Question about Bolzano-Weierstrass Theorem Proof

[shinobi20]

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.

 

[lavinia]

What don't you understand about the proof?

 

[shinobi20]

I don't understand the proof of the Bolzano-Weierstrass Theorem according to Serge Lang...

 

[lavinia]

Yes but what about it don't you understand?

 

[shinobi20]

How can Cn be increasing? How can we be sure that the Xn will less than Xn+1?

 

[lavinia]

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 $C_{n}$.

 

[shinobi20]

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.

 

[lavinia]

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.

 

[shinobi20]

Oh! Now I got it, because I was thinking that Cn's can overlap or surpass the Xn+1's and vice versa... Thanks!