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    X = [1,3]
    Y = (1,3]

    X - Y ={x-y | x ∈ X, y ∈ Y}

    I have two questions regarding this:

    a) Find the value of X-Y

    b) Are sup(X-Y) and inf(X-Y) elements of X-Y?


    In my work so far I realise that for a) I have 1-3=-2 by (X's smallest element) - (Y's largest value).

    But now for Y there is no infinum so the largest value of X-Y is not possible.

    So it the final answer just (2)?

    Also for part b) is it only that inf(X-Y) is not possible
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    N.B. It's "infimum" not "infinum"

    (Original post by ineffablemind)
    X = [1,3]
    Y = (1,3]

    X - Y ={x-y | x ∈ X, y ∈ Y}

    I have two questions regarding this:

    a) Find the value of X-Y
    Largest/smallest value of X-Y?


    b) Are sup(X-Y) and inf(X-Y) elements of X-Y?


    In my work so far I realise that for a) I have 1-3=-2 by (X's smallest element) - (Y's largest value).

    But now for Y there is no infinum so the largest value of X-Y is not possible.
    Y does indeed have an infimum; it's 1.

    The infimum is the greatest lower bound. If you take a number A that is larger than 1, but smaller than 3 i.e. A=1+\epsilon, 0 < \epsilon \le 2 then there is an element of Y that is less than A e.g. B=1+\epsilon/2. So no number bigger than 1 is a lower bound. However for all y \in Y, y > 1, so 1 is a lower bound. So 1 is the greatest lower bound.
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    (Original post by atsruser)
    N.B. It's "infimum" not "infinum"



    Largest/smallest value of X-Y?



    Y does indeed have an infimum; it's 1.

    The infimum is the greatest lower bound. If you take a number A that is larger than 1, but smaller than 3 i.e. A=1+\epsilon, 0 < \epsilon \le 2 then there is an element of Y that is less than A e.g. B=1+\epsilon/2. So no number bigger than 1 is a lower bound. However for all y \in Y, y > 1, so 1 is a lower bound. So 1 is the greatest lower bound.
    Thank you for your reply.

    Yes so for part a) I have 1-3=-2 by (X's smallest element) - (Y's largest value) which I consider to be correct. Is the answer simply -2, or do I need to bracket it off such as (2).

    But I have that for Y there is no smallest possible element as you suggested that 1 is the infimum. Hence there will be no largest element possible for X-Y.

    For the inf(X-Y) would I then have to calculate inf(X) and inf(Y) separatley. And then subrtract the difference?

    Thank you for clarifying that I am grateful for your help
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    (Original post by ineffablemind)
    Thank you for your reply.

    Yes so for part a) I have 1-3=-2 by (X's smallest element) - (Y's largest value) which I consider to be correct. Is the answer simply -2, or do I need to bracket it off such as (2).
    There seems to be a word missing from your part a) - is it largest or smallest value?

    But I have that for Y there is no smallest possible element as you suggested that 1 is the infimum. Hence there will be no largest element possible for X-Y.

    For the inf(X-Y) would I then have to calculate inf(X) and inf(Y) separatley. And then subrtract the difference?
    To do this, I would look at the endpoints. The max of Y is 3; it's easy to subtract this from X. The inf of Y is 1, so all elements are of the form 1+\epsilon, 0 < \epsilon \le 2.

    So using this endpoint, you will be forming elements in X-Y of the form x -(1+\epsilon), x \in X. What happens when you apply that to the max/min of X?
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    (Original post by atsruser)
    There seems to be a word missing from your part a) - is it largest or smallest value?
    It could be that the answer expected is a set. It is possible to write X - Y as an interval, after all...
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    (Original post by DFranklin)
    It could be that the answer expected is a set. It is possible to write X - Y as an interval, after all...
    I would find it weird to describe a set as a "value", but I guess it's not impossible.
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    So:
    inf(A) = 1.
    sup(A)=3
    inf(B) = infinity
    sup(B) = 3

    So A-B = 3-1 = 2?

    Then inf(A-B) is not possible

    And

    sup(A-B) = 0

    Have I got this correct now?
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    (Original post by ineffablemind)
    So:
    inf(A) = 1.
    sup(A)=3
    inf(B) = infinity
    sup(B) = 3

    So A-B = 3-1 = 2?

    Then inf(A-B) is not possible

    And

    sup(A-B) = 0

    Have I got this correct now?
    You seem to be talking about a different question. What are A and B?
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    I am ineffablemind I can't remember my password omg:

    Anyway I am at the stage:

    X−Y = [−2;2)

    So going back to my question that
    Are sup(X-Y) and inf(X-Y) elements of X - Y?


    inf(X) = 1 and sup(X) = 3
    Then:
    inf(Y) = 1 and sup(Y) = 3
    inf(X-Y) = inf X - sup Y = 1 - 3 = -2 sup(X-Y) = sup X - inf Y = 3 - 1 = 2


    So to solve my question:

    Since inf(X-Y) = -2 and sup(X-Y) = 2. Then from the interval notation 2 is not bounded in the interval notation, whereas -2 is. So then inf(X-Y) is an element of X-Y and sup(X-Y) is not an element of X-Y.

    Would people agree this is correct?
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    (Original post by alexgreyx)
    I am ineffablemind I can't remember my password omg:

    Anyway I am at the stage:

    X−Y = [−2;2)

    So going back to my question that
    Are sup(X-Y) and inf(X-Y) elements of X - Y?


    inf(X) = 1 and sup(X) = 3
    Then:
    inf(Y) = 1 and sup(Y) = 3
    inf(X-Y) = inf X - sup Y = 1 - 3 = -2 sup(X-Y) = sup X - inf Y = 3 - 1 = 2


    So to solve my question:

    Since inf(X-Y) = -2 and sup(X-Y) = 2. Then from the interval notation 2 is not bounded in the interval notation, whereas -2 is. So then inf(X-Y) is an element of X-Y and sup(X-Y) is not an element of X-Y.

    Would people agree this is correct?
    There's nothing wrong in what you wrote, but I would say the two statements:

    X-Y = [-2, 2)

    and

    inf(X-Y) = inf X - sup Y

    need justification. They are, basically, the two "meaty" parts of the question, and in both cases you've just said "this is true" without any justification.

    [If you've had a proof that inf(X-Y) = inf X - sup Y, then that's fine, you don't need to prove it in an exercise (but you probably would need to reproduce the proof if this was an exam question).]
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    (Original post by DFranklin)
    There's nothing wrong in what you wrote, but I would say the two statements:

    X-Y = [-2, 2)

    and

    inf(X-Y) = inf X - sup Y

    need justification. They are, basically, the two "meaty" parts of the question, and in both cases you've just said "this is true" without any justification.

    [If you've had a proof that inf(X-Y) = inf X - sup Y, then that's fine, you don't need to prove it in an exercise (but you probably would need to reproduce the proof if this was an exam question).]
    Yes you are correct, I have my full solution -althought I didn't define the proof that you stated.

    Thanks
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    (Original post by DFranklin)
    There's nothing wrong in what you wrote, but I would say the two statements:

    X-Y = [-2, 2)

    and

    inf(X-Y) = inf X - sup Y

    need justification. They are, basically, the two "meaty" parts of the question, and in both cases you've just said "this is true" without any justification.

    [If you've had a proof that inf(X-Y) = inf X - sup Y, then that's fine, you don't need to prove it in an exercise (but you probably would need to reproduce the proof if this was an exam question).]
    hey sir can I just ask you - how would you justify how inf(X-Y) is an element etc. I asked my teacher about if how I was saying it like in terms of bounded, he said to express it by interval notation i.e. [-2,2) = -2<= x-y < 2 hence -2 is an element and 2 is not. What would you say is better
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    (Original post by alexgreyx)
    hey sir can I just ask you - how would you justify how inf(X-Y) is an element etc. I asked my teacher about if how I was saying it like in terms of bounded, he said to express it by interval notation i.e. [-2,2) = -2<= x-y < 2 hence -2 is an element and 2 is not. What would you say is better
    Do you mean you want to show that -2 is indeed an element of X-Y? If so, you just need to exhibit a value of x \in X and y\in Y such that x-y = -2 (then x-y = -2 \in X-Y by the definition of X-Y)

    x=1 and y = 3 does the job and you know that both those values are in their respective sets.
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    (Original post by Zacken)
    Do you mean you want to show that -2 is indeed an element of X-Y? If so, you just need to exhibit a value of x \in X and y\in Y such that x-y = -2 (then x-y = -2 \in X-Y by the definition of X-Y)

    x=1 and y = 3 does the job and you know that both those values are in their respective sets.
    Yes excellent thanks
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    (Original post by alexgreyx)
    Yes you are correct, I have my full solution -althought I didn't define the proof that you stated.
    I just feel I should make it clear that by not actually providing detail on the two points I've raised, in my opinion you'd lose almost all the marks for this question if it was actually in an exam. It's not "oh, you've missed a couple of details", it's "there are two key things I expect to see shown in a solution, and you haven't done either of them".

    It's perhaps worth emphasiing that with questions like these, there are very few marks for getting the right answers - the marks are for proving they are right.
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    (Original post by DFranklin)
    I just feel I should make it clear that by not actually providing detail on the two points I've raised, in my opinion you'd lose almost all the marks for this question if it was actually in an exam. It's not "oh, you've missed a couple of details", it's "there are two key things I expect to see shown in a solution, and you haven't done either of them".

    It's perhaps worth emphasiing that with questions like these, there are very few marks for getting the right answers - the marks are for proving they are right.
    I understand that is good advice - thank you for your help
 
 
 
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