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 \(\displaystyle x_k\) satisfy \(\displaystyle a + \delta \lt x_k\), while there are infinitely many \(\displaystyle x_k\) with \(\displaystyle a - \delta \lt x_k\) ... ... "Can someone please explain (very slowly and simply if you will ... :) ... ) how/why only a finite number of \(\displaystyle x_k\) satisfy \(\displaystyle a + \delta \lt x_k\), while there are infinitely many \(\displaystyle x_k\) with \(\displaystyle a - \delta \lt x_k\) ... ... 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.

 

[math771]

Hi Peter,

I'll do my best to explain this in a simple and clear manner.

First, let's define what we mean by the supremum of a set. The supremum of a set S is the smallest number that is greater than or equal to all the numbers in S. In other words, it is the "least upper bound" of the set.

Now, in the proof of Theorem 1.6.2, we are dealing with a set of real numbers, let's call it X. We know that X is bounded above by a, which means that a is an upper bound for X. This is because the definition of a supremum states that it must be greater than or equal to all the numbers in the set.

Next, we introduce the concept of a neighborhood of a point a. A neighborhood of a is a set of numbers that are close to a, within some distance \delta. In other words, if we take any number x in the neighborhood, it will be within \delta distance from a.

Now, in the proof, we want to show that there are infinitely many numbers in X that are close to a, specifically within \delta distance. We do this by considering two cases: a + \delta and a - \delta.

For the first case, a + \delta, we know that there are only a finite number of numbers in X that are greater than a + \delta, since a is the supremum of X. This means that there are only a finite number of numbers in X that satisfy a + \delta \lt x_k.

On the other hand, for the second case, a - \delta, we know that there are infinitely many numbers in X that are less than a - \delta, since a is the supremum of X. This means that there are infinitely many numbers in X that satisfy a - \delta \lt x_k.

I hope this explanation helps to clarify why there are only a finite number of x_k satisfying a + \delta \lt x_k, while there are infinitely many x_k satisfying a - \delta \lt x_k.