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Let $x$ and $y$ be unique real numbers. How do you prove that there exists a neighborhood $P$ of $x$ and a neighborhood $Q$ of $y$ such that $P \cap Q = \emptyset$?

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Hint: Consider $d = \dfrac{|x - y|}{2}$. – JavaMan Sep 3 '12 at 23:32
One usually writes "Let $x$ and $y$ be distinct real numbers." – André Nicolas Sep 4 '12 at 0:34

Hint: open intervals of length $e$ centered at $x$ and $y$ are such neighborhoods, and if $e$ is small enough, they will not intersect. How small does $e$ have to be for this to happen?

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Without loss of generality, assume $x < y$. then $x\in(-\infty, (x + y)/2)$ and $y \in ((x+y)/2,\infty)$.

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As an alternative approach, prove (or reference the fact) that for any two distinct real numbers $x,y$, there is a rational number $q$ lying between them. If we assume (without loss of generality) that $x<y$, then $(x-1,q)$ and $(q,y+1)$ do the trick.

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