# Is every metric space compact? [duplicate]

This question already has an answer here:

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

• 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 – J. W. Tanner May 15 at 2:52
• The real line isn't compact. What is a finite subcover of $\{(-n,n)|n\in\mathbb{Z}^+\}$? – saulspatz May 15 at 2:53
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.