1. ## inequality proof

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

so far all I have is

$\displaystyle |a-b|=2\epsilon$

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

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

$\displaystyle -\epsilon<||a|-|b|| < \epsilon \rightarrow ||a|-|b|| < 2\epsilon \rightarrow ||a|-|b|| < |a-b|$ ?

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

RonL

3. Originally Posted by CaptainBlack

Do you want to show that:

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

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

$\displaystyle U \bigcap V = \emptyset$

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

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

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

4. 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:

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

$\displaystyle U \bigcap V = \emptyset$

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

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

If anyone could clarify this, it would help me out a lot.
The bit that is missing is:

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

Now by the triangle inequality:

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

etc.

RonL

5. Originally Posted by CaptainBlack
The bit that is missing is:

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

Now by the triangle inequality:

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

etc.

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

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

$\displaystyle |x-y|<2 \epsilon$

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

$\displaystyle |x-y|<|x-y|$