Question

Given a sequence {a_n} with a_n part of X and X is a metric space. Prove that if a_n is convergent, then the limit is unique.

Answer 1

Question 1. Given a sequence $\{a_n\}$ with $a_n$ part of $X$ and $X$ is a metric space. Prove that if $a_n$ is convergent, then the limit is unique.

Solution. Let $\rho$ denote the metric on $X$. Suppose there are $a, b \in X$ such that $a_n \to a$ and $a_n \to b$. Take an arbitrary $\varepsilon > 0$. By definition there are $N_1, N_2 \in \mathbb{N}$ such that $\rho(a_n, a) < \varepsilon$ for all $n > N_1$ and $\rho(a_n, b) < \varepsilon$ for all $n > N_2$. Choose $n > \max\{N_1, N_2\}$. Then by triangle inequality

$$\rho(a, b) < \rho(a, a_n) + \rho(a_n, b) < \varepsilon + \varepsilon = 2\varepsilon.$$

Since $\varepsilon$ is an arbitrary positive number, we conclude that $\rho(a, b) = 0$. By identity of indiscernibles $a = b$.