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This question comes from Advanced Calculus second edition by Patrick M. Fitzpatrick.

For each natural number $n$, let $I_n$ be a closed bounded interval. Suppose that $\{I_n\}_{n=1}^\infty$ covers the compact set consisting of the closed bounded interval $[0,1]$. Is it true that this cover has a finite subcover?

My intuition is telling me that there must be a finite subcover since I have been unable to think of a counterexample. I am working with the understanding that for subsets of the reals that compact, sequentially compact, and closed and bounded imply one another so there is quite a bit to work with.

Walt
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1 Answers1

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Take $I_1=[0,0]=\{0\}$ and $I_n=\left[\frac{1}{n},\frac{1}{n-1}\right]$ for $n\ge 2$. Then $\cup I_n=[0,1]$ and obviously $[0,1]$ does not have a finite cover.

Momo
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