# Real Numbers - Real Anaylsis

• Sep 3rd 2008, 10:20 AM
hockey777
Real Numbers - Real Anaylsis
Show that if $\displaystyle a,b$ $\displaystyle \in$$\displaystyle \Re, and \displaystyle a$$\displaystyle \neq$$\displaystyle b$, then there exist $\displaystyle \epsilon$-neighborhoods $\displaystyle U$ of $\displaystyle a$ and $\displaystyle V$ of $\displaystyle b$ such that $\displaystyle U$ $\displaystyle \cap$ $\displaystyle V$ = $\displaystyle \emptyset$
• Sep 3rd 2008, 10:24 AM
Plato
Let $\displaystyle \varepsilon = \frac{{\left| {a - b} \right|}}{4}$.
• Sep 3rd 2008, 10:35 AM
hockey777
any reason why you divided by 4?
• Sep 3rd 2008, 10:38 AM
Plato
Quote:

Originally Posted by hockey777
any reason why you divided by 4?

Two reasons: it works & I like 4.
• Sep 3rd 2008, 10:40 AM
hockey777
Sorry one more question. What else was throwing me off in this problem was the fact I don't think it's clear for a beginner like me. I think the problem suggest that $\displaystyle \epsilon$ should be the same for both, but I'm not sure. Is that true?
• Sep 3rd 2008, 10:54 AM
Plato
No there is no requirement the neighborhoods have the same radius.
But I gave you a value that will work for both.

Let $\displaystyle U = \left\{ {x:\left| {a - x} \right| < \varepsilon } \right\}\,\& \,V = \left\{ {x:\left| {b - x} \right| < \varepsilon } \right\}$.

Now suppose that $\displaystyle y \in U \cap V \Rightarrow \quad \left| {a - b} \right| \le \left| {a - y} \right| + \left| {y - b} \right| < \varepsilon + \varepsilon = \frac{{\left| {a - b} \right|}}{2}$.
That is a clear contradiction.
So U & V must have nothing in common.