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:

  1. If a sequence $(x_n)$ converges, then it is Cauchy.
  2. 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 $n$.
  3. 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!

  • 1
    $\begingroup$ There is something not right with your argument. Let $x_m$ be the sequence having all $1$ except for the $m$th spot in which it has $2$. Then for any $m \neq n$, $d(x_n,x_m) = 1/\min(m,n)$ so it is certainly Cauchy (and it does converge to the all ones sequence), but notice that $x_m \neq x_n$ ever. $\endgroup$
    – Suugaku
    Feb 21, 2015 at 23:09

1 Answer 1


You must use the full power of the $\forall\epsilon>0$ in the definition of being Cauchy. So, take $\epsilon=\frac{1}{n}$. Then, for $r,s>N(n)$, $d(x_r,x_s)<\frac{1}{n}$. This implies that $x_r,x_s$ are equal for all terms up to $n$. Define $x$ to have the term $x(n)$ equal to the $n$-th common entry of any such $x_r, x_s$.

Now you can show that the sequence (of sequences) $x_r$ converges to $x$. In fact, for $\epsilon>0$ there is $n$ such that $\epsilon>1/n$ then take $N(\epsilon)$ from the definition of Cauchy and we see that if $r>N(\epsilon)$ then $x_r$ and $x$ coincide up to the $n$-th term. Therefore their distance is smaller than $1/n<\epsilon$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.