In a metric space $(M,d)$, any Cauchy sequence $\{a_n\}_{n \in \mathbb{N}}$ in $M$ is convergent?
This is a question I was given. However, it does not specify whether the metric space is "complete?" Does that matter? In the set of rational numbers, for example, it is not true that every cauchy sequence is convergent. It is true for the real numbers. Is it not that the rational numbers is an incomplete metric space and the real numbers a complete one?
Best Answer
A metric space $M$ is called complete if every convergent sequence $(x_n)$ converges to a limit $x$ such that $x \in M.$ Otherwise, $M$ is incomplete. The rational numbers $\mathbb{Q}$ is an incomplete metric. Consider, for example, the sequence $(x_n)$ whose $n^{th}$ term is the approximation of $\sqrt{2}$ such that the first $n$ decimal places of $\sqrt2$ are kept and the rest discarded. The sequence converges to $\sqrt{2},$ yet $\sqrt{2} \not\in \mathbb{Q}.$ Now, a Cauchy sequence is convergent if and only if the metric space is complete. Considering the sequence $(x_n)$ just defined, $(x_n) \subset \mathbb{R}$ is Cauchy and convergent. However, $(x_n) \subset \mathbb{Q}$ is Cauchy but not convergent, for the reason that $L_{(x_n)} = \sqrt2$ and $\sqrt2 \not\in \mathbb{Q}.$
It is simple to show that if for all $\epsilon > 0,$ $$d(x_n - x_m) < \epsilon \quad \textrm{for all $n,m \geq N'$},$$ then for all $\epsilon > 0,$ $$d(x_n, L) < \epsilon \quad \textrm{for all $n \geq N'$}.$$ Simply use the nested interval property, which says that for any sequence of nested intervals, $I_1 \supset I_2 \supset \cdots,$ there exists a unique element $L \in \bigcap I_n.$ It is the case that for all $\epsilon >0,$ $L \in (x_n - \epsilon, x_n + \epsilon)$ for all $n \geq N'.$ You can also use the fact that every Cauchy sequence is bounded and every bounded sequence has a convergent subseqence $(x_{n_k}) \rightarrow x.$ It will follow from the triangle inequality that $(x_n) \rightarrow x.$