# inequality proof

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• September 30th 2008, 09:25 PM
lllll
inequality proof
Using the triangle inequality show $a$ and $b$ are in $\mathbb{R}$ with $a \neq b$, then $\exists$ open intervals $U$ centered at $a$ and $V$ centered at b, both with radius $\epsilon=\frac{1}{2}|a-b|$

so far all I have is

$|a-b|=2\epsilon$

$||a|-|b|| \leq |a-b|$

now I'm not sure if what I'm doing is right:

$-\epsilon<||a|-|b|| < \epsilon \rightarrow ||a|-|b|| < 2\epsilon \rightarrow ||a|-|b|| < |a-b|$ ?
• September 30th 2008, 11:09 PM
CaptainBlack
Quote:

Originally Posted by lllll
Using the triangle inequality show $a$ and $b$ are in $\mathbb{R}$ with $a \neq b$, then $\exists$ open intervals $U$ centered at $a$ and $V$ centered at b, both with radius $\epsilon=\frac{1}{2}|a-b|$

You do realise that as this stands it is not the question that you think it is, don't you?

Do you want to show that:

For all $a,b \in \mathbb{R},\ a \ne b$, then the two disjoint open intervals $U$ centered at $a$ and $V$ centered at $b$, both with radius $\epsilon=\frac{1}{2}|a-b|$ are disjoint?

RonL
• October 3rd 2008, 11:03 PM
lllll
Quote:

Originally Posted by CaptainBlack

Do you want to show that:

For all $a,b \in \mathbb{R},\ a \ne b$, then the two disjoint open intervals $U$ centered at $a$ and $V$ centered at $b$, both with radius $\epsilon=\frac{1}{2}|a-b|$ are disjoint?

RonL

I completely overlooked that part of the question. But I did found a prove for this statement but am still lost as to how it was done, even after asking my prof.

This is what I found:

$U=(x-\epsilon, \ x+\epsilon), \ V = (y-\epsilon, \ y+\epsilon)$

$U \bigcap V = \emptyset$

suppose $z$ is in $U \bigcap V$

$|x-y| \leq |x-z|+|z-y| < \epsilon + \epsilon = 2\epsilon = |x-y|$

which is apparently a contradiction.

If anyone could clarify this, it would help me out a lot.
• October 3rd 2008, 11:08 PM
CaptainBlack
Quote:

Originally Posted by lllll
I completely overlooked that part of the question. But I did found a prove for this statement but am still lost as to how it was done, even after asking my prof.

This is what I found:

$U=(x-\epsilon, \ x+\epsilon), \ V = (y-\epsilon, \ y+\epsilon)$

$U \bigcap V = \emptyset$

suppose $z$ is in $U \bigcap V$

$|x-y| \leq |x-z|+|z-y| < \epsilon + \epsilon = 2\epsilon = |x-y|$

which is apparently a contradiction.

If anyone could clarify this, it would help me out a lot.

The bit that is missing is:

$|x-y|=|x-z+z-y|=|(x-z)-(y-z)|$

Now by the triangle inequality:

$|x-y|=|(x-z)-(y-z)| \le |x-z|+|y-z|$

etc.

RonL
• October 4th 2008, 03:22 PM
lllll
Quote:

Originally Posted by CaptainBlack
The bit that is missing is:

$|x-y|=|x-z+z-y|=|(x-z)-(y-z)|$

Now by the triangle inequality:

$|x-y|=|(x-z)-(y-z)| \le |x-z|+|y-z|$

etc.

RonL

That solves a part of the proof, but where is the contradiction?

It lies is the fact that $z$ is assumed to be in $U \cap V$ so lies in $U$ and in $V$ so $|x-z|<\epsilon,$ and $|y-z|<\epsilon$, so:

$|x-y|<2 \epsilon$

But $|x-y|=2 \epsilon$, so we have:

$|x-y|<|x-y|$

which is a contradiction.

RonL
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