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I am a student majoring engineering.

I am studying real analysis with textbook 'Measure and Integral' by Wheeden and Zygmund.

This book defined compact like the following:

$E$ is compact if every open cover of $E$ has a finite subcover.

By the definition $[0, 1]$ is not compact.

However, by the Heine-Borel theorem, $[0, 1]$ is compact.

Let me prove why $[0, 1]$ is not compact by the definition.

According to the definition, it is enough to show an open cover of $E$ having infinite subcover.

If $C=\{U_\alpha:\alpha\in \mathbb{N}\}$ is an indexed family of sets $\displaystyle U_\alpha=\left(-1+\frac{1}{n}, 2\right)$, then $C$ is a cover of $[0, 1]$ because $\displaystyle [0,1] \subseteq \bigcup\limits_{\alpha \in \mathbb{N}} {\mathop U\nolimits_\alpha }$.

This $C$ has infinite subcovers like $C=\{U_{2\alpha}:\alpha\in \mathbb{N}\}$

Can someone teach me what is my fault?

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    $\begingroup$ It is only sufficient to show that you can find a covering that admits no finite subcovering. $\endgroup$ – Danu Mar 10 '16 at 12:12
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    $\begingroup$ There exists a finite subcover, say $[0,1]\subseteq U_2$. (This existence is the definition in your book.) $\endgroup$ – Did Mar 10 '16 at 12:12
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    $\begingroup$ As a now-deleted answer said, you are trying to prove that every subcover is finite, but that is not what you need to prove. You only need that some subcover is finite. $\endgroup$ – Carl Mummert Mar 10 '16 at 12:15
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    $\begingroup$ Your question was already answered fully and I can't really add anything. But I just want to say I like your thinking. I like that at the end, you ask "where is my fault". It shows the kind of humble curiosity that acchieves much. It's always better to assume YOU are wrong, and not some 100-or-more year old proof. $\endgroup$ – 5xum Mar 10 '16 at 12:47
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    $\begingroup$ @5xum thank you for your good advice. $\endgroup$ – Danny_Kim Mar 10 '16 at 12:50
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By the definition it's compact. You misunderstood and the proof is incorrect.

You should try to prove that if you take an arbitrary infinite cover of E, you can find within that cover a finite sub-cover. What you did was: you picked one particular infinite cover of E and then you showed that it has an infinite sub-cover. That doesn't do the work here.

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  • $\begingroup$ Is this wrong? I got downvoted 1-2 times so I am asking if it's wrong. Let me know, if wrong I will delete it. It's been a while since I've played with these concepts. $\endgroup$ – peter.petrov Mar 10 '16 at 12:21
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    $\begingroup$ Ah... I totally understand what is wrong. Thank you. Before reading answer, I had understood that all open cover should have finite number of sets. My bad English so frequently makes me misunderstand some notions. $\endgroup$ – Danny_Kim Mar 10 '16 at 12:34
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    $\begingroup$ @peter.petrov The anwer is otherwise perfect but I feel your emphasis is in the wrong place. The problem is not with "one particular infinite cover of $E$", it's with "it has an infinite sub-cover". (When showing a counterexample to the claim of the definition, finding a particular infinite cover of $E$ is what you should do, but then you should proceed show that there are no finite subcovers.) $\endgroup$ – JiK Mar 10 '16 at 20:37
  • $\begingroup$ @peter.petrov Your comment "you should try to prove that ... finite sub-cover" is probably the source of the downvotes. He's "trying" to prove it's not compact, and so these steps are not what he should "try to prove". What he should be trying to do is take an infinite cover and show that it has no finite subcover. Of course, this will fail. $\endgroup$ – Joel Mar 11 '16 at 3:19
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Your "proof" is wrong.

The definition says that for every cover there exists a finite subcover. In your post there is an example of an infinite subcover of $\{U_\alpha\}$.

By following your reasoning every subset $U$ of a topological space would be non-compact: pick a "redundant" infinite open cover of $U$ (you can always find it) and remove one set. You have an infinite subcover.

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  • $\begingroup$ Thank you for answering. I misunderstood the definition. $\endgroup$ – Danny_Kim Mar 10 '16 at 12:36
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As said in the other answers, you misunderstood a point of the definition.

It is actually tricky to prove that $[0,1]$ is compact. You have to consider an arbitrary cover of $[0,1]$, let's call it $\mathcal{U}$. Then, consider the set $A=\{x\in [0,1]$ such that $\mathcal{U}$ has a finite subcover covering $[0,x]\}$. The proof itself is done in three steps: First, prove that $A$ is non empty. Then prove that sup$A\in A$. And finally, prove that sup$A=1$. Maybe it doesn't look so hard, but the third step is not trivial.

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To show a set is not compact, you have to exhibit an open cover where every subcover is infinite (not just one). Alternatively, you have to exhibit an open cover where every finite subset fails to be a cover (so every subcover, if it exists, must by infinite).

So, for example, $(0,1)$ is not compact, because it has the infinite cover $\{ (1/n, 1) : n \in \mathbb{N}\}$; but if I pick any finite subset of that cover $$\{(1/n_{0}, 1), (1/n_{1}, 1), \ldots, (1/n_{m}, 1)\},$$ and I let $$m = \sup\{n_{0}, n_{1}, \ldots, n_{m}\},$$ then $$\bigcup \{(1/n_{0}, 1), (1/n_{1}, 1), \ldots, (1/n_{m}, 1)\} = (1/m, 1)$$ which does not include the whole set $(0,1)$, so my finite set is not a subcover.

To prove $[0,1]$ is not compact, you would have to show similarly that every finite subset of your cover is not a subcover; but since your cover does have (some) finite subcovers (as well as infinite ones), you can't do that.

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