Prove that for all $a,b \in \Bbb R$ with $a\neq b$ there exists $\epsilon, \delta \in \Bbb R^+$ such that $\{x \in \Bbb R : |x-a|<\epsilon\} \cap \{x \in \Bbb R : |x-b|<\delta\} = \emptyset$.
Usually I would at least show an attempt of the proof, but for this one I really have no idea how to start it.
Does anybody have any suggestions/ideas?
Hint: Let $d=|a-b|$. Take $\varepsilon =\delta =\dfrac d2$.
Edit: For the sake of argument, assume $\{x \in \Bbb R : |x-a|<\varepsilon\} \cap \{x \in \Bbb R : |x-b|<\delta\} \neq \emptyset,$ where $\varepsilon =\delta =\dfrac d2$. Then there exists an element $x_0\in\Bbb R $ such that $|x_0-a|<\varepsilon$ and $|x_0-b|<\delta$. Now using triangle inequality, we have $$|a-b|\leq|a-x_0|+|x_0-y|\lt\varepsilon+ \delta.$$ That is $$|a-b|\lt \dfrac d2+\dfrac d2=d=|a-b|.$$ In other words, $|a-b|\lt|a-b|$, a contradiction. Hence $\{x \in \Bbb R : |x-a|<\dfrac d2\} \cap \{x \in \Bbb R : |x-b|<\dfrac d2\}= \emptyset.$
I'm leaving a picture below to give a better understanding of what I've done.Replace $x$ and $y$ in the picture with $a$ and $b$ respectively. Note that the point in between $a $ and $b $ is $\dfrac {b-a}{2}=\dfrac {|a-b|}{2}$. I think you can complete other details.
Also observe that any $\varepsilon $(and $\delta$) satisfying $0\lt\varepsilon\leq \dfrac {|a-b|}{2}$ will do the work.