Problem 8 - Calculus Early Transcendentals

Question

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?

Step-by-Step Solution

Answer

Answer: (0,0) is a boundary point of R, and R is an open set.

Step 1: Understand the key concepts

Unit disk: A unit disk is a set of all points in a plane whose distance from the origin is less than or equal to 1. Open set: A set is called open if every point in the set has an open circle around it that is entirely contained within the set. Closed set: A set is called closed if it contains all its boundary points.

Step 2: Determine if (0,0) is a boundary point of R

A boundary point of a set is a point that can be "approached" or "approximated" by the points in the set. To determine if (0,0) is a boundary point of R, we will consider an open circle around (0,0) and see if it has any intersection with the set R or not. Let's consider an open circle with radius r (where r > 0) around (0,0). Every open circle of radius r around (0,0) will intersect R as long as r > 0 because there will be points in R which are at a distance less than r from (0,0). Thus, (0,0) is a boundary point of R.

Step 3: Determine if R is open or closed

Now that we have determined that (0,0) is a boundary point of R, we need to check if R contains this boundary point. By definition, our set R is the unit disk with (0,0) removed. Therefore, R does not contain (0,0). Since R does not contain one of its boundary points, it cannot be a closed set. Furthermore, every other point in R has an open circle around it that is entirely contained within R since the points in R is away from the origin. Thus, R is an open set. In conclusion, (0,0) is a boundary point of R, and R is an open set.

Key Concepts

Boundary Point

A boundary point is a concept in topology, which helps us understand the 'edges' of a set. For any given set, a boundary point can be reached by points within the set as well as points outside the set. The interesting part about boundary points is that they don't necessarily have to be part of the set themselves. In the context of the unit disk example given, we examined whether the origin, (0,0), is a boundary point of the set \(R\), which is the unit disk minus the point (0,0).

To explore this, you can imagine drawing a circle with any positive radius around (0,0). This circle will inevitably include some points that belong to \(R\) and some points that are not in \(R\) (since (0,0) is excluded from \(R\)). These circles will always cross the boundary defined by \(x^2 + y^2 = 1\). As a result, (0,0) satisfies the criteria to be a boundary point of the set \(R\).

This example illustrates that even points not included in the set itself can still be boundary points, providing an intriguing insight into the boundary properties of geometrical sets.

Open Set

An open set is a foundational concept in topology. Simply put, a set is open if, for every point within the set, there's a small "room" around it, which is entirely contained within the set. This "room" is usually described as an open circle or ball around the point.

In the context of the unit disk \(R\) (excluding the origin), consider any point in this set (other than the origin). We find that we can draw an open circle around every such point with small enough radius so that the circle is completely within \(R\). As a result, \(R\) effectively satisfies the condition for being an open set.

The set \(R\) doesn't include its boundary point (0,0), which is a key characteristic that distinguishes it as an open set. It's important to note that being an open set implies that its boundary points lie outside the set itself.

Closed Set

A closed set is somewhat the opposite of an open set in topology. A set is deemed closed if it includes all its boundary points. More formally, if you can list every boundary point and they're all part of the set, then the set is closed.

In examining our set \(R\), the unit disk without the point (0,0), we find that while (0,0) is a boundary point, it is not included in \(R\). This immediately tells us that \(R\) is not closed. Remember, a closed set must include all its boundaries.

Closed sets often include their "edges" or "limits," but \(R\) in this scenario does not include the crucial boundary point of the disk, thus disqualifying it from being a closed set. This property is inherently linked with the idea that closed sets in ext{general topological } spaces are complete and contain all limit points.