Sequences and Limits in Metric Spaces Review

We will now review some of the recent content regarding sequences and limits of sequences in metric spaces.

Let $(M, d)$ be a metric space.

• Recall from the Limits of Sequences in Metric Spaces page that an (infinite) Sequence in a metric space is an infinite ordered list $(x_n)_{n=1}^{\infty}$ of points where $x_k \in M$ for each $k \in \{ 1, 2, ... \}$.

• We said that a sequence $(x_n)_{n=1}^{\infty}$ is Converges to a point $p \in M$ which we write $\lim_{n \to \infty} x_n = p$ if the distance between $x_n$ and $p$ can be made arbitrarily close to $0$, that is, $\lim_{n \to \infty} d(x_n, p) = 0$. If such a $p \in M$ exists, then we say that $p$ is the Limit of this sequence, and if not such $p \in M$ exists then we said that the sequence $(x_n)_{n=1}^{\infty}$ Diverges.

• On The Uniqueness of Limits of Sequences in Metric Spaces page we saw that if a sequence $(x_n)_{n=1}^{\infty}$ converges its limit is unique, that is, if $\lim_{n \to \infty} x_n = p$ and $\lim_{n \to \infty} x_n = q$ then $p = q$.

• We then saw on The Boundedness of Convergent Sequences in Metric Spaces page that if $(x_n)_{n=1}^{\infty}$ is a convergent sequence, then the set of elements $\{ x_1, x_2, ..., x_n, ... \}$ is a bounded set in $M$. We proved this by noting that since $(x_n)_{n=1}^{\infty}$ is a convergent sequence that then for $\epsilon_0 = 1 > 0$ there exists an $N(\epsilon_0) \in \mathbb{N}$ such that if $n \geq N(\epsilon_0)$ then $d(x_n, p) < \epsilon_0 = 1$, so if $M^* = \max \{ d(x_1, p), d(x_2, p), ..., d(x_{N(\epsilon_0) - 1}) \}$ then for $1 \leq n < N(\epsilon_0)$ we have that $d(x_n, p) \leq M^*$. So letting $M = \max \{ M^*, 1 \}$ shows us that for all $n \in \mathbb{N}$, $d(x_n, p) \leq M$ so $\{x_1, x_2, ..., x_n, ... \} \subseteq B(p, M + 1)$ which shows that $\{ x_1, x_2, ..., x_n, ... \}$ is bounded in $M$.

• On the Adherent Points and Convergent Sequences in Metric Spaces page we saw that if $(x_n)_{n=1}^{\infty}$ converges to $p \in M$ then $p$ is an adherent point to the set $\{ x_1, x_2, ..., x_n, ... \}$.

• We then proved that if $S \subseteq M$ and $p$ is an adherent point of $S$ then there exists a sequence $(x_n)_{n=1}^{\infty}$ contained in $S$ that converges to $p$.

• On the Cauchy Sequences in Metric Spaces page we said that sequence $(x_n)_{n=1}^{\infty}$ is a Cauchy Sequence if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then $d(x_m, x_n) < \epsilon$ - in other words, given any $\epsilon > 0$ we can find a tail to the sequence $(x_n)_{n=1}^{\infty}$ such that the distance between any two points in this tail of the sequence is less than $\epsilon$.

• We immediately proved that every convergent sequence is a Cauchy sequence. Of course, the converse is not necessarily true - not all Cauchy sequences are convergent.

• Regardless of whether a Cauchy sequence is convergent or not, on The Boundedness of Cauchy Sequences in Metric Spaces we saw that every Cauchy sequence is bounded by using a similar proof to that of showing that every convergent sequence is bounded.

• On the Complete Metric Spaces page we defined a special type of metric space regarding Cauchy sequences. We said that the metric space $(M, d)$ is Complete if every Cauchy sequence in $M$ converges in $M$. We saw that if $M = [0, 1)$ and $d$ is the usual metric on $\mathbb{R}$ then $(M, d)$ is not a complete metric space because the sequence $\left ( 1 - \frac{1}{n} \right )_{n=1}^{\infty}$ is contained in $[0, 1)$ but converges to $1 \not \in [0, 1)$!