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Chapter 12: Problem 8

Let \(R\) be the unit disk \(\left\\{(x, y): x^{2}+y^{2} \leq 1\right\\}\) with (0,0) removed. Is (0,0) a boundary point of \(R ?\) Is \(R\) open or closed?

Short Answer

Expert verified
A: (0,0) is a boundary point of the set R. The set R is open but not closed.

Step by step solution

01

Understanding the definition of a boundary point

Recall that a point \(p\) is a boundary point of a set \(A\) if every neighborhood around \(p\) contains a point in \(A\) and a point not in \(A\). To check whether (0,0) is a boundary point of the set \(R\), we will investigate if every neighborhood around (0,0) satisfies this condition.
02

Investigate neighborhoods around (0,0)

Consider a neighborhood around (0,0) with radius \(r > 0\). There exist points within this neighborhood that are both in the set \(R\) and not in \(R\). For example, take a point \((x, y)\) such that \(0 < x^{2} + y^{2} < r^{2}\) and \(x^{2} + y^{2} \leq 1\). This point belongs to \(R\), because it is in the unit disk and not equal to (0,0). A point outside of \(R\) is (0,0) itself, since it has been removed from the set. Thus, every neighborhood around (0,0) contains a point in \(R\) and a point not in \(R\).
03

Determine if (0,0) is a boundary point of R

Since every neighborhood around (0,0) contains a point in \(R\) and a point not in \(R\), (0,0) is a boundary point of the set \(R\).
04

Understanding open and closed sets

Recall that a set is open if every point in the set is an interior point, meaning that there exists a neighborhood around the point that is entirely contained within the set. A set is closed if it contains all its boundary points.
05

Check if R is an open set

To check if the set \(R\) is open, we need to find if each point in \(R\) is an interior point. Consider any point \((x, y) \in R\) with \(0 < x^{2} + y^{2} \leq 1\). We can find a neighborhood around this point with radius \(r = \frac{1}{2}\min\{x^{2}+y^{2}, 1 - x^{2} - y^{2}\}\). This neighborhood is entirely contained in the unit disk because all points in the neighborhood have a distance to \((x, y)\) less than \(r\) and thus also satisfy \(0 < x^{2} + y^{2} \leq 1\). Therefore, \((x, y)\) is an interior point. As this holds for all \((x, y) \in R\), the set \(R\) is open.
06

Check if R is a closed set

To check if the set \(R\) is closed, we must verify if it contains all its boundary points. However, we have already shown that (0,0) is a boundary point of \(R\), and it has been removed from the set. Therefore, \(R\) does not contain all its boundary points and is not a closed set.
07

Conclusion

The point (0,0) is a boundary point of the unit disk \(R\) with the center removed. The set \(R\) is open but not closed.

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