Justify: In a metric space every bounded sequence has a convergent subsequence.

  • My Attempt: False: Consider the metric space $(X,d)$ where $X=\mathbb R$ and $d$ is the discrete metric on $X.$ Consider the sequence $f=\{n\mapsto n\}_{n\in\mathbb N}$ in $X.$ Then $f$ is bounded since for $k\in\Im(f),$$d(0,k)=1<2$$\implies k\in B(0,2)\implies\Im(f)\subset B(0,2).$

    Let $g=\{x_{r_n}\}$ be a subsequence of $f$ and $p\in\mathbb R.$ From the definition of subsequence $g=f\sigma$ for some strictly increasing $\sigma:\mathbb N\to\mathbb N.$ To show $\{x_{r_n}\}$ doesn't converge to $p.$ If not, $\{p\}$ being a open set containing $p,~g(n)=p~\forall~n\ge k$ (for some $k\in\mathbb N$) i.e. in particular $f(\sigma(k))=f(\sigma(k+1))\implies \sigma(k)=\sigma(k+1),$ a contradiction since $\sigma$ is strictly incresing.

    Thus $f$ doesn't have any convergent subsequence.

Am I correct?

  • 1
    $\begingroup$ You'll find that this property is something called "sequential compactness". It is is equivalent to usual compactness (every open cover has a finite subcover) for metric spaces. So, the problem is asking whether every metric space is compact, and as you have shown, this is definitively not true. $\endgroup$ Aug 2 '13 at 3:30

+1: Nicely done!${}{}{}{}{}{}{}$


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.