The Uniqueness of Limits of Sequences in Metric Spaces

# The Uniqueness of Limits of Sequences in Metric Spaces

Recall from the Limits of Sequences in Metric Spaces that if $(M, d)$ is a metric space then a sequence of elements from $M$ is of the form $(x_n)_{n=1}^{\infty}$ where $x_k \in M$ for all $k \in \{ 1, 2, ... \}$.

We will now show that if the sequence $(x_n)_{n=1}^{\infty}$ is convergent then its limit $p \in M$ is unique.

Theorem 1: Let $(M, d)$ be a metric space and let $(x_n)_{n=1}^{\infty}$ be a convergent sequence in $M$. Then the limit of $(x_n)_{n=1}^{\infty}$ is unique. |

- Let $(M, d)$ be a metric space and $(x_n)_{n=1}^{\infty}$ be a convergent sequence in $M$. Suppose that $\lim_{n \to \infty} x_n = p$ and $\lim_{n \to \infty} x_n = q$ for some $p, q \in M$. We want to show that $p = q$ are equal - but this happens if and only if $d(p, q) = 0$. So, we will prove that $d(p, q) = 0$.

- Now since $\lim_{n \to \infty} x_n = p$ then $\lim_{n \to \infty} d(x_n, p) = 0$. So for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists an $N_1 \in \mathbb{N}$ such that if $n \geq N_1$ then:

\begin{align} \quad \mid d(x_n, p) - 0 \mid = d(x_n, p) < \epsilon_1 = \frac{\epsilon}{2} \quad (*) \end{align}

- Similarly, since $\lim_{n \to \infty} x_n = q$ then $\lim_{n \to \infty} d(x_n, q) = 0$. So for $\epsilon_2 = \frac{\epsilon}{2} > 0$ there exists an $N_2 \in \mathbb{N}$ such that if $n \geq N_2$ then:

\begin{align} \quad \mid d(x_n, q) - 0 \mid = d(x_n, q) < \epsilon_2 = \frac{\epsilon}{2} \quad (**) \end{align}

- Let $N = \max \{ N_1, N_2 \}$ so that $(*)$ and $(**)$ both hold for $n \geq N$, and consider $d(p, q)$. Since $d$ is a metric, we have that the triangle inequality holds on $d $ ]] and so for [[$ n \geq N$ we have that:

\begin{align} \quad d(p, q) \leq d(p, x_n) + d(x_n, q) < \epsilon_1 + \epsilon_2 = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}

- So for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ we have that $d(p, q) < \epsilon$ which implies that $d(p, q) = 0$. So $p = q$. Hence if $(x_n)_{n=1}^{\infty}$ is convergent then its limit is unique. $\blacksquare$