I understand the proof but I want to confirm one. So in discrete metric space, every Cauchy sequence is constant sequence and that way every Cauchy sequence is convergent sequence. Thus we conclude the discrete metric space is complete. Am I understanding correctly?
[Math] Prove that discrete metric space is complete
metric-spacessequences-and-series
Related Solutions
Example -1: Any set endowed with the discrete metric is complete: every Cauchy sequence is eventually constant, hence convergent.
Example 0: A subset $Y$ of a complete metric space $(X,d)$ is complete with the inherited metric if and only if it is closed.
Example 1: The real numbers $\mathbb{R}$ with $d(x,y) = |x-y|$. (Some people regard using $\mathbb{R}$ as an early example of a metric space to be circular; I am not one of them.)
Example 2: Any compact metric space. (More generally, one has the characterization of compact metric spaces as those which are complete and totally bounded.)
Example 3: a) For any positive integer $n$, if $(X_1,d_1),...,(X_n,d_n)$ are complete metric spaces, and we endow the Cartesian product $X = \prod_{i=1}^n X_i$ with any of several reasonable metrics -- e.g. $d(x,y) = \max_{1 \leq i \leq n} d(x_i,y_i)$ -- then $(X,d)$ is a complete metric space.
b) If $\{(X_n,d_n)\}_{n=1}^{\infty}$ is a sequence of complete metric spaces, and we endow $X = \prod_{i=1}^{\infty} X_i$ with the metric $d(x,y) = \sum_{i=1}^{\infty} \frac{1}{2^i} \frac{ d_i(x_i,y_i)}{1+d_i(x_i,y_i)}$, then $(X,d)$ is a complete metric space.
Example 4: For any metric space $X$, let $C_b(X)$ be the set of bounded, continuous functions $f: X \rightarrow \mathbb{R}$, endowed with the metric $d(f,g) = \sup_{x \in X} |f(x) - g(x)|$. This is a complete metric space and indeed a Banach space.
Example 5: The completion of any metric space. For instance, completing the rational numbers with respect to the $p$-adic metric one gets the field $\mathbb{Q}_p$ of p-adic numbers.
I thought about taking seriously the idea of formalizing "no holes" as a definition of a complete metric space. Here is what I came up with:
Proposition: For a metric space $(X,d)$, the following are equivalent:
(i) For any isometric embedding $\iota: (X,d) \rightarrow (Y,d)$ of $X$ into another metric space $Y$ and any sequence $\{x_n\}$ in $X$, if $\iota(x_n)$ converges in $Y$ then
$x_n$ converges in $X$.
(ii) $X$ is complete.
Proof: The basic observations here are that if $\iota: (X,d) \rightarrow (Y,d)$ is an isometric embedding and $\{x_n\}$ is a sequence in $X$, then:
$\bullet$ $\{x_n\}$ is Cauchy iff $\{\iota(x_n)\}$ is Cauchy, hence also
$\bullet$ if $\{ \iota(x_n)\}$ is convergent, then $\{x_n\}$ is Cauchy.
Then (ii) $\implies$ (i) is immediate; to show (i) $\implies$ (ii) look at the completion $\iota: X \rightarrow \tilde{X}$ of $X$.
Thus the "holes" in $X$ are detected by embeddings into larger spaces. I am skeptical though that this definition would be helpful for beginning students: aside from relying on the existence of the completion of a metric space, the idea of considering all possible embeddings of one metric space into another seems relatively abstract and sophisticated.
You're probably confused about what you have to prove.
The task is proving that every Cauchy sequence converges. So, let $(x_n)$ be a Cauchy sequence. By definition of Cauchy sequence, there exists $N$ such that, for $m,n\ge N$, $d(x_m,x_n)<1/2$.
Since the metric is discrete, this implies that
for every $m,n\ge N$, $x_m=x_n$.
Therefore the sequence is eventually constant, hence convergent. Indeed, given $\varepsilon>0$, we have that, for every $n\ge N$, $d(x_n,x_N)=0<\varepsilon$.
Why $1/2$? Because it's good for the proof. Any positive number less than $1$ would have done as well.
Your confusion possibly comes from the dual usage of $\varepsilon$. Just not mentioning it in the first part should be sufficient for clearing up the matter.
Best Answer
If $x_n$ is a cauchy sequence then, for every $\epsilon>0$ exists an $N \in \mathbb{N}$ such that if $n,m$ are greater than $N$ you have $d(x_n,x_m)<\epsilon$. Now take $\epsilon=1/2$ then, it exists an $N$ such that $d(x_n,x_m)<1/2$ because d is the discrete metric, this is only possible if $x_n$ is a constant for $n$ greater than $N$