1. ## 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?

2. Originally Posted by kathrynmath
So, we have (x-epsilon, x+epsilon) and (y-epsilon, y+epsilon).
Let $\epsilon = |x-y|/2$.

3. Originally Posted by ThePerfectHacker
Let $\epsilon = |x-y|/2$.
How did you even get that?

4. 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.

5. 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.

6. 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.

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

8. 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?

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

10. 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

11. 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 $\left| {x - y} \right| < \left| {x - y} \right|$.

12. 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?