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Show that if $a,b \in \mathbb{R}$, and $a \not= b$, then there exist e-neighborhoods $U$ of $a$ and $V$ of $b$ such that $U \cap V = \varnothing$

I am thinking that I may need to use the triangle inequality but I am not sure really how to get started. Is this the appropriate thing to do? Thanks!

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Assume, without loss of generality that $a<b$. Let $m := \frac{1}{2}(b-a)$ be half the distance from $a$ to $b$. The neighbourhoods could be $U_a := \{ x \in \mathbb{R} : a-m < x < a+m\}$ and $U_b := \{ x \in \mathbb{R} : b-m < x < b+m\}$. It's clear that $U_a \cap U_b = \emptyset$. Better still, you could define

$$U_a := \{ x \in \mathbb{R} : a-m/2 < x < a+m/2 \} \, , $$

$$U_b := \{ x \in \mathbb{R} : b-m/2 < x < b+m/2 \} \, , $$

so that $\overline{U_a} \cap \overline{U_b} = \emptyset.$

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