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Chapter 13: Problem 29

Determine whether or not the given set is (a) open, (b) connected, and (c) simply-connected. \(\left\\{(x, y) | 1 \leqslant x^{2}+y^{2} \leqslant 4, y \geqslant 0\right\\}\)

Short Answer

Expert verified
The set is not open, connected, and not simply-connected.

Step by step solution

01

Understand the Set

The set is defined as \( \{(x, y) | 1 \leqslant x^{2}+y^{2} \leqslant 4, y \geqslant 0\} \). This represents the area between two concentric circles with radii 1 and 2, including the boundaries, and only the top half-plane (\(y \geqslant 0\)).
02

Check if the Set is Open

A set is open if all points within it can be enclosed within an open ball that does not touch the boundary. Here, the points satisfy \( \leqslant \) or \( \geqslant \), which include the boundary lines. Therefore, this set is not open as it includes its boundary.
03

Check if the Set is Connected

A set is connected if it is in one piece and there is a path between any two points within it. The region described is a half-annulus above the x-axis, without any separations or isolated parts, so the set is connected.
04

Check if the Set is Simply-Connected

A set is simply-connected if it is connected and any loop within the set can be continuously contracted to a single point without leaving the set. The set has a hole (the smaller circle of radius 1), so it is not simply-connected.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Open Sets
In topology, an open set is defined as a set where for any point inside the set, you can find a small open ball around it, with all the points of this ball still within the set itself. The intuitive idea is that for all points in an open set, you are not near a boundary.

For the given set: \( \{(x, y) | 1 \leq x^{2}+y^{2} \leq 4, y \geq 0\} \), it includes the boundaries of the concentric circles with radii 1 and 2. Since they include the boundary, meaning \( \leq \) and \( \geq \) are used in the inequalities, you cannot find an open ball entirely inside the set for points on these boundaries.

Because of this, the set is not considered open. Here are a few characteristics of open sets for better understanding:
  • An open set in a Euclidean space does not include its boundary.
  • Any union of open sets is open.
  • If you remove a closed boundary from a closed set, what's left might be open.
Connected Sets
Connected sets are, in simplest terms, sets that are all in one piece. If you can draw a path from any point of the set to any other point within the set without exiting the set, then the set is connected.

In the example provided, the set represents a half-annulus above the x-axis, essentially the area between two concentric circles with all points maintained above the x-axis. This region does not break or separate, meaning there鈥檚 no isolated part. You can draw a continuous path between any two points in this area without needing to leave the set to "break through" or cross gaps.

Key properties of connected sets include:
  • They cannot be divided into two disjoint open subsets.
  • In two dimensions, connected regions can often be "traced" without lifting a pencil.
  • Every simply-connected set is connected, but not every connected set is simply-connected.
Simply-Connected
A set is deemed simply-connected if it has no "holes." This means that any closed loop you draw within the set can shrink to a single point without having to leave the set.

In the given set, \( \{(x, y) | 1 \leq x^{2}+y^{2} \leq 4, y \geq 0\} \), even though it's connected, it is not simply-connected due to the presence of a hole鈥攖he region inside the smaller circle of radius 1. If you were to try and shrink a loop around this hole to a point, the loop would have to "jump out" of the set at some point to become a single point, which is not allowed.

Consider these traits for simply-connected sets:
  • A simply-connected space should be a single piece and free of holes.
  • Any loop drawn in a simply-connected set can shrink to a point.
  • The fundamental group of a simply-connected space is trivial (i.e., consists only of the identity element).

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Most popular questions from this chapter

(a) Use Stokes' Theorem to evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r},\) where \(\mathbf{F}(x, y, z)=x^{2} y \mathbf{i}+\frac{1}{3} x^{3} \mathbf{j}+x y \mathbf{k}\) and \(C\) is the curve of intersection of the hyperbolic paraboloid \(z=y^{2}-x^{2}\) and the cylinder \(x^{2}+y^{2}=1\) oriented counterclockwise as viewed from above. (b) Graph both the hyperbolic paraboloid and the cylinder with domains chosen so that you can see the curve \(C\) and the surface that you used in part (a). (c) Find parametric equations for \(C\) and use them to graph \(C .\)

\(5-15\) " Use the Divergence Theorem to calculate the surface integral \(\iint_{S} \mathbf{F} \cdot d \mathbf{S} ;\) that is, calculate the flux of \(\mathbf{F}\) across \(S .\) $$\mathbf{F}=|\mathbf{r}|^{2} \mathbf{r},\( where \)\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}$$ \(S\) is the sphere with radius \(R\) and center the origin

Maxwell's equations relating the electric field \(\mathbf{E}\) and mag- netic field \(\mathbf{H}\) as they vary vith time in a region containing no charge and no current can be stated as follows: $$\begin{array}{ll}{\operatorname{div} \mathbf{E}=0} & {\operatorname{div} \mathbf{H}=0} \\ {\operatorname{curl} \mathbf{E}=-\frac{1}{c} \frac{\partial \mathbf{H}}{\partial t}} & {\operatorname{curl} \mathbf{H}=\frac{1}{c} \frac{\partial \mathbf{E}}{\partial t}}\end{array}$$ where \(c\) is the speed of light. Use these equations to prove the following: $$\begin{array}{l}{\quad\left(\text { a) } \nabla \times(\nabla \times \mathbf{E})=-\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{E}}{\partial t^{2}}\right.} \\ {\text { (b) } \nabla \times(\nabla \times \mathbf{H})=-\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{H}}{\partial t^{2}}} \\\ {\text { (c) } \nabla^{2} \mathbf{E}=\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{E}}{\partial t^{2}} \quad[\text {Hint} \text { : Use Exercise } 27 .]} \\\ {\text { (d) } \nabla^{2} \mathbf{H}=\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{H}}{\partial t^{2}}}\end{array}$$

\(5-15\) " Use the Divergence Theorem to calculate the surface integral \(\iint_{S} \mathbf{F} \cdot d \mathbf{S} ;\) that is, calculate the flux of \(\mathbf{F}\) across \(S .\) $$\mathbf{F}(x, y, z)=x^{4} \mathbf{i}-x^{3} z^{2} \mathbf{j}+4 x y^{2} z \mathbf{k}$$ \(S\) is the surface of the solid bounded by the cylinder \(x^{2}+y^{2}=1\) and the planes \(z=x+2\) and \(z=0\)

\(33-44=\) Find the area of the surface. The part of the plane with vector equation \(\mathbf{r}(u, v)=\langle u+v, 2-3 u, 1+u-v\rangle\) that is given by \( 0 \leq u \leq 2,-1 \leq v \leq 1\)
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