# 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?

• 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

• wow.${}{}{}{}{}$ – Michael Greinecker Oct 18 '12 at 10:06
The union of the compact intervals $[\frac{1}{n},1]$ is the non-compact interval $(0,1]$.