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In general topology, a topological space is said to be compact, if every one of its open cover has a finite subcover. However, I cannot see the compactness of the close interval [0,1] from the above definition. To be a little specific,let us consider the following open cover for [0,1]: $C= \{[0,1/2),(1/3,3/4), (2/3,1]\}$. Now, the open interval (1/3,3/4) itself has at least one open cover (let's call it P) which does not have a finite subcover. We use P to cover the open interval (1/3,3/4). This gives a new open cover C' for the interval [0,1]. It looks like C' does not have an finite subcover, since C' includes P which does not have a finite subcover.

Of course I misunderstood something here. If somebody can catch my error it will be very helpful.

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    $\begingroup$ 'It looks like C' does not have an finite subcover, since C' includes P which does not have a finite subcover.' Not necessarily true, as all but finitely many sets in P may be contained in [0,1/2) or (2/3,1] , rendering them unnecessary. $\endgroup$ – Tim The Enchanter Aug 26 '17 at 5:32
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    $\begingroup$ Go to the Wikipedia page for the Heine-Borel theorem, and look at the proof section. A bit down, they prove that $[-a,a]^n\subseteq \Bbb R^n$ is compact. You can follow that proof. $\endgroup$ – Arthur Aug 26 '17 at 6:24
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    $\begingroup$ No finite subset of P will cover $(1/3,3/4).$ But a finite subset of $P$ will cover [$2/5, 5/7]$ and the rest of $[0,1]$ is covered by $\{[0,1/2),(2/3,1]\}$. $\endgroup$ – DanielWainfleet Aug 26 '17 at 7:47
  • $\begingroup$ Maybe some of these posts might help: How to prove every closed interval in R is compact? and Showing that $[0,1]$ is compact $\endgroup$ – Martin Sleziak Aug 26 '17 at 10:50
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So you get an open cover by retaining $[0,1/2)$ and $(2/3,1]$ but replacing $(1/3,3/4)$ by a bunch $P$ of open sets where no finite collection covers $(1/3,3/4)$.

You can do this.

But it is still the case that this new covering $C'$ has a finite subcovering. Don't forget that $[0,1/2)$ and $(2/3,1]$ are still available. If we used both of these, all we have to do is find a finite subset of $P$ that covers $[1/2,2/3]$. (We don't need it to cover all of $(1/3,3/4)$.) As $[1/2,2/3]$ is compact, then there will be such a finite subset.

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    $\begingroup$ This is circular: how would you know that $[1/2,2/3]$ is compact? $\endgroup$ – Arthur Aug 26 '17 at 6:39
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    $\begingroup$ @Arthur, it seems that the OP wanted to understand the flaw in his example of an open covering with no finite subcovering. We were not looking for a proof of compactness of a closed interval. $\endgroup$ – Ted Shifrin Aug 26 '17 at 6:43
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    $\begingroup$ It's not circular; it'd be circular if he used the face that $[0, 1]$ is compact. Perhaps Lord knows that $[1/2, 2/3]$ is compact because Lord's has read a proof of the HB-theorem. It may not be helpful to OP, but it's not circular. $\endgroup$ – John Hughes Aug 26 '17 at 6:44
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I will present you a proof different from Heine-Borel which i hope it will help you.

We'll use this corollary of Cantor's intersection theorem.

Corollary:Let $(X,d)$ be a complete metric space.Then for every decreasing sequence $F_n$ of closed subsets of $X$ such that $diam(F_n) \rightarrow 0$ we have that $\bigcap_{n=1}^{\infty}F_n =\{x\}$ for some $x \in X$.

Note that $diam(A)=\sup\{d(x,y)|x,y \in A \}$

Now instead of the diameter of a set i will use the notion of length deonting by $l$..You can see that in the real line (with the usual metric $d(x-y)=|x-y|$)

that $diam(I)\leq l(I)$ where $I$ is an interval.

$Proof$

Now let $I_0=[0,1]$ and assume that $I_0$ is not compact.Thus exists a collection $P$ of open sets which does not contain a finite subfamily of open sets that cover $I_0$

Now we divide $I_0$ in two intevals of equal length each namely $[0,1/2],[1/2,1]$

At least one of this intervals cannot be covered by finitely many elements of $P$.Call this interval $I_1$.

Now divide $I_1$ into two intervals of length $1/4$.At least one of these intervals cannot be covered my finitely elements of $P$(Why?)

Call this inteval $I_2$.

We continue this argument inductively and we find a decreasing sequence $I_0 \supseteq I_1 \supseteq I_2 \supseteq....$ of closed intervals.

Now we know that $[0,1]$ is complete that the intersection of these intervals is nonempty from the theorem.

Also notice that $l(I_n)=\frac{1}{2^n} \rightarrow 0$ and we know that forall $n \in \mathbb{N} \cup \{0\}$ none of the intervals can be covered by a finite elements of $P$

Thus exists a $x \in [0,1]$ such that $x$ belongs to any of the intervals $I_0,I_1,I_1...$ .

Now because $P$ covers $I_0$ we have that exists an open set $A$ as element of $P$ where $x \in A$

Now because $A$ is open exists $\epsilon>0$ such that $x \in (x- \epsilon,x+\epsilon) \subseteq A $

Now exists $n_0 \in \mathbb{N}$ such that $l(I_{n_0})< \epsilon$ and $x \in I_{n_0}$

Thus $I_{n_0} \subseteq (x- \epsilon,x+\epsilon)$

We arrived to a contradiction because $I_{n_0}$ cannot be covered by finitely elements of $P$ and at the same time we found the element $A$ that covers $I_{n_0}$.

Thus $[0,1]$ is compact.

So from this you can see that an open interval say $(a,b)$(it can also be a subset of $[0,1]$) fails to be compact because it s not complete.

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  • $\begingroup$ @Shalop you are right..thanks for the comment...the theorem is correct if the sets $F_n$ are compact..but it can be fixed if we let $diam(F_n) \rightarrow 0$ so in this cas we need only the closeness $\endgroup$ – Marios Gretsas Aug 26 '17 at 8:23
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You can also use the theorem that

Let $\mathcal{S}$ be a subbase for $X$. Then $X$ is compact iff every open cover with elements of $\mathcal{S}$ has a finite subcover.

This is Alexander's subbase theorem and for any ordered space like $[0,1]$ has the following subbase $\mathcal{S}= \{[0,a): a \in [0,1]\} \cup \{(b,1]: b \in [0,1]\}$

Then the compactness of $[0,1]$ follows, using the order completeness: suppose $\mathcal{O} \subseteq \mathcal{S}$ is an open cover of $[0,1]$.

As $0$ is only covered by a set of the form $[0,a)$ there must be at least one such member in $\mathcal{O}$. So define $A = \{a \in [0,1]: [0,a) \in \mathcal{O}\}$ which is non-empty and bounded above (by $1$) so $a_0 = \sup A$ exists in $[0,1]$. Then $a_0$ is covered by some member of $\mathcal{O}$ and it cannot be of the form $[0,b)$ (as then $b \in A$ and $b >a_0$, so $a_0$ would not be an upperbound of $A$) and so it is covered by some $(b,1] \in \mathcal{O}$. As $b < a_0$ this means that $b$ is not an upperbound for $A$, as $a_0$ is the least upperbound of $A$, so for some $a_1 \in A$ we have $a_1 > b$. But then $[0,a_1)\in \mathcal{O}$ and the two sets $[0,a_1), (b,1]$ from $\mathcal{O}$ clearly cover $[0,1]$. By the Alexander subbase theorem, $[0,1]$ is compact. In fact, any ordered space with a minimum and maximum which is order-complete is compact (same proof essentially).

The Alexander theorem is proved using AC, so this does not make for a fully "constructive" proof. But it is a useful theorem, e.g. also to show Tychonoff's theorem for compactness of products (product also have a natural subbase), or the compactness of the hyperspace of a compact space in the Vietoris topology.

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The intervals $(\frac{1}{2},\frac{1}{3})$ and $(\frac{2}{4},\frac{3}{4})$ will make infinitely many of the sets required to cover $P$ redundant leaving only finitely many necessary to cover the gap.

Lets consider an example. For simplicity I'll use the set $[-2,2]$ $C=\{[-2,\frac{1}{2}),(0,1),(\frac{1}{2},2]\}$ and I cover $P=(0,1)$ with sets of the form $(0,1-\frac{1}{n})$ for $n \in \mathbb{N}$, $n\geq3.$ Clearly no finite subcover of $P$ would do because we need to get arbitrarily close to $1$ but we can in fact select a single interval $(0,1-\frac{1}{n}$), discard the rest and be left with an open covering of $[-2,2]$.

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  • $\begingroup$ I understand that, the cover you chose has a finite subcover. In my mind I have different kind of cover which I could not formally write but it is the following sequence of covers: [0,1]= [0,1/2+$\epsilon$) $\cup$ (1/2+$\epsilon$,1] = [0,1/4+$\epsilon$)$\cup$ (1/4-$\epsilon$, (1/2+$\epsilon$)$\cup$ (1/2-$\epsilon$, 3/4+$\epsilon$)$\cup$(3/4-$\epsilon$,1]=$\cdots$. We can go on like that up to the limit where the cover has infinite number of open sets. Clearly if we omit any one of the members of the cover at any point of the above sequence, it will no longer cover [0,1]. $\endgroup$ – Tuhin Subhra Mukherjee Aug 28 '17 at 17:12
  • $\begingroup$ If we can indeed continue the sequence up to infinite terms, it seems that the above cover does not even have a infinite subcover since we need all of the open sets to cover [0,1]. I can't seem to figure out where does my argument go wrong but I know it IS wrong. $\endgroup$ – Tuhin Subhra Mukherjee Aug 28 '17 at 17:17
  • $\begingroup$ @TuhinSubhraMukherjee Did you mean $[0,1/2+\epsilon) \cup (1/2-\epsilon,1]$? In any case the issue is when you took the limit you didn't ensure that the sets were still open. Since the length of each open interval in your construction can easily be seen to converge to zero, they are not open intervals in the limit. I believe your construction is equivalent to the binary representation of the reals on that interval. Trying writing out your scheme in base 2. Or it could be that you forgot $\epsilon > 0$ so there exists a limit on how small the balls can be, namely $2\epsilon$. $\endgroup$ – CyclotomicField Aug 28 '17 at 21:30
  • $\begingroup$ @ CyclotomicField Sorry, I meant [0,1/2+ϵ)∪(1/2−ϵ,1]. Yes, there is a limit on the size of the balls as you have pointed out; it is 2$\epsilon$. But $\epsilon$ can be arbitrarily small. May be I have to think about it a little more. $\endgroup$ – Tuhin Subhra Mukherjee Aug 29 '17 at 2:04
  • $\begingroup$ @TuhinSubhraMukherjee $\epsilon$ can be arbitrarily small but Archimedes's principle tells us that there exist an $M \in \mathbb{N}$ such that $2\epsilon M > 1$ which means we will still end up with a finite number of them to cover the interval. Can you see how the length of the intervals going to zero in the limit means that in the limit we have singletons covering the interval instead of open set? $\endgroup$ – CyclotomicField Aug 29 '17 at 3:37
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This kind of thing (compactness of $[0,1]$) is typically proved via an appeal to the Heine-Borel theorem, which says that a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded. The subset $[0,1]$ is closed and bounded and therefore compact.

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    $\begingroup$ This is absurdly circular $\endgroup$ – mathworker21 Aug 26 '17 at 5:51
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    $\begingroup$ @mathworker21 That may be, although parts of the proof of the HB theorem, as written in that Wikipedia article, is exactly what we need. So while this answer doesn't answer the question, it does inadvertently link to an answer. $\endgroup$ – Arthur Aug 26 '17 at 6:27
  • $\begingroup$ This doesn't address the OP. $\endgroup$ – Ted Shifrin Aug 26 '17 at 6:44
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    $\begingroup$ In either case it would benefit the OP to read and understand the Heine-Borel thm, which was my intention, since it's an important thm in itself and the OP can answer his question himself by studying the proof. $\endgroup$ – g.s Aug 26 '17 at 7:07

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