# Thread: Infinum and surpemum

1. ## Infinum and surpemum

If i have a set S = {x| |x-2| + |3-x| = 1 }

and i have to check what the sup and inf are for it, is this correct way of working it out:

-1 < x-2 + 3-x < 1 or -2 < 0 < 0

Since x cancels out can i say that S has no supremum not infinum?

Thank you

2. ## Re: Infinum and surpemum

Originally Posted by iva
If i have a set S = {x| |x-2| + |3-x| = 1 }
and i have to check what the sup and inf are for it, is this correct way of working it out:
Well $S=[2,3]$

3. ## Re: Infinum and surpemum

Thanks then that would make sup S = 3 and inf S = 2, is that right?
But how did you get to S=[2,3] as with my equation above i couldn't get a sensible result for x

Thank you

4. ## Re: Infinum and surpemum

Why not consider each term separately?

Since |3-x| is non-negative, |x-2|<=1. Solve this to get 1<=x<=3. Since 3 is in the set, it's the supremum. Now do something similar with |3-x|.

5. ## Re: Infinum and surpemum

Originally Posted by LoblawsLawBlog
Why not consider each term separately?
Since |3-x| is non-negative, |x-2|<=1. Solve this to get 1<=x<=3. Since 3 is in the set, it's the supremum. Now do something similar with |3-x|.
The solution set for $|x-2|+|x-3|=1$ is $2\le x\le 3$

$|x-3|$ is the distance from x to 3.

$|x-2|$ is the distance from x to 2.

So if $2\le x\le 3$ we have $|x-2|+|x-3|=1$.

6. ## Re: Infinum and surpemum

I didn't mean that 1<=x<=3 is the solution set, but it's still true. |3-x|<=1 then gives 2<=x<=4, which gives the infinum. I didn't see your solution immediately so I went with the slightly longer one.