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Proof: Let $\mathscr{U}$ := {$U_n$}$_{n \geq 1}$ = {$(n-1,n+1)$}$_{n \geq 1}$ be an open cover of $\mathbb{R}$. Consider finitely many of them, say: $U_{n_1} \ldots U_{n_k}$. Setting $N := \max(n_1 \ldots n_k)$, we get that: $U_{n_i} \subset U_N \implies \bigcup_{i=1}^{k} U_{n_i} = U_N = (N-1,N+1) \subsetneq \mathbb{R}$.

Hence, $\mathscr{U}$ cannot be reduced to a finite subcover of $\mathbb{R}$ so that it is not compact. QED.

How did I do? Thanks.

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    $\begingroup$ You did good. (this text added to satisfy comment length requirement) $\endgroup$ Dec 9, 2016 at 3:45
  • $\begingroup$ Thank you so much for taking a look and providing me with feedback!! $\endgroup$
    – Javier
    Dec 9, 2016 at 3:47
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    $\begingroup$ Yep, good job. All you needed to do was come up with some open cover that has no finite subcover. This is one of the two main nonexamples for real numbers. The Heine-Borel theorem says that a closed and bounded subset of $\mathbb{R}^n$ is compact, and the reason we need "boundedness" is due to the type fo open cover you came up with. $\endgroup$
    – stochasm
    Dec 9, 2016 at 3:53
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    $\begingroup$ With the sets $U_n $ you have not an open cover of R ! $\endgroup$
    – Fred
    Dec 9, 2016 at 4:22
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    $\begingroup$ You probably have $\mathscr{U}=\{(-n-1,n+1)\}$ $\endgroup$
    – egreg
    Dec 9, 2016 at 11:03

1 Answer 1

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As pointed out in the comments by Fred and Levent, $\mathcal U$ is not an open cover of $\mathbb R$ since $\bigcup_{n\geqslant 1}(n-1,n+1)=(-1,+\infty)$. However, this can be fixed easily: define instead $U_n:=(-n,n)$. Then $\bigcup_{n\geqslant 1}U_n=\mathbb R$ and a finite subcover is contained in an interval of the form $(-R,R)$.

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  • $\begingroup$ Ahh I see my error now. Thank you so much for pointing this out to me!! $\endgroup$
    – Javier
    Dec 9, 2016 at 15:52
  • $\begingroup$ According to Wikipedia ,(n-1,n+1) for n$\in$ Z cover R. How about the fact that R is uncountable? $\endgroup$
    – user960654
    Dec 1, 2021 at 15:55

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