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I've been having a hard time solving this problem that I was given in class. The problem states " Give an example of an infinite open cover of the interval (0,1) that has no finite subcover."

I know that the set has to be compact and that both 0 and 1 are limit points. Aside from those two known factors I'm at a complete loss as to how to go about solving this problem.

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    $\begingroup$ What set has to be compact? $\endgroup$
    – user64687
    Feb 4, 2015 at 17:06
  • $\begingroup$ A set that is closed an is a subset of a compact metric space. $\endgroup$
    – BesMath
    Mar 27, 2020 at 15:34

5 Answers 5

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Actually, if you haven't already learned this, you will learn that in $\Bbb R^{n}$, a set is compact if any only if it is both closed and bounded.

The set $(0,1)$ is bounded, but it is not closed, so it can't be compact.

The solution to your exercise of finding an open cover with no finite subcover proves that $(0,1)$ is not compact, because the definition of a set being compact is that every open cover of the set has a finite subcover.

So, the whole trick to finding an open cover with no finite subcover is this:

$(0,1)$ is not closed, and so it doesn't contain all of its limit points (it's easy to see that the two it doesn't contain are $0$ and $1$). Well, can you construct an open cover whose open sets get closer and closer and closer to at least one (or maybe both) of these end points, but never quite reaches them? Hint: since we are in $\Bbb R$, think about how you would get closer and closer to a real number $r$ without ever actually being $r$. If you want to get closer from above, then $\{r + \frac{1}{n} \}_{n = 1}^{\infty}$ is a sequence that gets closer to $r$ from above. What would be a sequence that gets close to $r$ from below? You should hopefully say $\{r - \frac{1}{n} \}_{n = 1}^{\infty}$.

Anyway, so maybe we can construct our open intervals so that the end points get closer and closer to the limit points $0$ and $1$, but finitely many of the sets would always have gaps between them and $0$ and $1$... hmm...

Well, we could do $(0 + \frac{1}{3}, 1 - \frac{1}{3}) \cup (0 + \frac{1}{4}, 1 - \frac{1}{4}) \cup (0 + \frac{1}{5}, 1 - \frac{1}{5}) \cup \ldots$.

These open sets are in $(0,1)$ and have end points getting closer and closer to $0$ and $1$, but any finite number of them won't contain all of $(0,1)$ (why?). If we use a formula to define these open intervals, it would be $\{ (0 + \frac{1}{n}, 1 - \frac{1}{n}) \}_{n = 3}^{\infty}$.

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    $\begingroup$ Of the four answers so far, this is the only one that answers the question of “how to go about solving this problem”. $\endgroup$
    – MJD
    Feb 4, 2015 at 17:48
  • $\begingroup$ @user46944 thank you so much for explaining this. He said that this one would make us think for a while. I assume he said this because we haven't fully touched upon this yet. Thanks for the indepth and comprehensive answer $\endgroup$ Feb 4, 2015 at 18:59
  • $\begingroup$ @ChrisMillet You're welcome! I hope it makes sense. $\endgroup$
    – layman
    Feb 4, 2015 at 20:33
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    $\begingroup$ This answer really cleared a lot of doubts +1 $\endgroup$
    – Siddhartha
    Sep 17, 2019 at 14:59
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    $\begingroup$ WOW, thank you. such a clear great explanation. I really appreciate. $\endgroup$
    – BesMath
    Mar 27, 2020 at 15:47
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Think about... $$ \bigcup_{n=1}^{\infty}\left(0+\frac{1}{n},1\right) $$

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  • $\begingroup$ would the open cover $\{(0+\frac{1}{n},\frac{n+1}{n})\}_{n \in \Bbb{Z}^+}$ work? $\endgroup$
    – anonymous
    Dec 22, 2022 at 0:04
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Since you have the real analysis tag, I'm assuming you are using the standard topology on $\Bbb{R}$, so open sets look like intervals of the form $(a,b)$. An open cover of $(0,1)$ could look something like $$\bigcup_{i=1}^\infty \left(0,1-\frac{1}{i} \right)$$ But you need to verify this is an open cover. That is, show $$(0,1) \subseteq \bigcup_{i=1}^\infty \left(0,1-\frac{1}{i} \right)$$ and further show that if $K \subset \Bbb{N}$ is any finite subset of $\Bbb{N}$ that $$(0,1) \not\subseteq \bigcup_{i\in K} \left(0,1-\frac{1}{i} \right)$$ which demonstrates that the open cover has no finite subcover.

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  • $\begingroup$ gotcha this makes sense too. Thanks a bunch $\endgroup$ Feb 4, 2015 at 19:00
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You can take the open cover $\{(0+\frac{1}{n}, 1-\frac{1}{n})\mid n>2\} $. It has no finite subcover because if it does simply take the largest $n$ and $\frac{1}{2n}$ is not included.

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  • $\begingroup$ I never thought of it like that. I assume it's because we havent fully finished the topic yet and I wasn't sure whether or not to keep reading thanks! $\endgroup$ Feb 4, 2015 at 19:03
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In case you are still interested. Abbot's has this problem as Example 3.3.7. For each point $x\in(0,1)$ consider the open interval $0_x = (x/2,1)$. The infinite union of $O_x$'s forms an open cover of $(0,1)$. Take any finite subset of such $O_x$, and set $x'=\operatorname{min}(x_1,x_2,\dots,x_n)$. You can still find a number $0<y\leq x'/2$ not in any subfinite union.

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