Sequences and Limits in Metric Spaces Review
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 $1 \not \in [0, 1)$!