in-exercises-3-and-4-determine-whether-each-set-is-open-or-closed-or-neither-open-nor-closed-a-leftx

Question

In Exercises 3 and $4,$ determine whether each set is open or closed or neither open nor closed.
a. $\left\{(x, y) : x^{2}+y^{2}=1ight\}$
b. $\left\{(x, y) : x^{2}+y^{2} > 1ight\}$
c. $\left\{(x, y) : x^{2}+y^{2} \leq 1 	ext { and } y > 0ight\}$
d. $\left\{(x, y) : y \geq x^{2}ight\}$
e. $\left\{(x, y) : y < x^{2}ight\}$

James Chok · Accepted Answer

Okay, So this question we want to see if each set is either open or closed or neither. For one, the 1st 1 is closed. It is closed. The very simple reason why it&#x27;s place to do that. It contains all of its boundary points because there&#x27;s no inequality. It&#x27;s straight equality like you itself of this one. It is open because it is not striking a quality. And if you actually brought out scene, you should be cool and it does not contain it&#x27;s boundary points to defaults every more. In this one we have neither, because is closed on this one is pleasant, this one that is open on this step. So when you combine them, it&#x27;s closed and rightly so. Neither. This one is very easily close because it&#x27;s a strictly politics here. Well, this case is open because there is a strict quality very definitely doesn&#x27;t contain. Well, it was bound

Frank Lin · Answer

determining this open, connected or simply connected is this. No, this is our scores between one and four. Ours between one to ours, a radiance. Let&#x27;s draw Surkov Radius E Force One and this drew a circle. Radius equals two, and we includes a boundary point. I also want y to be greater or equal to zero. So it&#x27;s only the upper part. There will be only be this part because in this case we do include a hundred point. And that means is it is. It&#x27;s not open. We pick any point in Hungary. Any neighborhood off this point will not be a subset of the original seven, so this is not open. I guess it, Chrissy cannot be is path connection. So their voice, their voice connected and it&#x27;s simply connected because we find any usual, any clothes, curving sides, Great green region. You always shrinking to a point. No, this is not possible to cross the course a curve here in order to draw a curve inside of Green Region. It should be the case where I can be shrunk to the point. So it&#x27;s a simply conducted

Megan Mcfarland · Answer

in this problem, you&#x27;re given the shown information and morass is tell if this set of points is an open, regional connected vision in a simply connected region. So first we&#x27;re going to graph it, and now we know that one is less cynical to extra. Plus y Squared is US medical to four. Now I scripless y spirit forms a circle, and we know the extra plus y squared equals r squared. So arsed word is going to be one is going to be four. So therefore our radius is going to be one is going to be too And so we&#x27;re going to have a radius of one of the circle and we&#x27;re gonna have a radius of two for a circle. And then we&#x27;re also told that why is greater than or equal to zero, which is all this area up here, including the X axis, because it&#x27;s equal to zero. So we can see that because one is less or equal to X scripless. Why screws? Listen able to four that gets the area in here in this whole circle, but because why is greater than or equal to zero? It&#x27;s just going to be the upper half right here, including the boundaries because it is equal to is less than or equal to, um for the circle and his girl unable to for why And so is it open? So, like I said, it includes the boundaries. And so if a region includes the boundaries, that is not an open region. And so no, it is not open. Is it connected? We know region is connected if you can choose to points in the region and connects them with a path in the region itself. So any two points and so I can choose the point down here I can choose. Oh, Sarah, nothing there cause that&#x27;s not in the region right up here and over here that I can connect those with the path in the region. Any two points and have that path still be in the region. And so, yes, it is connected No for a reason to be simply connected. There can be no holes, and the region cannot be in pieces are region does not have holes, and it is not in pieces. And so, yes, it is simply connected. And that is our final answer.

Frank Lin · Answer

so determines the study&#x27;s open. They are similar. So is the entire swipe at one point two three. So we&#x27;re basically excluding this point. I have everything else. So this says Open a single point, said. He&#x27;s close and the compliment all close. Our eyes open, its eyes connected, way just removed. One point so we can still is. Still even past conduct is because any two point we just have to avoid this point to draw past of the other. It&#x27;s not simply connected because here&#x27;s an example for a closed curve, a draw within this set. But because off the whole here, we cannot shrink in tow. One point I can assure you this curve to one point, so this is not simply connected.

Frank Lin · Answer

So this No, Tell me whether is open on that You simply connected so well. Draw picture. So we y strictly between zero and three. So why ho zeros? Absences? It&#x27;s true. Some light represent wife or three over everything between without a boundary point because we don&#x27;t include the Hungary Point. This sense open on for every point is set. You can always find a small neighborhood that is contained is set. So that&#x27;s wanted me to be open notice if we include any points our boundary happy open because those Hungary point does not satisfy the criteria to be open and connect is pretty clear A simple con f It is also true because if you draw any perv inside this region, you know always trends of curve to one point.

Megan Mcfarland · Answer

in this problem, you&#x27;re given the shown information and morass determine if the set of points is open, connected and simply connected. And so first we should Grabsch the set of points. And so zero is us some wise, less than three. Now you notice that there is no equal sign and so therefore we&#x27;re gonna draw dotted lines are Val injuries. So for zeros, listen wise. Listen, my wise and our top boundary is going to be a wine. But it&#x27;s also going to be a dotted line. And so is that wise Lesson three. And so we determine if something is open by whether it&#x27;s boundaries are included or not. Now we can see that we have dotted lines in the boundaries. And so the founders were not included. And so therefore, this PSA is open. So yes, it is open. We determine if it&#x27;s connected. Fine. If we can do the following we look at if we can join any two points in this region, any two points by a paths that it&#x27;s Onley in the region itself. And we can do that in this case. And so yes, it is connected. No, determine if something is simply connected. We check for two things. We check for holes and we check for if the region is in to or more pieces. Now. This region is in one piece and there are no holes. And so therefore it isn&#x27;t simply connected and we have bound our answers.

Problem

In Exercises 5 and $6,$ determine whether or not …

Problem 4 Easy Difficulty

Answer

No Answer

Related Courses

Linear Algebra and Its Applications

Related Topics

Discussion

Top Calculus 3 Educators

Lily A.

Catherine R.

Caleb E.

Michael J.

Calculus 3 Courses

Vectors Intro

Vector Basics - Example 1

Recommended Videos

Watch More Solved Questions in Chapter 8

Video Transcript

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

Lily A.

Catherine R.

Caleb E.

Michael J.

Vectors Intro

Vector Basics - Example 1

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

No Answer

Linear Algebra and Its Applications

Lily A.

Catherine R.

Caleb E.

Michael J.

Vectors Intro

Vector Basics - Example 1

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Lily A.

Catherine R.

Caleb E.

Michael J.

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications