Thread: Real Numbers - Real Anaylsis

1. 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 $U$ $\cap$ $V$ = $\emptyset$

2. Let $\varepsilon = \frac{{\left| {a - b} \right|}}{4}$.

3. any reason why you divided by 4?

4. Originally Posted by hockey777
any reason why you divided by 4?
Two reasons: it works & I like 4.

5. 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?

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