# Neighborhoods

• Oct 22nd 2008, 07:08 PM
kathrynmath
Neighborhoods
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?
• Oct 22nd 2008, 07:22 PM
ThePerfectHacker
Quote:

Originally Posted by kathrynmath
So, we have (x-epsilon, x+epsilon) and (y-epsilon, y+epsilon).

Let $\displaystyle \epsilon = |x-y|/2$.
• Oct 22nd 2008, 07:27 PM
kathrynmath
Quote:

Originally Posted by ThePerfectHacker
Let $\displaystyle \epsilon = |x-y|/2$.

How did you even get that?
• Oct 22nd 2008, 07:31 PM
ThePerfectHacker
Quote:

Originally Posted by kathrynmath
How did you even get that?

Draw a picture. Draw a real number line and put $\displaystyle x$ and $\displaystyle y$. The distance between those points is $\displaystyle |x-y|$. Thus, if you draw a interval around them half the length $\displaystyle |x-y|/2$. Then those two intervals would not intersect.
• Oct 22nd 2008, 07:34 PM
kathrynmath
Quote:

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

Ok, thanks. I get that.
• Nov 8th 2008, 07:27 AM
kathrynmath
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.
• Nov 8th 2008, 10:46 AM
Plato
Suppose that $\displaystyle \varepsilon = \frac{{\left| {x - y} \right|}}{2} > 0$ and that $\displaystyle z \in N_\varepsilon (x) \cap N_\varepsilon (y)$.
$\displaystyle \left| {x - y} \right| \leqslant \left| {x - z} \right| + \left| {z - y} \right| < \varepsilon + \varepsilon = \left| {x - y} \right|$
• Nov 8th 2008, 11:27 AM
kathrynmath
Quote:

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

Ok, I think I understand this. So, is N the neighborhood?
Is this like a triangle inequality?
• Nov 8th 2008, 11:52 AM
Plato
$\displaystyle {\color{blue}\left| {x - y} \right|} \leqslant \left| {x - z} \right| + \left| {z - y} \right| {\color{blue}<} \varepsilon + \varepsilon = {\color{blue}\left| {x - y} \right|}$
Do you see the contradiction in the blues?
• Nov 8th 2008, 12:09 PM
kathrynmath
Quote:

Originally Posted by Plato
$\displaystyle {\color{blue}\left| {x - y} \right|} \leqslant \left| {x - z} \right| + \left| {z - y} \right| {\color{blue}<} \varepsilon + \varepsilon = {\color{blue}\left| {x - y} \right|}$
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
• Nov 8th 2008, 12:25 PM
Plato
Quote:

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 $\displaystyle \left| {x - y} \right| < \left| {x - y} \right|$.
• Nov 8th 2008, 12:32 PM
kathrynmath
Quote:

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

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