Prove something is an open set

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In summary, to prove that (a,b) x (c,d) is open in R^2, we need to show that for any point (x,y) in the set, there exists a radius r such that the open ball around (x,y) with radius r is a subset of (a,b) x (c,d). By considering the distance of (x,y) from the edges of the rectangle, we can choose a minimum radius and show that the open ball around (x,y) is contained in the rectangle. Therefore, (a,b) x (c,d) is open in R^2. Additionally, in a metric space containing more than one point, it is not possible to have only the empty set and
  • #1
l888l888l888
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Homework Statement



prove (a,b) x (c,d) is open in R^2

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The Attempt at a Solution


given an x that belongs to (a,b)x(c,d) there exists a delta such that given any y that belongs to R^2 whose d2(x,y)< epsilon, y also belongs to (a,b)x(c,d). Need to prove this.
 
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  • #2
Look as R = (a,b)x(c,d) as an open rectangle. Fix y in (a,b)x(c,d). How do you think we can find an open ball(centered at y) that's entirely inside the rectangle?

Hint: Consider the distance of y from a,b,c, and d.
 
  • #3
to find a ball that is in the rectangle with center y, we can imagine the ball as having radius b-y, y-a, d-y or y-c...
 
  • #4
Are all of those numbers small enough to guarantee that the open ball around y is a subset of the rectangle?

Your suggest in the attempt part of post #1 mentions both a delta and an epsilon. No need for that. Just let y be an arbitrary point in the rectangle and find an r such that B(y,r) (the open ball around y with radius r) is a subset of the rectangle.
 
  • #5
l888l888l888 said:
to find a ball that is in the rectangle with center y, we can imagine the ball as having radius b-y, y-a, d-y or y-c...

You've got the right idea, but if you take a point in (a,b)x(c,d) it's not just a number. It's a pair of numbers. Let's call it (x,y) instead. Now what are the choices of radii, and which radius do you choose?
 
  • #6
in an answer the first reply... anything bigger than those numbers would imply that the ball would have points inside it but not inside the rectangle... In an answer to the second reply... using (x,y) instead then a possible radius would be b-x, x-a, d-y, or y-c... we could choose b-x as the radius
 
  • #7
l888l888l888 said:
in an answer the first reply... anything bigger than those numbers would imply that the ball would have points inside it but not inside the rectangle... In an answer to the second reply... using (x,y) instead then a possible radius would be b-x, x-a, d-y, or y-c... we could choose b-x as the radius

You wouldn't want to choose b-x as the radius if it were larger than x-a, would you? Wouldn't that make the ball contain points outside of the rectangle, as you said. How about choosing a minimum of those numbers?
 
  • #8
so letting z= min{b-x, x-a, d-y, y-c} now what?. Using this with the info from first replay. we can find a ball with radius z st the ball is contained in the rectangle.
 
  • #9
oh. have we completed the proof?
 
  • #10
l888l888l888 said:
so letting z= min{b-x, x-a, d-y, y-c} now what?. Using this with the info from first replay. we can find a ball with radius z st the ball is contained in the rectangle.

Well, the ball around (x,y) of radius z would be contained in the open rectangle, right? It's radius is less than or equal to distance to any edge, yes?
 
  • #11
l888l888l888 said:
so letting z= min{b-x, x-a, d-y, y-c} now what?. Using this with the info from first replay. we can find a ball with radius z st the ball is contained in the rectangle.

Since your open ball centered at (x,y) with radius z is entirely contained in (a,b)x(c,d), what can you conclude?
 
  • #12
I have another question about open sets. Is it possible in a metric space containing more than one point to have only the empty set and the the metric space itself as it's open sets? I say no because if you pick a point that point itself with the area around it is a subset of the metric space and u can always find a ball around it with radius small enough such that that ball is in the subset... What do you guys think>
 
  • #13
l888l888l888 said:
I have another question about open sets. Is it possible in a metric space containing more than one point to have only the empty set and the the metric space itself as it's open sets? I say no because if you pick a point that point itself with the area around it is a subset of the metric space and u can always find a ball around it with radius small enough such that that ball is in the subset... What do you guys think>

Absolutely right.

For any open set, you can always find a smaller set that's open. Just pick an arbitary point (call it x) in the set. By the definition of open, x is an interior point. Choose a radius that's small enough and create an open ball around x. This open ball is a subset of the entire set and it is not equal to the entire set.
 
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FAQ: Prove something is an open set

What is an open set?

An open set is a subset of a metric space in which all points within the set have a neighborhood that is also contained within the set. In simpler terms, it is a set in which all points are surrounded by other points in the set.

How do you prove that something is an open set?

To prove that something is an open set, you must show that for every point in the set, there exists a neighborhood around that point that is entirely contained within the set. This can be done by using the definition of an open set and mathematical proofs.

Can you give an example of proving something is an open set?

Yes, for example, we can prove that the set (0,1) is an open set. Let x be any point in the set. Then, there exists a neighborhood around x, such as (x-0.5, x+0.5), which is entirely contained within the set. Therefore, (0,1) is an open set.

What is the importance of open sets in mathematics?

Open sets are important in mathematics because they allow us to define and study continuity, connectedness, and convergence in a metric space. They also play a crucial role in topology, which is a branch of mathematics that deals with the properties of spaces that are preserved under continuous deformations.

Can a set be both open and closed?

Yes, in some cases, a set can be both open and closed. For example, in a discrete metric space, where every point has a neighborhood of size 1, every set is both open and closed. However, in most metric spaces, a set cannot be both open and closed simultaneously.

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