Show that a subset in E^2 is open.

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In summary: Solving I get c is equal to sqrt2. (sorry for the format, I am replying from my phone. But what if I use the distance formula for n-tuples in 2 space? Then I get that the distance is 1...2 space is just the plane, and n-tuples are just ordered pairs. So the distance formula is just the Pythagorean theorem in this case. Therefore, the distance between (1,0) and the line x=y is sqrt(1^2+1^2)= sqrt(2). Does that make sense? So then the radius of the ball centered at (1,0) would be sqrt(2)-sqrt(2)=0. Sorry
  • #1
jlemus85
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Homework Statement



Show that the subset of E^2 given by { (x1,x2) belonging to E^2 : x1>x2} is open.

Homework Equations



d(x,y)= sqrt[ (x1-y1)^2 +(x2-y2)^2 ]

The Attempt at a Solution




Ok so I would like someone to tell me if I'm on the right track or if I'm way off.

So to show that a subset is open I need to show that if I choose an arbritrary point in my subset S, that I can form a ball around this point p and all the points near p will be within S.

So what I want to do is first draw a ball with center x and radius r. Then I want to draw another ball that is within the first ball with center y and radius r-d(x,y). If this set is open then y and the radius r-d(x,y) will belong to S.

SO far this is about as far as I can get without confusing myself and probably making you guys cringe me plugging in the distance formula and all that.

Any help would be very appreciated.
 
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  • #2
jlemus85 said:

Homework Statement



Show that the subset of E^2 given by { (x1,x2) belonging to E^2 : x1>x2} is open.

Homework Equations



d(x,y)= sqrt[ (x1-y1)^2 +(x2-y2)^2 ]

The Attempt at a Solution




Ok so I would like someone to tell me if I'm on the right track or if I'm way off.

So to show that a subset is open I need to show that if I choose an arbritrary point in my subset S, that I can form a ball around this point p and all the points near p will be within S.
jlemus85 said:
So what I want to do is first draw a ball with center x and radius r. Then I want to draw another ball that is within the first ball with center y and radius r-d(x,y). If this set is open then y and the radius r-d(x,y) will belong to S.
The radius is not a point, so it isn't a point in the set.

How does the set itself enter this problem? You are given that S = {(x, y) | x > y}. Seems to me you need to use this information.
jlemus85 said:
SO far this is about as far as I can get without confusing myself and probably making you guys cringe me plugging in the distance formula and all that.

Any help would be very appreciated.
 
  • #3
Thanks for your reply.

I know what you're saying about using the given set, I just didn't want to put it in yet so as not to confuse people lol. But here goes.

I know that d(p,q)= sqrt[ (x1-y1)^2 + (x2-y2)^2], given that p= (x1, y1) and q=(x2,y2).

So if I draw a ball around Po (the center) this ball will have radius r.

If I draw another point in S, called q, then it will have radius r-d(Po, q). So now I have to prove that every point that is part of this ball with center q and radius r-d(Po,q) is inside of S. I know I will have to pick an arbritrary point in the q-ball set and then apply the traingle inequality, but I am having trouble figuring out how to calculate the radius? Correct me if I'm wrong, but since I am given a set and a distance equation, I should/need to be able to calculate the radius correct? I'm thinking maybe it's x1-x2>0 = r?
 
  • #4
Did you try and draw a sketch of the set in the plane? Your set has a boundary described by the equation x1=x2. That's a straight line, can you use that to make your sketch? If you pick r=x1-x2 that's going to be too big. Just for practice pick x1=1, x2=0. How large can r be?
 
  • #5
Dick,

Thanks for your reply.

I didn't realize I could be thinking about this on the x,y plane, duh. So the boundary line in other words is the line that goes through the origin aka the line y=x. So if I use x=1 and y=0 I get my r to be sqrt(2)? So then is it that r= sqrt(x1+x2)?
 
  • #6
jlemus85 said:
Dick,

Thanks for your reply.

I didn't realize I could be thinking about this on the x,y plane, duh. So the boundary line in other words is the line that goes through the origin aka the line y=x. So if I use x=1 and y=0 I get my r to be sqrt(2)? So then is it that r= sqrt(x1+x2)?

No, I don't think r=sqrt(2). r is the distance from (1,0) to the line x=y. Just draw some triangles and use the pythagorean theorem to find the distance.
 
  • #7
Hmmm... so I got sqrt(2) by applying pythagorean th. If my point is 1,0 and I drop the perpendicular from the line x equals y, then the sides of my triangle will be of length 1 and 1. Putting this into pyth eqn I get 1 plus 1 equals c squared. Solving I get c is equal to sqrt2. (sorry for the format, I am replying from my phone.

But what if I use the distance formula for n-tuples in 2 space? Then I get that the distance is 1
 
  • #8
jlemus85 said:
Hmmm... so I got sqrt(2) by applying pythagorean th. If my point is 1,0 and I drop the perpendicular from the line x equals y, then the sides of my triangle will be of length 1 and 1. Putting this into pyth eqn I get 1 plus 1 equals c squared. Solving I get c is equal to sqrt2. (sorry for the format, I am replying from my phone.

But what if I use the distance formula for n-tuples in 2 space? Then I get that the distance is 1
2 space is just the plane, and n-tuples are just ordered pairs.
 
  • #9
jlemus85 said:
Hmmm... so I got sqrt(2) by applying pythagorean th. If my point is 1,0 and I drop the perpendicular from the line x equals y, then the sides of my triangle will be of length 1 and 1. Putting this into pyth eqn I get 1 plus 1 equals c squared. Solving I get c is equal to sqrt2. (sorry for the format, I am replying from my phone.

But what if I use the distance formula for n-tuples in 2 space? Then I get that the distance is 1

Your point is (1,0). It is the corner of a right triangle with the other two points (0,0) and (1,1) which are on the boundary x=y. The length of the hypotenuse is sqrt(2). The distance of (1,0) from the boundary is not sqrt(2). It's the distance to the hypotenuse, not the length of the hypotenuse. It's shorter. How much shorter?
 

Related to Show that a subset in E^2 is open.

1. What is a subset in E^2?

A subset in E^2 is a collection of points that are contained within the two-dimensional Euclidean space. This space is made up of an x-axis and a y-axis, and any point in E^2 can be described by its coordinates (x, y).

2. What does it mean for a subset in E^2 to be open?

An open subset in E^2 is one that does not contain any of its boundary points. In other words, every point in an open subset has a small surrounding region that is also contained within the subset. This concept is important in topology and analysis, as it allows us to define continuity and convergence of functions.

3. How can you show that a subset in E^2 is open?

In order to show that a subset in E^2 is open, we need to prove that every point in the subset has a small enough neighborhood that is also contained within the subset. This can be done by showing that for any point (x, y) in the subset, we can find a positive number r such that all points within a distance r from (x, y) are also in the subset.

4. What is the importance of open subsets in E^2?

Open subsets in E^2 are important because they allow us to define concepts such as continuity and convergence of functions. In topology, they are used to define open sets, which are the building blocks for more complex topological spaces. In analysis, they are used to define open intervals, which are crucial for the definition of derivatives and integrals.

5. Can a subset in E^2 be both open and closed?

Yes, it is possible for a subset in E^2 to be both open and closed. This is known as a clopen subset, and it means that the subset contains all of its boundary points. One example of a clopen subset in E^2 is the entire space itself, as it contains all of its boundary points (i.e. the entire x-axis and y-axis).

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