Neighborhoods

**kathrynmath** · October 22nd 2008, 07:08 PM

Let x and y be distinct real numbers. Prove that there is a nbhood P of x and a nbhood Q of y such that PintersectQ=null set.
So, we have (x-epsilon, x+epsilon) and (y-epsilon, y+epsilon).
I have an idea of how to do this but I'm getting stuck. So, I think I need to define epsilon, right because these are two different epsilons. This is where I get stuck. Can someone give me pointers on defining epsilon?

**ThePerfectHacker** · October 22nd 2008, 07:22 PM

Originally Posted by kathrynmath

So, we have (x-epsilon, x+epsilon) and (y-epsilon, y+epsilon).

Let $\epsilon = |x-y|/2$ .

**kathrynmath** · October 22nd 2008, 07:27 PM

Originally Posted by ThePerfectHacker

Let $\epsilon = |x-y|/2$ .

How did you even get that?

**ThePerfectHacker** · October 22nd 2008, 07:31 PM

Originally Posted by kathrynmath

How did you even get that?

Draw a picture. Draw a real number line and put $x$ and $y$ . The distance between those points is $|x-y|$ . Thus, if you draw a interval around them half the length $|x-y|/2$ . Then those two intervals would not intersect.

**kathrynmath** · October 22nd 2008, 07:34 PM

Originally Posted by ThePerfectHacker

Draw a picture. Draw a real number line and put $x$ and $y$ . The distance between those points is $|x-y|$ . Thus, if you draw a interval around them half the length $|x-y|/2$ . Then those two intervals would not intersect.

Ok, thanks. I get that.

**kathrynmath** · November 8th 2008, 07:27 AM

I'm not quite sure how to prove it,though. I understand why it works and all, but I know a picture doesn't qualify as a proof.

**Plato** · November 8th 2008, 10:46 AM

Suppose that $\varepsilon = \frac{{\left| {x - y} \right|}}{2} > 0$ and that $z \in N_\varepsilon (x) \cap N_\varepsilon (y)$ .
$\left| {x - y} \right| \leqslant \left| {x - z} \right| + \left| {z - y} \right| < \varepsilon + \varepsilon = \left| {x - y} \right|$

**kathrynmath** · November 8th 2008, 11:27 AM

Originally Posted by Plato

Suppose that $\varepsilon = \frac{{\left| {x - y} \right|}}{2} > 0$ and that $z \in N_\varepsilon (x) \cap N_\varepsilon (y)$ .
$\left| {x - y} \right| \leqslant \left| {x - z} \right| + \left| {z - y} \right| < \varepsilon + \varepsilon = \left| {x - y} \right|$

Ok, I think I understand this. So, is N the neighborhood?
Is this like a triangle inequality?

**Plato** · November 8th 2008, 11:52 AM

${\color{blue}\left| {x - y} \right|} \leqslant \left| {x - z} \right| + \left| {z - y} \right| {\color{blue}<} \varepsilon + \varepsilon = {\color{blue}\left| {x - y} \right|}$
Do you see the contradiction in the blues?

**kathrynmath** · November 8th 2008, 12:09 PM

Originally Posted by Plato

${\color{blue}\left| {x - y} \right|} \leqslant \left| {x - z} \right| + \left| {z - y} \right| {\color{blue}<} \varepsilon + \varepsilon = {\color{blue}\left| {x - y} \right|}$
Do you see the contradiction in the blues?

I'm trying to see the contradiction. Why is < epsilon a contradiciton? I'm not getting it....
and the x-y in the end appears correct because 2epsilon = x-y

**Plato** · November 8th 2008, 12:25 PM

Originally Posted by kathrynmath

I'm trying to see the contradiction. Why is < epsilon a contradiciton? I'm not getting it....
and the x-y in the end appears correct because 2epsilon = x-y

A number cannot be less than itself.
There in blue you see $\left| {x - y} \right| < \left| {x - y} \right|$ .

**kathrynmath** · November 8th 2008, 12:32 PM

Originally Posted by Plato

A number cannot be less than itself.
There in blue you see $\left| {x - y} \right| < \left| {x - y} \right|$ .

Oh ok. So then this is a a contradicition and leads to the fact that P intersect Q = null set?

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