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Proving that the intersection of any two intervals is an interval

  1. Oct 31, 2011 #1
    The question is as follows:

    Prove that if I1, I2 are intervals and J = I1[itex]\cap[/itex]I2 then J is an interval.

    To be honest I don't even know where to start. There's a "hint" that suggests that I first write out the definitions of I1, I2, J as intervals and of the intersection between I1 and I2, but that hasn't really enlightened me...

    So I just have:

    A subset In of ℝ is an interval if [itex]\forall[/itex] x,y,z [itex]\in[/itex] ℝ , x[itex]\in[/itex]In, z[itex]\in[/itex]In and x<y<z then y[itex]\in[/itex]In (I used n instead of 1 and 2 because I am too lazy to write it out twice. Also, substitute J in as appropriate xD )

    I1[itex]\cap[/itex]I2 = {x: x[itex]\in[/itex]I1 and x[itex]\in[/itex]I2}

    I don't know where to go from there basically. If someone could even so much as nudge me in the right direction I would be very appreciative :D
  2. jcsd
  3. Oct 31, 2011 #2


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    how would you go about deciding if x was in (a,b)∩(c,d)?

    i'll give an example, which you can play with:

    what is (2,4)∩(3,5)? that is:

    if 2 < x < 4 AND

    3 < x < 5, can you express just one inequality that covers both at once?
  4. Oct 31, 2011 #3
    Hmm, but is it mathematically rigorous enough to, in a proof, say something along the lines of "given I1 is the interval (I1a,I1b) and I2 is the interval (I2a,I2b) , I1[itex]\cap[/itex]I2
    is the interval (I1a,I2b) and given J=I1[itex]\cap[/itex]I2=and interval, J is an interval" ?

    Because I don't have any definitions or theorems to work with that say anything remotely similar, at least that I'm aware of...

    Actually you probably couldn't tell me whether that was sufficient seeing as you did not set the questions nor are you marking them... ¬.¬
  5. Oct 31, 2011 #4


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    that's true...but, go ahead, answer the question of my earlier post:

    what is (2,4)∩(3,5)?
  6. Oct 31, 2011 #5
    2<x<5 ...right? Now I'm worried I've done something unbelievably stupid...
  7. Oct 31, 2011 #6


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    try drawing a picture.
  8. Oct 31, 2011 #7
    right, I HAVE done something unbelievably stupid then...so 3<x<4 ¬.¬ so I would actually ahve to revise my previous interval suggestion as (I1b,I2a)=J as well...
    Last edited: Oct 31, 2011
  9. Oct 31, 2011 #8


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    hmm, it looks like 3 = max{2,3} and 4 = min{4,5}.

    can you generalize?

    (hint: suppose max{a,c} > min{b,d}. what is (a,b)∩(c,d) in this case).
  10. Oct 31, 2011 #9
    well if the max of {a,c} > {b,d} then...would it be c>b, or would there be no interval?

    as for generalizing...hmm...I have a kind of vague idea as to what I might say, btu I have no idea how to express it in a mathematical manner -_-

    something along the lines of the interval being between greatest element of I1 that is also [itex]\in[/itex]I2 and least element of I2 that is also [itex]\in[/itex]I1
    Last edited: Oct 31, 2011
  11. Oct 31, 2011 #10


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    i've given you a hint as to "the general formula".

    suppose L = max{a,c} < M = min{b,d}.

    try to describe (a,b)∩(c,d) in terms of L and M.

    now, to consider "half-open" and "closed intervals", you'll have to do a careful case-by-case analysis. make sure you don't overlook the situation:

    L = M.
  12. Oct 31, 2011 #11


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    To show J is an interval, you want to show that if x, z ∈ J, then if x<y<z, then y ∈ J. The key thing here is to know what it means for a number to be in J.
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