Proof-- Limits (Missed Class, please help)

**amm345** · October 22nd 2009, 05:53 PM

Let

x, y be elements ofR such that x not= y. Prove that there exists epsom > 0 such that y not element of[x − epsom, x + epsom]

**artvandalay11** · October 22nd 2009, 06:03 PM

Originally Posted by amm345

Let

x, y be elements ofR such that x not= y. Prove that there exists epsom > 0 such that y not element of[x − epsom, x + epsom]

Since $x\not = y$ , distance between x and y, denoted $d(x,y)\not = 0$

Define $\epsilon=d(x,y)/2$

The distance function is never negative so epsilon is greater than 0

Then....

can you finish it from here?

**amm345** · October 22nd 2009, 06:21 PM

I'm not sure, I had kinda had that based on the lecture notes he posts online, but I'm still not sure what to do next.

**artvandalay11** · October 22nd 2009, 06:48 PM

Originally Posted by artvandalay11

Since $x\not = y$ , distance between x and y, denoted $d(x,y)\not = 0$

Define $\epsilon=d(x,y)/2$

The distance function is never negative so epsilon is greater than 0

Then....

can you finish it from here?

Assume $y\in[x-\frac{d(x,y)}{2}, x+\frac{d(x,y)}{2}]$

Then $x-\frac{d(x,y)}{2}\leq y\leq x+\frac{d(x,y)}{2}$

So $-\frac{d(x,y)}{2}\leq y-x\leq\frac{d(x,y)}{2}$

On the real line, $d(x,y)=|x-y|$ so

$-\frac{1}{2}|x-y|\leq y-x\leq \frac{1}{2}|x-y|$

So $-\frac{1}{2}|x-y|\leq y-x$ and

$y-x\leq\frac{1}{2}|x-y|$

Multiplying by -2 and 2 respectively yields

$-2(y-x)\geq |x-y|$ and

$2(y-x)\geq |x-y|$

These two facts imply x=y, but that is a contradiction with the given information

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