Let $X$ be the collection of all sequences of positive integers. If $x=(n_j)_{j=1}^\infty$ and $y=(m_j)_{j=1}^\infty$ are two elements of $X$, set
$$k(x,y)=\inf\{j:n_j\neq m_j\}$$ and
$$d(x,y)= \begin{cases} 0 & \text{if $x=y$} \\ \frac{1}{k(x,y)} & \text{if $x \neq y$} \end{cases}$$
We know that $d$ is a metric on $X$. Now, I must prove that the metric space $(X,d)$ is complete.
By definition, a metric space $(X,d)$ is complete if every Cauchy sequence in $X$ is convergent. On the real line, this is trivial, but I have trouble applying the concept to metric spaces. We know that a sequence is Cauchy if: $$(\forall \epsilon>0)(\exists N>0)(\forall m,n \geq N)(d(x_m,x_n)<\epsilon)$$ A Cauchy sequence also has the following properties:
- If a sequence $(x_n)$ converges, then it is Cauchy.
- Every Cauchy sequence is bounded: for all $a \in X$, there exists $C_a>0$ such that $d(a,x_n)<C_a$ for all $n$.
- If a Cauchy sequence $(x_n)$ has a converging subsequence $(x_{n_k})$ such that $\lim_{k \to \infty}x_{n_k}=x$, then $(x_n)$ converges to $x$.
Here is my attempt thus far, although my reasoning feels wrong. Let $(x_n)_{n=1}^\infty$ be any Cauchy sequence in $X$. Take $\epsilon=\frac12$. Let $N$ be such that for $n,m \geq N$, we have $d(x_n,x_m)<\frac12$. So, $x_n=x_m$ for all $n,m \geq N$ and $x_n=x_N$ for $n \geq N$. Thus, $x_n$ is eventually constant and hence convergent, which proves that the metric space is complete.
Any corrections or help on how to prove this would be appreciated. Thank you!