1
$\begingroup$

I have no verified solution for this question.

I had a few questions regarding this:

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

$X-Y = \{ x-y| x\in X, y\in Y\}$

The two questions that I have are that:

$a)$ Find the value of: $X-Y$.

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

Firstly is the answer to $a)$ Simply; $1-1=0, 3-3=0\Rightarrow X-Y= (0,0]$

And for $b)$ I have: $\sup(X)=1, \inf(X)=3$. And $\sup(Y)=3, \inf(Y)$ not possible.

$\sup(X-Y) = 1-3 = -2$

$\inf(X-Y)$ not possible.

If anyone has any feedback it would be greatly appreciated.

  • 0
    Why do you think $\sup(X)=1$ and $\inf(X)=3$? – 2017-02-03
  • 0
    Thank you for your help with the latex, I will implement that in future. – 2017-02-03
  • 0
    Thank you for your reply @Lewis From my work I have adhered to say if we have Z = (1,3) then sup(Z)=3, and inf(Z)=1. As it is not bounded. However with these types I seemed unsure. Perhaps it does not exist? I take it my answer was wrong :l – 2017-02-03
  • 0
    Supremum of a set is the least upper bound, that is, supremum is the least element of a set that is that is greater than or equal to all elements of set. So $1$ cannot be greater than or equal to all elements of the set $[1,3]$. – 2017-02-03

4 Answers 4

1

Hint: To find $X-Y$, think about what the smallest element in that set is. For that you have to take the smallest element of $X$ and subtract the largest element of $Y$ (since negating it yields the smallest element). Hence, $1-3 = -2$ is the smallest element in $X-Y$.

  • 0
    Great thank you for your help here, would I then have to find the largest element in the set? – 2017-02-03
  • 0
    @princetongirl818 Yes, that's the next step. However, since $Y$ has no smallest element (only $1$ as the infimum), there is also no largest element in $X-Y$... – 2017-02-03
  • 0
    Oh of course. Thank you so much now it makes sense. So X-Y would reduce to (-2, infinity)? – 2017-02-03
  • 0
    @princetongirl818 Why infinity? Can $X-Y$ contain $3$? – 2017-02-03
  • 0
    No of course not, mb I was looking at another problem simultaneously. Rather I would say it is (-2) – 2017-02-03
1

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?

  • 0
    Do you mean in terms of X and Y I would think? – 2017-02-08
1

The solution to part a) is [-2,2) do you understand why?

0

The solution is [-2,2) let me know if you need further clarification.

  • 4
    This answers only a). – 2017-03-14
  • 0
    Well I don't think sup is possible, it is easier to see in interval notation. – 2017-03-14