# Thread: Use the triangle inequality to show...

1. ## Use the triangle inequality to show...

Use the triangle inequality to show that if a and be are in R, with a not equal to b, then there exists an open interval U centered at a and V centered at b, both of radius epsilon= 1/2 abs[a-b] , with U/\V= 0 (intersection)

I really don't know what I wanna do here. I'm imagining a line centered at A and another at b with radius epsilon something like this =>

----- (-e)-------a-------(e)------

-----(-e)--------b-------(e)-----

2. Originally Posted by mtlchris
Use the triangle inequality to show that if a and be are in R, with a not equal to b, then there exists an open interval U centered at a and V centered at b, both of radius epsilon= 1/2 abs[a-b] , with U/\V= 0 (intersection)

I really don't know what I wanna do here. I'm imagining a line centered at A and another at b with radius epsilon something like this =>

----- (-e)-------a-------(e)------

-----(-e)--------b-------(e)-----
We may assume wlog that $a. Thus the open interval around $a$ is $U=\left(a-\frac{1}{2}(b-a),a+\frac{1}{2}(b-a)\right)=\left(\frac{1}{2}(3a-b),\frac{1}{2}(a+b)\right)$

The open interval around $b$ is $V=\left(b-\frac{1}{2}(b-a),b+\frac{1}{2}(b-a)\right)=\left(\frac{1}{2}(a+b),\frac{1}{2}(-a+3b)\right)$

From this it's clear that $U$ and $V$ are disjoint, so $U\cap V=\emptyset$.

3. what about showing the interval for both U and V to be of radius epsilon= 1/2*absolute value[a-b]?

4. Originally Posted by mtlchris
what about showing the interval for both U and V to be of radius epsilon= 1/2*absolute value[a-b]?
It doesn't have to be; take any $c\leq\frac{1}{2}$, and taking the radius to be $c|a-b|$, will guarantee $U\cap V=\emptyset$.

5. I don't really follow...how come we can assume that a < b ?

6. Originally Posted by mtlchris
I don't really follow...how come we can assume that a < b ?
One of them has to be greater than the other. If it's not $a$, then it's $b$. The reason we can assume wlog is that if we assumed $b, we would still get the same result.

7. and we make this assumption so we can relate U and V?

8. I made the assumption so I didn't have to deal with the absolute value sign.

9. ok I follow this, but if I need to use the triangle property how do i go about using it?
A friend told me:
let u= (a-e, a+e) = absolute value(c-a) < e for c in R
let v= (b-e, b+e) = absolute value(d-a) < e for d in R

then absloutevalue(x-a) < e = 1/2absloute(a-b)
absloute value(x-b) < e = 1/2absloute(a-b)

is this correct? and if so....how does he go from abs(c-a) < e to
abs(x-a) < e
and then to = 1/2 abs(a-b)

10. Originally Posted by mtlchris
ok I follow this, but if I need to use the triangle property how do i go about using it?
A friend told me:
let u= (a-e, a+e) = absolute value(c-a) < e for c in R
let v= (b-e, b+e) = absolute value(d-a) < e for d in R
$U$ is a set; $|c-a|$ is a number. How can those be equal?

11. ok i should have written
{ c in R : abs(c-x) < e }

12. Originally Posted by mtlchris
ok I follow this, but if I need to use the triangle property how do i go about using it?
A friend told me:
let u= (a-e, a+e) = absolute value(c-a) < e for c in R
let v= (b-e, b+e) = absolute value(d-a) < e for d in R

then absloutevalue(x-a) < e = 1/2absloute(a-b)
absloute value(x-b) < e = 1/2absloute(a-b)

is this correct? and if so....how does he go from abs(c-a) < e to
abs(x-a) < e
and then to = 1/2 abs(a-b)
Your friend introduced more variables than are necessary (namely $c$ and $d$). If $U=\{x\in\mathbb{R}:|a-x|<\epsilon\}$ and $V=\{x\in\mathbb{R}:|x-b|<\epsilon\}$, then assume for a contradiction that $U\cap V\neq\emptyset$; that is, $\exists x$ such that $x\in U$ and $x\in V$. Then $|a-x|<\epsilon$ and $|x-b|<\epsilon$.

Then by the triangle inequality $|a-b|\leq|a-x|+|x-b|<\epsilon+\epsilon=2\epsilon=|a-b|$.

So we have $|a-b|<|a-b|$, which is a contradiction. So $U\cap V=\emptyset$.