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 $k<n$. $\endgroup$ – Stefan Hamcke Mar 14 '16 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. $\endgroup$ – Friedrich Philipp Mar 14 '16 at 13:58
  • $\begingroup$ You haven't shown how the convergence of the subsequence $(r_n)_n$ leads to the boundedness of $X$. $\endgroup$ – DanielWainfleet Mar 14 '16 at 14:32

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


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.