# Real Numbers - Real Anaylsis

• Sep 3rd 2008, 11:20 AM
hockey777
Real Numbers - Real Anaylsis
Show that if $a,b$ $\in$ $\Re$, and $a$ $\neq$ $b$, then there exist $\epsilon$-neighborhoods $U$ of $a$ and $V$ of $b$ such that $U$ $\cap$ $V$ = $\emptyset$
• Sep 3rd 2008, 11:24 AM
Plato
Let $\varepsilon = \frac{{\left| {a - b} \right|}}{4}$.
• Sep 3rd 2008, 11:35 AM
hockey777
any reason why you divided by 4?
• Sep 3rd 2008, 11:38 AM
Plato
Quote:

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

Two reasons: it works & I like 4.
• Sep 3rd 2008, 11:40 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 $\epsilon$ should be the same for both, but I'm not sure. Is that true?
• Sep 3rd 2008, 11:54 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 $U = \left\{ {x:\left| {a - x} \right| < \varepsilon } \right\}\,\& \,V = \left\{ {x:\left| {b - x} \right| < \varepsilon } \right\}$.

Now suppose that $y \in U \cap V \Rightarrow \quad \left| {a - b} \right| \le \left| {a - y} \right| + \left| {y - b} \right| < \varepsilon + \varepsilon = \frac{{\left| {a - b} \right|}}{2}$.