Use the triangle inequality to show...

**mtlchris** · October 22nd 2009, 09:02 PM

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)-----

**redsoxfan325** · October 22nd 2009, 09:47 PM

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 $a<b$ . 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 $U\cap V=\emptyset$ .

**mtlchris** · October 22nd 2009, 09:51 PM

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

**redsoxfan325** · October 22nd 2009, 09:55 PM

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 $U\cap V=\emptyset$ .

**mtlchris** · October 22nd 2009, 10:01 PM

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

**redsoxfan325** · October 22nd 2009, 10:04 PM

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 $b<a$ , we would still get the same result.

**mtlchris** · October 22nd 2009, 10:10 PM

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

**redsoxfan325** · October 22nd 2009, 10:11 PM

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

**mtlchris** · October 22nd 2009, 10:32 PM

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)

**redsoxfan325** · October 22nd 2009, 10:37 PM

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?

**mtlchris** · October 22nd 2009, 10:42 PM

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

**redsoxfan325** · October 22nd 2009, 11:06 PM

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 $U\cap V=\emptyset$ .

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