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Proving two sets are equal

  1. Jan 27, 2012 #1
    & means belong to and # not equal to : $ subsets of

    A={(t-1,1/t): t&R, t # 0}
    B= {(x,y) &R^2:y=1/(x+1), x#-1}

    i started by say A$B

    let x= t-1 and y=1/t
    so we have y= 1/(t-1)+
    = Y=1/t hence A$B
    to prove B$A
    is where i am stuck- as I think I have got my first part wrong anyway and I ma not sure if I have to make reference to x#-1 ot t#0

    this is a fairy new topic for me and I am finding it bit abstract!!!

    Thanks
     
  2. jcsd
  3. Jan 27, 2012 #2

    chiro

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    Hey foreverdream and welcome to the forums.

    For the B$A you just have to show that the range of the (x,y) in B lies in that of (x,y) of A. So for the x you know that for (t-1), then t != 0 which means t != -1 which is the same for B. Since the structure is the reals you have shown that since x is a real and the reals are a subset of the reals then you have done it for x. (Its a property that A is a subset of A for any set A and its only a couple of lines to show this)

    Based on the above hint, can you see how to prove it for the y coordinate?
     
  4. Jan 27, 2012 #3
    so if I understand it correctly:
    we must show that (x,y) belongs to B
    but lost completely on t! part
     
    Last edited: Jan 27, 2012
  5. Jan 27, 2012 #4
    let t-1=x and then y=1/(x+1) = 1/(t-1)+1 = 1/t
    so (x,y)=(t-1, 1/t) hence A=B ????? where does t! fit in-?
     
  6. Jan 27, 2012 #5

    chiro

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    You have to show that in the second B$A part of your 2-tuple (the y coordinate), that all the values for this are inside.

    I think you know intuitively what to do, but I don't think its formal enough. For your A set in the second part of your 2-tuple you have 1/t where t != 0 and for B you have 1/(x+1) where x != -1. So we know that t != 0 and x + 1 != 0 => x != -1 so that means there is nothing disjoint in those two sets. This is important because if there is anything disjoint then there is no way of subsets happening.

    After that you who that that both are subsets of R \ {0} and since A is always a subset of A you have proven it for the 2-tuple (the y part).

    In terms of making the above more formal you can show that both parts of the two tuple are the set R \ {0} (This just means the real numbers without 0). Once you do this then you use the subset argument listed above to complete the proof.

    You shouldn't have to get any more formal than that.
     
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