Prove or disprove:
$X$ is compact if and only if every closed ball in $X$ is compact.
$X$ is compact if and only if every closed ball in $X$ is compact?
2
$\begingroup$
real-analysis
compactness
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2$\Bbb R$ is not compact.. – 2017-01-07
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2If there is a closed ball B that is not compact than there is an open cover of B with no finite subcover. If you add the complement of B to that open cover it is an open cover of X which has no finite subcover. So X is not be compact.So X compact $\implies$ alll closed balls are compact.The converse is not true if X is not bounded as $\mathbb R $ is not compact as Open Ball pointed out. (But if X is bounded, you can cover it in a finite number of closed balls and every open cover will have a finite subcover covering each ball and the union of those finite covers will be finite cover of X) – 2017-01-07
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0@fleablood Slightly more efficient version of the parenthetical comment at the end of your answer: If $X$ is bounded then $X$ **is** a closed ball; just take a ball whose radius exceeds the diameter of $X$. – 2017-10-06
1 Answers
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Lets try with the first example we can think of: $\mathbb R$
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1So I just say every closed ball in $\mathbb{R}$ is closed and bounded, so it is compact. But $\mathbb{R}$ is not compact? – 2017-01-07
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1$\Bbb R=\bigcup_{n\in\Bbb Z}(n-0.6,n+0.6)$ but if you remove a single set from this cover, the union is not $\Bbb R$ anymore. – 2017-01-07
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1Yeah, exactly right OP – 2017-01-07