Let's choose an open covering for $\left [ 0,1 \right ]$. For example $$\left \{ \left ( \frac 1 n,1-\frac 1 n \right ) \mid n\in \{ 3,4,\dots\} \right \}.$$ How can one choose a finite open subcover to prove compactness?

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    $\begingroup$ If your collection of open sets is to be a cover of the interval, it has to cover 0 and 1, which - even given the fact that my browser is not interpreting your formula - seem to be missing from the union of your proposed covering sets. $\endgroup$ – Mark Bennet Aug 30 '12 at 21:59
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    $\begingroup$ Indeed, that is an open cover of $(0,1)$, not $[0,1]$. Since the former interval is not compact, it is not a surprise that there is no finite subcover. $\endgroup$ – Harald Hanche-Olsen Aug 30 '12 at 22:02
  • $\begingroup$ See here for some other proofs of this fact: math.stackexchange.com/questions/368108/… $\endgroup$ – Martin Sleziak Apr 23 '16 at 13:29
  • $\begingroup$ Read the statement of compactness theorem again. WORD BY WORD. $\endgroup$ – Subhasis Biswas Dec 19 '18 at 13:49

Here is a quick and elegant proof of the actual result that $[0,1]$ is compact based on real induction:

Let $\mathcal{O}$ be an (arbitrary!) open cover. Let $P$ be the set of points $x$ in $[0,1]$ such that $[0,x]$ can be covered by finitely many elements of $\mathcal{O}$. We have $0\in P$ and $P$ is bounded above by $1$. Therefore, $P$ has a supremum $s$.

We first show that $[0,s]$ can be covered by finitely many sets in $\mathcal{O}$. This is trivial when $s=0$, so assume $s>0$. Let $O_s\in\mathcal{O}$ be a set containing $s$. Then there is an $\epsilon \in (0, s)$ such that $(s-\epsilon,s]\subseteq O_s$. By assumption, there is a finite subcover of $[0,s-\epsilon/2]$. By adding $O_s$ to that finite subcovering, we get a finite subcovering of $[0,s]$.

We now show that $s=1$. Suppose $s<1$ and let $O_s\in\mathcal{O}$ be a set containing $s$. Then there is an $\epsilon>0$ such that $[s,s+\epsilon)\subseteq O_s$. So taking a finite subcover of $[0,s]$ and adding the set $O_s$ gives us a finite subcover of $[0,s+\epsilon/2]$, contradicting the construction of $s$.

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    $\begingroup$ I am just learning about compactness and I'm a bit confused by the fourth sentence in the third paragraph. How do I know $\exists \epsilon \in (0,s)$ such that $(s-\epsilon,s]\subseteq O_s$? Does this follow from the fact $[0,s]$ can be covered by finitely many sets in $\mathcal{O}$? If so, why? $\endgroup$ – Stan Shunpike Jul 9 '15 at 15:22
  • $\begingroup$ @StanShunpike That is basically the definition of an open set. If $O_s$ is open, there must be some "wiggle room" around each of its elements. In particular, if $s\in O_s$ then for some $\epsilon>0$, $(s-\epsilon, s+\epsilon)\subseteq O_s$. The result follows now from $(s-\epsilon,s]\subseteq (s-\epsilon,s+\epsilon)\subseteq O_s$. $\endgroup$ – Michael Greinecker Jul 10 '15 at 8:49
  • $\begingroup$ @Michael Greinecker May I ask a question: Does this argument work in the case of $[0,1]$ as a subspace of $(\mathbb{R},T)$ where $T$ is the topology with bases $\mathcal{B}=\{(a,b]:a,b\in\mathbb{R}\}$ ?? I think it does. Thanks in advence! $\endgroup$ – giannispapav Aug 23 at 19:05
  • $\begingroup$ @giannispapav No, it does not. Consider the open cover of $[0,1]$ consisting of $(-1,1/2]$ and sets of the form $(1/2+1/n,1]$. It has no finite subcover. If you go through the proof, you see that $s=1/2$. The problem is that the corresponding finite cover, consisting of $(-1,1/2]$ does not extend to a finite subcover of $[0,s+\epsilon]$ for any $\epsilon>0$. $\endgroup$ – Michael Greinecker Aug 23 at 19:53
  • $\begingroup$ @Michael Greinecker I see! Thanks! $\endgroup$ – giannispapav Aug 24 at 6:08

The collection of sets above is not an open cover of $[0,1]$. The points $0$ and the point $1$ are not in any of them.

Also you can not prove compactness of anything by taking an open cover and finding a subcover.

The definition of compactness is that for all open covers, there exists a finite subcover.

If you want to prove compactness for the interval $[0,1]$, one way is to use the Heine-Borel Theorem that asserts that compact subsets of $\mathbb{R}$ are exactly those closed and bounded subsets.

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    $\begingroup$ Borel actually proved the theorem by showing that each closed and bounded interval is compact... $\endgroup$ – Michael Greinecker Aug 31 '12 at 0:08

Answering your question would not prove compactness. The condition for compactness is that every open cover has a finite subcover, not just that a single given open cover has a finite subcover. A proof that $[0,1]$ is compact with respect to the Euclidean metric topology can be found here.

  • $\begingroup$ $EVERY$. A very important word. $\endgroup$ – Subhasis Biswas Dec 19 '18 at 13:48

I assume that you would like to prove that it is compact with respect to the usual topology. Since that topology is "generated" by a metric the topological compactness is equvalent with metrical compactness, i.e. the condition stating that every sequence has converging subsequence.

Observe that every sequence of elements of the interval $[0,1]$ is bounded. Therefore, by the Bolzano-Weierstrass theorem, it has a convergent subsequence.

Notice that the Bolzano-Weierstrass theorem can be proved by analytical arguments, so we are not in the vicious circle.

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    $\begingroup$ Proving the equivalence of compactness in terms of open coverings and proving the equivalence with sequential compactness is much, much harder than proving Heine-Borel in the one-dimensional case. $\endgroup$ – Michael Greinecker Aug 31 '12 at 0:05

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