Adherent Points and Convergent Sequences in Metric Spaces

# Adherent Points and Convergent Sequences in Metric Spaces

 Theorem 1: Let $(M, d)$ be a metric space and let $(x_n)_{n=1}^{\infty}$ be a convergent sequence in $M$ such that $\lim_{n \to \infty} x_n = p$. Then $p$ is an adherent point of $\{ x_1, x_2, ..., x_n, ... \}$.
• Proof: Recall that $p$ is an adherent point of a set $S = \{ x_1, x_2, ..., x_n, ... \}$ if every ball centered at $p$ contains elements of $S$, i.e., for all positive real numbers $r > 0$, $B(p, r) \cap S \neq \emptyset$
• Now let $(x_n)_{n=1}^{\infty} = (x_1, x_2, ..., x_n, ...)$ be a convergent sequence such that $\lim_{n \to \infty} x_n = p$. Then $\lim_{n \to \infty} d(x_n, p) = 0$ So, for all $\epsilon > 0$ there exists an $N(\epsilon) \in \mathbb{N}$ such that if $n \geq N$ then $d(x_n, p) < \epsilon$.
• So, for each $r > 0$, there exists an $N(r) \in \mathbb{N}$ such that if $n \geq N(r)$ then $d(x_n, p) < \epsilon$. So for each $x_n \in S$ where $n \geq N(r)$ we have that $x_n \in B(p, r)$. Hence $B(p, r) \cap S \neq \emptyset$, so $p$ is an accumulation point of $S = \{x_1, x_2, ..., x_n, ... \}$.
 Theorem 2: Let $(M, d)$ be a metric space, let $S \subseteq M$, and let $p \in M$ be an adherent point of $S$. Then there exists a sequence $(x_n)_{n=1}^{\infty}$ contained in $S$ such that $\lim_{n \to \infty} x_n = p$.
• Proof: Let $S \subseteq M$ and let $p \in M$ be an adherent point of $S$. Then every ball centered at $p$ contains elements from $S$, that is, for all positive real numbers $r > 0$, $B(p, r) \cap S \neq \emptyset$.
• Since $\frac{1}{n} > 0$ for all $n \in \mathbb{N}$ then for all $n \in \mathbb{N}$, $B \left ( p, \frac{1}{n} \right ) \cap S \neq \emptyset$. So, take $x_n \in B \left ( p, \frac{1}{n} \right ) \cap S$ to form a sequence $(x_n)_{n=1}^{\infty}$ that is fully contained in $S$. We claim that $(x_n)_{n=1}^{\infty}$ converges to $p$.
• Since $x_n \in B \left ( p, \frac{1}{n} \right )$ we have that $0 \leq d(x_n, p) < \frac{1}{n}$ for all $n \in \mathbb{N}$. Taking the limits of this inequality yield:
(1)
\begin{align} \quad \lim_{n \to \infty} 0 \leq \lim_{n \to \infty} d(x_n, p) \leq \lim_{n \to \infty} \frac{1}{n} \\ \quad 0 \leq \lim_{n \to \infty} d(x_n, p) \leq 0 \end{align}
• By the Squeeze Theorem we have that $\lim_{n \to \infty} d(x_n, p) = 0$ and so $\lim_{n \to \infty} x_n = p$, so $(x_n)_{n=1}^{\infty}$ converges to $p$. $\blacksquare$