Ordered field/Sequence/Unique limit/Fact/Proof

Proof

We assume that the sequence has two distinct limits ${}x,y$, ${}x\neq y$. Then ${}d:=\vert {x-y}\vert >0$. We consider ${}\epsilon :=d/3>0$. Because of the convergence to ${}x$ there exists an ${}n_{0}$ such that

$\vert {x_{n}-x}\vert \leq \epsilon {\text{ for all }}n\geq n_{0}$

and because of the convergence to ${}y$ there exists an ${}n_{0}'$ such that

$\vert {x_{n}-y}\vert \leq \epsilon {\text{ for all }}n\geq n_{0}'.$

hence both conditions hold simultaneously for ${}n\geq \max\{n_{0},n_{0}'\}$. Suppose that ${}n$ is as large as this maximum. Then due to the triangle inequality we arrive at the contradiction

${}d=\vert {x-y}\vert \leq \vert {x-x_{n}}\vert +\vert {x_{n}-y}\vert \leq \epsilon +\epsilon =2d/3\,.$

To fact