Compactness of the closed interval [0,1]

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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 $[-2,2]$.

Answer 5

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.