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From the definition of a compact set, a set $A$ is said to be compact if for every open cover of $A$, there exists a finite subcover such that $A$ is a subset of this finite subcover.

Here if we consider the open cover, is it necessary for all the elements (which are sets) of this open cover to be of finite?

Also should the finite subcover have elements which are finite?

If we consider the subcover to have the set $(-\infty, +\infty)$ as one of its sets, then we have that all sets of $R$ compact.

But while proving theorems related to compact sets (from Rudin), this property of finiteness of the sets is never used. So should the elements of the finite open cover be finite or that we should not include the set $(-\infty, +\infty)$ in specific.

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    $\begingroup$ To clarify this a little more than Thomas already has: by "finite", the definition simply means that there are a finite number of sets in the reduced cover, not that the sets themselves are finite (though the condition you describe is called "bounded", not "finite"). $\endgroup$ – Paul Sinclair Sep 19 '15 at 5:49
  • $\begingroup$ To clarify, what finite covering means: that a finite number of sets $V_1$, $V_2$, ..., $V_n$ that covers the set. Of course you are not required to have finite number of sets in the definition of compactness (or it would be trivial to find a finite covering). The definition requires this to be true no matter which open covering you're using. Next the concept of infinite set is not the same as bounded set. Infinite means that there are infinite number of elements (which is true for every open interval in $\mathbb R$) which doesn't require unboundedness. $\endgroup$ – skyking Sep 19 '15 at 7:34
  • $\begingroup$ Not finite set like $\{-1,0,7,23\}$, also not bounded set like $(-25,16)$, no no, but finite... COVER like $\{(-\infty,-2),[-3,4),(1,5],(2,\infty)\}$. $\endgroup$ – C-Star-W-Star Sep 19 '15 at 12:33
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No, finiteness of the individual sets is not requested nor is it useful.

$(-\infty,\infty)$ is one special cover of $\mathbb{R}$. The definition of compactness however demands that every open cover has a finite subcover. if you take, e.g., the cover of $\mathbb{R}$ consisting of the intervals $(n-1, n+1)$ with $n\in \mathbb{Z}$, then you will not find a finite subcover.

(The general definition of compact set you are citing takes some getting used to. For the beginning and for subsets of $\mathbb{R}$ or $\mathbb{R}^n$ the property which is easiest to grasp is that a set is compact if and only if it is bounded and closed).

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  • $\begingroup$ That makes it clear! Thank you. $\endgroup$ – Almond Pie Sep 19 '15 at 5:58
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As you say, a set is called compact if every open cover admits a finite subcover. More specificly, given any collection of open sets (no further restrictions) such that $A$ is contained in the union of them, you should be able to find finitely many of these open sets such that $A$ is still contained in the union of these finitely many open sets.

In case of the real numbers, $(-\infty, +\infty)$ is one (finite) open cover for any subset $A$ of $\mathbb{R}$, but in order to conclude that $A$ is compact, you should have that all open covers admit a finite subcover. Example: the open interval $(0,1)$ can be covered by the intervals $(1/n,1)$, $n\in\mathbb{N}$, but this cover does not have a finite subcover. So although you can find finite open covers easily, the condition for compactness is in this case not satisfied.

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