Let $(X, \mathcal{T})$ be a topological space. Suppose $\{ U_{\alpha} \}_{\alpha \in \Lambda}$ is a family of non-compact subsets of $X$. I want to argue that $\bigcup \limits_{\alpha \in \Lambda} U_{\alpha}$ is non-compact, but I don't know if this is true...

Here is what I thought the argument would be: Let $A_{\alpha}$ be an open cover for $U_{\alpha}$ with no finite subcover, for each $\alpha \in \Lambda$. Then $\bigcup \limits_{\alpha \in \Lambda} A_{\alpha}$ is an open cover of $\bigcup \limits_{\alpha \in \Lambda} U_{\alpha}$.

Suppose by contradiction that $\bigcup \limits_{\alpha \in \Lambda} U_{\alpha}$ is compact. Then there is a finite subcover of $\bigcup \limits_{\alpha \in \Lambda} A_{\alpha}$ (not necessarily finitely many $A_{\alpha}$, but finitely many sets inside of the $A_{\alpha}$'s). I want to say that this implies we have a finite subcover of $U_{\beta}$ for some $\beta \in \Lambda$, but I don't think this is necessarily true...

If the $U_{\alpha}$'s are pairwise disjoint, and their open covers are also pairwise disjoint from each other, then this clearly holds. But if there is overlap between the open covers, somehow that presents a problem.

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    $\begingroup$ $\bigcup_{x\in[0,1]}((1/2,1)\cup\{x\})=[0,1]$. $\endgroup$ – Pp.. Feb 5 '15 at 14:43
  • $\begingroup$ @Pp.. Doesn't your example cheat by including in the union a compact set? $\{x \}$ is compact. $\endgroup$ – layman Feb 5 '15 at 14:44
  • $\begingroup$ The $U_x=(1/2,1)\cup\{x\}$ are non-compact. $\endgroup$ – Pp.. Feb 5 '15 at 14:45
  • $\begingroup$ @Pp.. But we can rewrite this as a union with compact sets in it, as $(1/2, 1) \cup (\bigcup \limits_{x \in [0,1]} \{x\})$. So maybe my proposition is true if my union cannot be rewritten as a union with compact sets in it? $\endgroup$ – layman Feb 5 '15 at 14:46
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    $\begingroup$ You can always rewrite anything with a union of a compact set. A singleton is always compact. $\endgroup$ – Pp.. Feb 5 '15 at 14:46

Consider $\mathbb{R}$ with the Euclidean Topology.

Let $U_1=[0,1)$ and $U_2=(0,1]$.

They are obviousely non compact because thay are not closed but theirs union is $[0,1]$ that is compact.

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  • $\begingroup$ Ok, so how do I argue that an open subset of $\Bbb R^{n}$ is not compact? I wanted to say it is a union of open balls which are not compact, but clearly that argument is not true, since the union of non-compact sets can be compact. $\endgroup$ – layman Feb 5 '15 at 14:48
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    $\begingroup$ If you want (non-empty) open sets, then it is true. Arbitrary union of open sets is open while compact sets in $\mathbb{R}^n$ are closed. $\endgroup$ – Pp.. Feb 5 '15 at 14:49
  • $\begingroup$ @Pp.. So? Open and closed are not mutually exclusive (i.e., a set can be both open and closed). $\endgroup$ – layman Feb 5 '15 at 14:49
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    $\begingroup$ But $\mathbb{R}^n$ is connected... $\endgroup$ – Sabino Di Trani Feb 5 '15 at 14:50
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    $\begingroup$ @user46944 In $\mathbb{R}^n$ the only such sets are $\emptyset$ and $\mathbb{R}^n$. $\endgroup$ – Pp.. Feb 5 '15 at 14:51

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