Real Numbers - Real Anaylsis

**hockey777** · September 3rd 2008, 10:20 AM

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 $U$ $\cap$ $V$ = $\emptyset$

**Plato** · September 3rd 2008, 10:24 AM

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

**hockey777** · September 3rd 2008, 10:35 AM

any reason why you divided by 4?

**Plato** · September 3rd 2008, 10:38 AM

Originally Posted by hockey777

any reason why you divided by 4?

Two reasons: it works & I like 4.

**hockey777** · September 3rd 2008, 10:40 AM

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?

**Plato** · September 3rd 2008, 10:54 AM

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

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