now I got it, because I was thinking that Cn's can overlap or surpass the Xn+1's and vice versa... Thanks!
 

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