Question

Sketch the set \(X=\left\\{(x, y) \in \mathbb{R}^{2}: 1 \leq x^{2}+y^{2} \leq 4\right\\}\) on the plane \(\mathbb{R}^{2} .\) On a separate drawing, shade in the set \(\bar{X}\).

Short answer

The set \(X\) is the area between two circles of radii 1 and 2, centered at the origin. The set \(\bar{X}\) is the closure of \(X\), and it covers the same area as \(X\), but also includes the boundary of the outer circle.

Step-by-step solution

Identify the circles

The inequality \(1 \leq x^{2} + y^{2} \leq 4\) defines a region between two circles. The first circle (inside) has radius \(r_1 = \sqrt{1} = 1\) and the second circle (outside) has radius \(r_2 = \sqrt{4} = 2\). Both are centered at the origin (0,0).

Sketch the set X

Draw the first circle of radius 1 and the second circle of radius 2 on the same graph. The set \(X\) is the area enclosed between these two circles, including the inner circle's boundary and excluding the outer circle's boundary.

Sketch the closed set \(\bar{X}\)

The closure of \(X\) denoted as \(\bar{X}\) is the smallest closed set containing \(X\), meaning it includes the boundary points. Thus, the set \(\bar{X}\) includes the same region as \(X\) and also the boundary of the second circle. This set should be shaded on separate drawing.

Key concepts

Cartesian Plane

The Cartesian plane, named after the French mathematician René Descartes, is a fundamental concept in mathematics and is particularly crucial when sketching the regions of inequalities. It is a two-dimensional surface formed by two number lines: the x-axis (horizontal) and the y-axis (vertical), which intersect at a point called the origin. Each point on the plane can be identified by an ordered pair of numbers \( (x, y) \) which correspond to its coordinates on the x-axis and y-axis respectively.

When dealing with inequalities such as \( x^{2} + y^{2} \), it is important to note that they describe regions in this plane. In the given exercise, we're looking at a set \( X \) defined by an inequality that captures a range of points whose combined squares of x and y coordinates are between 1 and 4. Continuously, the Cartesian plane allows us to visualize this as a ring or an annulus centered around the origin with specified inner and outer radii defined by the given inequality.

Circle Equations

A circle equation is a way to represent all the points \( (x, y) \) that are at a certain distance (radius) from a central point (the center). In the Cartesian plane, the equation of a circle with center \( (h, k) \) and radius \( r \) is given by the formula \( (x - h)^{2} + (y - k)^{2} = r^{2} \). If the circle is centered at the origin, as is the case with the exercise, the equation simplifies to \( x^{2} + y^{2} = r^{2} \).

In the context of the exercise, we use this form to define two circles—the first with radius 1 (\( r_{1} \)) and the second with radius 2 (\( r_{2} \)). Sketching these circles on the Cartesian plane gives us visual boundaries for the inequality region \( X \) which includes all the points lying on the first circle and those placed within the annulus, up to but not including the points lying on the second circle.

Closed Sets in Mathematics

In mathematical analysis, closed sets play a significant role in understanding the behavior of points and their limits. A set is closed if it contains all its boundary points or, in other words, if it contains its closure. The closure of a set includes the set itself and all its limit points—the points that can be approached by elements within the set, even if those points do not actually lie within the set.

In our exercise, \( \bar{X} \) (the closure of \( X \) ) is the smallest closed set that contains all points of \( X \), including the boundary which is the outer circle of radius 2. To visualize \( \bar{X} \) is to shade the entire region enclosed by the outer circle, thereby showing that we are including all boundary points of \( X \) within this set. This inclusion transforms \( X \) from being an open set (excluding its outer boundary) to a closed set, making \( \bar{X} \) inclusive of the periphery where \( x^{2} + y^{2} = 4 \) stands.