# Proof

Assume, $$x\neq x'$$. In the metric space $$(X,d)$$, set $$\epsilon=d(x,x')/2$$. From the definition of convergence,:

1. it follows from $$\lim_{n\rightarrow\infty} x_n=x$$ that there is an $$N(\epsilon)\in\mathbb N$$ with $$d(x_n,x) < \epsilon$$ for all $$n > N(\epsilon)$$.
2. it follows from $$\lim_{n\rightarrow\infty} x_n=x'$$ that there is an $$M(\epsilon)\in\mathbb N$$ with $$d(x_n,x') < \epsilon$$ for all $$n > M(\epsilon)$$.

Therefore, for all $$n > \max(N(\epsilon),M(\epsilon))$$ we have $d(x_n,x) < \epsilon\quad\quad\text{and}\quad\quad d(x_n,x') < \epsilon.$

It follows that

$d(x,x')=\le d(x,x_n) + d(x_n,x') < 2\epsilon=d(x , x'),$

which would mean that $$d(x,x') < d(x,x')$$. This is a contradiction. Therefore, the assumption $$x \neq x'$$ must be wrong and both limits are identical, or $$x=x'$$.

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### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983
2. Forster Otto: "Analysis 2, Differentialrechnung im $$\mathbb R^n$$, Gewöhnliche Differentialgleichungen", Vieweg Studium, 1984