# Is there a quicker, nicer way to show that the union of compact sets is not necessarily compact?

Working on this real analysis problem set, and the problem is:

Is every union of compact sets compact?

I decided to try to construct a counterexample, but it seems convoluted and hard to follow (maybe not). Here's my answer, which I intend to use as my answer, but I'm wondering if there is a way I can answer this using the properties of compactness rather than constructing an explicit counterexample.

Let $S_n=[n,n+1]$. Then for any $i\in\mathbb{N}$, $S_i$ is compact (this is a theorem so I'm good here). Now suppose $S=\bigcup_{i=1}^\infty S_i$ is compact and let ${G_n}$ be the open interval $(0,2^n)$. Then $\bigcup_{i=1}^{\infty}G_i$ is an open cover of $S$ (is this obvious? Should I try to show this explicitly?), so there exist finitely many indices $\{\alpha_1,...,\alpha_n\}$ such that $G=\bigcup_{i=1}^nG_{\alpha_i}\supseteq S$. Since there are finitely many indices $\{\alpha _i\}$, there exists some index $\alpha_i^*$ with the greatest least upper bound $2^{\alpha_i^*}$. Now let $x=2^{\alpha_i^*}+1$. Then $x\in S_x\subset S$ but $x\notin G$. But under the hypothesis that $S$ is compact, $G\supseteq S$. Hence $S$ is not compact. $\square$

So yeah, it seems convoluted and I'm not sure if it's even airtight—there might be some blanks that I need to fill in. Is there a faster way?

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A subset of $\mathbb{R}$ is compact iff it is closed and bounded. That is a standard theorem. If you do not have it yet, then you need to proceed from the definition. – André Nicolas Oct 18 '12 at 6:55
Why not take $S_n = \{ n \}$? Then $S$ is just an infinite discrete set, which is obviously not compact. – Zhen Lin Oct 18 '12 at 6:58

## 3 Answers

Every point set is compact. Every set is a union of points. If the union of compact sets is compact, every set is compact. Give any example of a noncompact set you have seen in class.

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wow. That's all I wanted to say, but it won't let me just say "wow.". – crf Oct 18 '12 at 7:10
wow.${}{}{}{}{}$ – Michael Greinecker Oct 18 '12 at 10:06
Now that is truly elegant. – Brian M. Scott Oct 18 '12 at 12:26
I also liked this answer! Are points always compact, in any topological space? It seems so since given an open cover one could just select any one of the open set in the cover, and cover the given point. – coffeemath Oct 18 '12 at 13:43

Pay attention: one thing is the union of a finite number of sets, another thing the union of an infinite number of sets. The union of an infinite number (countable or more) of compact sets might be non compact, as the previous answer shows. On the other hand, the union of a finite number of compact sets, is compact (the finite subcover being just the union of the finite subcovers of the single sets)

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The union of the compact intervals $[\frac{1}{n},1]$ is the non-compact interval $(0,1]$.

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