1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Prove something is an open set!

  1. Jun 22, 2011 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations



    3. 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.
     
  2. jcsd
  3. Jun 22, 2011 #2

    gb7nash

    User Avatar
    Homework Helper

    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.
     
  4. Jun 22, 2011 #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...
     
  5. Jun 22, 2011 #4

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
  6. Jun 22, 2011 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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?
     
  7. Jun 22, 2011 #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
     
  8. Jun 22, 2011 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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?
     
  9. Jun 22, 2011 #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.
     
  10. Jun 22, 2011 #9
    oh. have we completed the proof?
     
  11. Jun 22, 2011 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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?
     
  12. Jun 23, 2011 #11

    gb7nash

    User Avatar
    Homework Helper

    Since your open ball centered at (x,y) with radius z is entirely contained in (a,b)x(c,d), what can you conclude?
     
  13. Jun 24, 2011 #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>
     
  14. Jun 24, 2011 #13

    gb7nash

    User Avatar
    Homework Helper

    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.
     
    Last edited: Jun 24, 2011
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Prove something is an open set!
  1. Proving a set is open. (Replies: 5)

  2. Prove a set is open. (Replies: 8)

Loading...