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Chapter 16: Problem 31

\(29-32\) Determine whether or not the given set is (a) open, (b) connected, and (c) simply-connected. $$ \\{(x, y) | 1

Short Answer

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

Step by step solution

01

Identifying the Set

The set \( \{(x, y) | 1<x^2+y^2<4\} \) represents the region between two circles centered at the origin with radii \(1\) and \(2\). It includes all points \((x, y)\) such that the sum of their squared coordinates is greater than \(1\) and less than \(4\), excluding the boundaries of both circles.
02

Check if the Set Is Open

A set is open if each of its points is an interior point, meaning that for each point, there is a small radius such that the entire ball inside the radius is contained in the set. In our set, points more than a distance \(1\) away from the origin and less than a distance \(2\) away can have small circles fully contained within the larger annular region. Hence, the set is open.
03

Check if the Set Is Connected

A set is connected if it is in one piece and there are no separate or isolated parts. The region \(1 < x^2 + y^2 < 4\) forms a continuous annular region around the origin without any breaks or separations. Therefore, the set is connected.
04

Check if the Set Is Simply-Connected

A set is simply-connected if it is connected and every simple loop in the set can be continuously contracted to a point without leaving the set. Due to the hole in the region (center \(0,0\) where \(x^2 + y^2 \leq 1\), it is not possible to contract a loop surrounding this hole to a point without leaving the set. Thus, the set 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, open sets are fascinating because they include all possible interior points, but none of the boundary points. Imagine a playground with no edges to stop you鈥攁nywhere you go, you can still freely move in any direction without leaving the playground. To put it mathematically, a set is open if, for every point in the set, there exists a small "neighborhood" around it that's completely contained within the set. You can think of this neighborhood as a small disk or circle around a point in the set.

For instance, consider the set \[\{(x, y) | 1
Connected Sets
Connected sets in topology refer to sets that are all in "one piece." Imagine a rubber band鈥攊f you can lift the whole band without it falling apart, it's connected. Mathematically, a set is considered connected if there are no separate chunks; in other words, it's possible to travel within the set from any point to any other point, all without leaving the set.

In our example, the set \[\{(x, y) | 1
Simply-Connected Sets
A simply-connected set is a bit trickier than a merely connected one. Think of simply-connected sets like a pancake without holes. In topology, a set is simply-connected if it's connected and any loop drawn within the set can continuously shrink down to a single point. This means you can 'lasso' any part of the set and pull the loop tight without leaving the set.

In contrast, consider the set \[\{(x, y) | 1

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Most popular questions from this chapter

\(37-47\) Find the area of the surface. The surface with parametric equations \(x=u^{2}, y=u v\) \(z=\frac{1}{2} v^{2}, 0 \leqslant u \leqslant 1,0 \leqslant v \leqslant 2\)

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Find, to four decimal places, the area of the part of the surface \(z=\left(1+x^{2}\right) /\left(1+y^{2}\right)\) that lies above the square \(|x|+|y| \leqslant 1 .\) Illustrate by graphing this part of the surface.

\(19-30\) Evaluate the surface integral \(\iint_{s} \mathbf{F} \cdot d \mathbf{S}\) for the given vector field \(\mathbf{F}\) and the oriented surface \(S\) . In other words, find the flux of \(\mathbf{F}\) across \(S .\) For closed surfaces, use the positive (outward) orientation. $$\mathbf{F}(x, y, z)=y \mathbf{j}-z \mathbf{k},$$ \(S\) consists of the paraboloid \(y=x^{2}+z^{2}, 0 \leqslant y \leqslant 1\) and the disk \(x^{2}+z^{2} \leq 1, y=1\)

Let $$\mathbf{F}(x, y)=\frac{-y \mathbf{i}+x \mathbf{j}}{x^{2}+y^{2}}$$ (a) Show that \(\partial P / \partial y=\partial Q / \partial x\) (b) Show that \(\int_{c} \mathbf{F} \cdot d \mathbf{r}\) is not independent of path.[Hint: Compute \(\int_{c} \mathbf{F} \cdot d \mathbf{r}\) and \(\int_{c} \mathbf{F} \cdot d \mathbf{r},\) where \(C_{1}\) and \(C_{2}\) are the upper and lower halves of the circle \(x^{2}+y^{2}=1\) from \((1,0)\) to \((-1,0) . ]\) Does this contradict Theorem 6\(?\)
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