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I am referring to Rudin's definition 2.32 of compactness here: A subset K of a metric space X is said to be compact if every open cover of K contains a finite subcover.

Obviously X is a subset of X itself. We also knew that any metric space is both open and closed relative to itself. Then {X} is an open cover of X that contains a finite subcover which is also {X}. It seems to me that {X} is the only open cover X has.

If it really is the case that X is compact, it would follow that X is closed and bounded. Closed indeed, but being bounded would be troublesome.

I am sure one of the above statement is false, but I don't know which one.


marked as duplicate by Randall, Community May 15 at 3:03

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    $\begingroup$ Let $X=\mathbb R$ and take, as open cover, intervals $(n-1,n+1), n\in\Bbb Z$; there are many open covers of $\mathbb R$, contrary to what you said $\endgroup$ – J. W. Tanner May 15 at 2:52
  • $\begingroup$ The real line isn't compact. What is a finite subcover of $\{(-n,n)|n\in\mathbb{Z}^+\}$? $\endgroup$ – saulspatz May 15 at 2:53
  • $\begingroup$ Thank you guys. This is kinda a silly question due to my lack of imagination. So it was the statement that "It seems to me that {X} is the only open cover X has" was false. $\endgroup$ – Caus May 15 at 2:59

The requirement is that any open cover have a finite sub cover. Consider $\Bbb R.$ The set of intervals $(n,n+2)$ for $n$ an integer covers $\Bbb R$ but there is no finite sub cover.


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