Prove that closed balls are closed sets in the standard topology on R?.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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![**Prove that closed balls are closed sets in the standard topology on \(\mathbb{R}^2\).**
In this task, you are asked to demonstrate that in the context of the standard topology on the Euclidean plane \(\mathbb{R}^2\), a closed ball is a closed set.
A **closed ball** of radius \(r\) centered at a point \(c = (x_0, y_0)\) in \(\mathbb{R}^2\) is defined as the set of points \((x, y)\) that satisfy the inequality:
\[
\sqrt{(x - x_0)^2 + (y - y_0)^2} \leq r
\]
Mathematically, this can be expressed as:
\[
B(c, r) = \{ (x, y) \in \mathbb{R}^2 \mid \sqrt{(x - x_0)^2 + (y - y_0)^2} \leq r \}
\]
### Explanation of the Proof:
To prove the closedness of a set in topology, you can demonstrate that its complement is open. In the standard topology, a set is open if for every point in the set, there exists a neighborhood entirely contained in the set.
**Steps in the Proof:**
1. **Complement of the Closed Ball**: Consider the set of all points \((x, y) \in \mathbb{R}^2\) such that \(\sqrt{(x - x_0)^2 + (y - y_0)^2} > r\). This set represents the complement of the closed ball.
2. **Openness of the Complement**: For any point \((x, y)\) in this complement, construct an epsilon-neighborhood around it that is fully contained within this complement. This involves ensuring that the smallest distance from any point in the epsilon-neighborhood to the center \(c\) is greater than \(r\).
3. **Conclusion**: If the complement of the closed ball satisfies the criteria for an open set in the standard topology, then the original closed ball must be closed.
This concept is central to understanding topology's basic properties on metric spaces like \(\mathbb{R}^2\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff610f60e-c097-44c0-9c2f-23e045cb2fc2%2F8d9a42ef-6b59-4006-b2ec-253518acf029%2Fzns8y44_processed.png&w=3840&q=75)
Transcribed Image Text:**Prove that closed balls are closed sets in the standard topology on \(\mathbb{R}^2\).**
In this task, you are asked to demonstrate that in the context of the standard topology on the Euclidean plane \(\mathbb{R}^2\), a closed ball is a closed set.
A **closed ball** of radius \(r\) centered at a point \(c = (x_0, y_0)\) in \(\mathbb{R}^2\) is defined as the set of points \((x, y)\) that satisfy the inequality:
\[
\sqrt{(x - x_0)^2 + (y - y_0)^2} \leq r
\]
Mathematically, this can be expressed as:
\[
B(c, r) = \{ (x, y) \in \mathbb{R}^2 \mid \sqrt{(x - x_0)^2 + (y - y_0)^2} \leq r \}
\]
### Explanation of the Proof:
To prove the closedness of a set in topology, you can demonstrate that its complement is open. In the standard topology, a set is open if for every point in the set, there exists a neighborhood entirely contained in the set.
**Steps in the Proof:**
1. **Complement of the Closed Ball**: Consider the set of all points \((x, y) \in \mathbb{R}^2\) such that \(\sqrt{(x - x_0)^2 + (y - y_0)^2} > r\). This set represents the complement of the closed ball.
2. **Openness of the Complement**: For any point \((x, y)\) in this complement, construct an epsilon-neighborhood around it that is fully contained within this complement. This involves ensuring that the smallest distance from any point in the epsilon-neighborhood to the center \(c\) is greater than \(r\).
3. **Conclusion**: If the complement of the closed ball satisfies the criteria for an open set in the standard topology, then the original closed ball must be closed.
This concept is central to understanding topology's basic properties on metric spaces like \(\mathbb{R}^2\).
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