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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\text{Im}(f),$$d(0,k)=1<2$$\implies k\in B(0,2)\implies\text{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?

hbghlyj
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Sriti Mallick
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    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. – Alex Youcis Aug 02 '13 at 03:30
  • @AlexYoucis I thought "sequential compactness" was "every sequence has a convergent subsequence". The real line is not compact but it does have the property that every bounded sequence has a convergent subsequence. – user14111 Dec 24 '23 at 02:52
  • You don't have to go so far to find a counterexample. Consider the space $\mathbb R\setminus{0}$ with the usual metric, and the bounded sequence $1,1/2,1/3,1/4,\dots$. – user14111 Dec 24 '23 at 02:55

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

Cameron Buie
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