Answer to Question #217106 in Real Analysis for Prathibha Rose

Question

Prove that a sequence in a metric space cannot converge more than one limit

Answer 1

Let us prove that a sequence $(x_n)_{n\in\N}$ in a metric space $(X,d)$ cannot converge more than one limit using the method by contradiction. Suppose that the sequence $(x_n)_{n\in\N}$ converges to $x$ and also it converges to $y\ne x.$ Let $\rho=d(x,y).$ Since $x\ne y,$ we conclude that $\rho>0.$ Let $\varepsilon=\frac{\rho}{2}.$ By definition of limit there exists $n_x\in\N$ such that $d(x_n,x)<\varepsilon$ for all $n\ge n_x.$ Also there exists $n_y\in\N$ such that $d(x_n, y)<\varepsilon$ for all $n\ge n_y.$ Let $n_0=\max\{n_x,n_y\}.$ Then $d(x,y)<d(x,x_n)+d(x_n,y)<\varepsilon+\varepsilon=2\varepsilon=\rho$ for all $n\ge n_0,$ a contradiction with $d(x,y)=\rho.$ We conclude that a sequence in a metric space cannot converge more than one limit.