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Find a formula for the intersection math

  1. Sep 8, 2006 #1
    Let I=(a,b) and J=(c,d) with I and J having a nonempty intersection. Find a formula for the intersection of I and J and prove it.

    When a<c I have found that the intersection is [c,b]. Now I need to prove that it is. The way I intend to prove it is by showing that [c,b] and the intersection of I and J are both subsets of each other. I've done one, [c,b] being a subset of I intersect J.

    I intersect J being a subset of [c,b] is a bit trickier. I've broken it down to:

    let x be an element of I intersect J then x is an element of I and x is an element of J. Then:

    a<=x<=b and c<=x<=d.

    Now this is where I'm stumped.
     
  2. jcsd
  3. Sep 8, 2006 #2

    JasonRox

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    Are I and J sets? Like on the real line?
     
  4. Sep 8, 2006 #3

    JasonRox

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    If a<c, how do you know if c<b to make an intersection of [c,b]? And, how can an intersection of two open sets become closed?

    If a<c and b<c, then there is no intersection.
     
  5. Sep 8, 2006 #4
    Ooops. Yes, I and J are intervals. Intersection is nonempty.
     
  6. Sep 8, 2006 #5

    JasonRox

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    There's a formula for solving this?

    Note: The intersection of two open intervals is an open interval, so I don't know why you represent the intersection as a closed interval.

    You said if a<c, then the intersection is NOT [c,b] or necessarily (c,b) because what if d<b, then the intersection is in fact (c,d) and not (c,b) like you're assuming.
     
  7. Sep 9, 2006 #6

    HallsofIvy

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    If you want to do well in Analysis, you are going to have to make an effort to at least write things correctly!. (a, b) is not the same as [a,b] and you are confusing the two.

    No, you haven't- for two reasons. First, as others said, the intersection of the two open intervals, (a,b) and (c,d) cannot be the closed interval (c,b)- you are being careless with your notation. More importantly, what if a= 0, b= 10, c= 1, d= 2- that is, I= (0, 10) and J= (1, 2). Then the intersection is (1, 2), not (1, 10).
     
  8. Sep 9, 2006 #7

    benorin

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    There are only four possibilities that produce a non-empty intersection, namely:

    a < c < b < d
    a < c < d < b
    c < a < b < d
    c < a < d < b

    now convince yourself that [tex]I\cap J = (a,b)\cap (c,d) = \left( \max \{ a,c\} , \min \{b,d\} \right)[/tex]
     
  9. Sep 9, 2006 #8

    JasonRox

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    I never said that!
     
  10. Sep 9, 2006 #9

    JasonRox

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    That's what I was thinking.
     
  11. Sep 9, 2006 #10

    HallsofIvy

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    Never said what?

    In your first post you definitely did say
     
  12. Sep 9, 2006 #11

    matt grime

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    No he didn't. Zoix wrote it. Just scroll upwards....
     
  13. Sep 9, 2006 #12
    I wasn't being careless in notation. I do know the difference between closed and open intervals, and their specific notation. I did indeed use notation for both open and closed intervals in my OP. I made the mistake that I thought the intersection would be a closed interval (which is why I used closed interval notation)...since they overlap they would have both, but I realize the error in this. I hope you realize your error too.
     
  14. Nov 22, 2007 #13
    Just falling on this I thought it was funny. Actually, I think there are 9 possibilities of nonempty intersection:

    a < c < b < d
    a < c < d < b
    c < a < b < d
    c < a < d < b
    a = c < b < d
    a = c < d < b
    a < c < b = d
    c < a < b = d
    a = c < b = d

    Still get [tex]I\cap J = (a,b)\cap (c,d) = \left( \max \{ a,c\} , \min \{b,d\} \right)[/tex]
     
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