I'm sure that this holds for $\mathbb{R}^n$ and for $L^p$ spaces. Is it true in general?
Is it true that a set is compact iff it is closed, bounded, and has finite measure?
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0Thanks! I understand what $L^p$-spaces are, but I am unaware of a notion of finite measure in them that guarantees compactness of closed and bounded sets. That's what I was curious about. – 2012-10-29
2 Answers
Here’s a counterexample. For $A\subseteq\Bbb N$ let $\mu(A)=\sum_{n\in A}2^{-n}$, and for $m,n\in\Bbb N$ let $d(m,n)=\begin{cases}1,&\text{if }m\ne n\\0,&\text{if }m=n\end{cases}\;.$
Then $\langle\Bbb N,d\rangle$ is a metric space of diameter $1$ with the discrete topology, so all subsets of $\Bbb N$ are bounded and closed. $\langle\Bbb N,\wp(\Bbb N),\mu\rangle$ is a measure space in which every measurable set has finite measure. Thus, every $A\subseteq\Bbb N$ is closed, bounded, and of finite measure, but only the finite sets are compact.
No, it's not true in general. Consider the following topology on $\mathbb{N}$: A set is open if it's of the form $\{n, n + 1, \dots\}$. All open sets are compact and unbounded in this topology. Furthermore, if you equip this space with the counting measure, you'll have compact sets with infinite measure.