I know this is usually done by contradiction but I'm trying out something a bit different:

Let $\mathbb{X}$ be a sequentially compact metric space. Let $(s_n)$ be a sequence in the metric space such that $d(s_n, s_{n+1}) > 1$. Let $(r_n) \subset (s_n)$. Then sequential compactness $\implies$ $\lim_{n\to\infty} (r_n) \in \mathbb{X}$ which implies that the limit exists. But it can only exist if $\mathbb{X}$ is bounded.

  1. Is the essence of the proof salvageable or completely wrong?
  2. Do I need to prove the existence of $(s_n)$?
  3. How do I turn it into a more mathematically rigorous proof?
  • $\begingroup$ Note that the sequence $(1,-1,1,-1,1,-1,\dots)$ satisfies $d(s_n,s_{n+1})>1$, but it is bounded. You want to make use of a sequence whose existence essentially relies on the unboundedness of $X$. Maybe use the condition $d(s_n,s_k)\ge1$ for all $k<n$. $\endgroup$ Mar 14, 2016 at 13:58
  • $\begingroup$ In what you write you say that an arbitrary subsequence of $(s_n)$ (which could be $(s_n)$ itself!) converges. This is false. $\endgroup$ Mar 14, 2016 at 13:58
  • $\begingroup$ You haven't shown how the convergence of the subsequence $(r_n)_n$ leads to the boundedness of $X$. $\endgroup$ Mar 14, 2016 at 14:32

1 Answer 1


To show that a sequentially compact metric space is bounded, show that an unbounded metric space $X$ is not sequentially compact, as follows:

When $\phi\ne F\subset X$ and $F$ is finite, take $p\in F$, and let $M=\max \{d(x,y):x,y\in F\} .$ Then $B_d(p,1+M)\ne X$ because $X$ is unbounded. So there exists $q\in X \backslash B_d(p,1+M).$

By def'n of $M,$ we have $F\subset B_d(p,1+M)$ so $q\not \in F.$ By the triangle equality, for any $r\in F$ we have $$d(r,q)\geq d(p,q)-d(p,r)>(1+M)-d(p,r)\geq (1+M)-M=1.$$

Using this, we can take $\{p_0\}=F_0\subset X$ and for each $n\in N,$ find $p_{n+1}\in X$ with $p_{n+1}\not \in F_n,$ such that $d(p_{n+1},p)>1$ for every $p\in F_n.$

Consider the sequence $(p_n)_{n\in N}.$ For $n\ne m$ we have $d(p_n,p_m)>1.$ So no subsequence of it is a $d$-Cauchy sequence so $X$ is not sequentially compact.

Remark: For metrizable spaces, sequential compactness is equivalent to compactness


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