Proof of Bolzano-Weierstrass on R .... .... D&K Theorem 1.6.2 .... ....

[Math Amateur]

I am reading "Multidimensional Real Analysis I: Differentiation" by J. J. Duistermaat and J. A. C. Kolk ...

I am focused on Chapter 1: Continuity ... ...

I need help with an aspect of the proof of Theorem 1.6.2 ...

Duistermaat and Kolk"s Theorem 1.6.2 and its proof read as follows:View attachment 7708In the above proof we read the following:

" ... ... By the definition of supremum, only a finite number of $$x_k$$ satisfy $$a + \delta \lt x_k$$, while there are infinitely many $$x_k$$ with $$a - \delta \lt x_k$$ ... ... "Can someone please explain (very slowly and simply if you will ... :) ... ) how/why only a finite number of $$x_k$$ satisfy $$a + \delta \lt x_k$$, while there are infinitely many $$x_k$$ with $$a - \delta \lt x_k$$ ... ... Help will be much appreciated ... ...

Peter
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MHB members reading the above post may be assisted by having access to Theorem 1.6.1 and the notes succeeding the Theorem ... ... so I providing the same ... as follows ... ...https://www.physicsforums.com/attachments/7709Hope that helps ...

Peter

 

[Euge]

Given $\delta > 0$, $a - \delta$ is not an upper bound for $A$. So there must be a point $x\in A$ such that $a - \delta < x$. Since $x\in A$, $x < x_k$ for infinitely many $k$. Therefore $a - \delta < x_k$ for infinitely many $k$.

To see that only finitely many $x_k$ are greater than $a + \delta$, suppose otherwise. Then $x_k > a + \delta$ for infinitely many $k$, which implies $a + \delta \in A$. Since $a$ is an upper bound for $A$, we would obtain $a + \delta \le a$, a contradiction.