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

In summary: Your Name]In summary, in the proof of Theorem 1.6.2, we use the definition of supremum to show that 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. This is because a is the supremum of the set X, and a + \delta is a neighborhood of a, while a - \delta is not.
  • #1
Math Amateur
Gold Member
MHB
3,998
48
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
==========================================================================================

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
 
Physics news on Phys.org
  • #2
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.
 
  • #3


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.

 

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

1. What is the Proof of Bolzano-Weierstrass on R?

The Proof of Bolzano-Weierstrass on R, also known as the D&K Theorem 1.6.2, is a mathematical proof that states that every bounded sequence in the real numbers has a convergent subsequence. This means that if a sequence has values that are limited to a certain range, there will always be a subsequence that approaches a specific limit within that range.

2. Who discovered the Proof of Bolzano-Weierstrass on R?

The proof was first introduced by mathematicians Bernard Bolzano and Karl Weierstrass in the 19th century. However, it has been refined and expanded upon by many other mathematicians since then.

3. Why is the Proof of Bolzano-Weierstrass on R important?

The proof is important because it is a fundamental theorem in real analysis and has numerous applications in other areas of mathematics, including calculus, differential equations, and topology. It also helps to establish the completeness of the real numbers, which is essential in many mathematical proofs.

4. What is the difference between the Proof of Bolzano-Weierstrass on R and the Bolzano-Weierstrass Theorem?

There is no difference between the Proof of Bolzano-Weierstrass on R and the Bolzano-Weierstrass Theorem. They both refer to the same mathematical concept and proof. The only difference is the way in which it is referred to or presented by different mathematicians or textbooks.

5. Can the Proof of Bolzano-Weierstrass on R be extended to other number systems?

Yes, the proof can be extended to other complete metric spaces, such as the complex numbers. However, the proof may need to be modified slightly to account for the differences in the properties of these number systems.

Back
Top