I am understanding proof of theorem stated in title from Spivak's calculus. It is as below.

(0) Let $\mathcal{O}$ be an open cover of $[0,1]$.

(1) Let $A=\{x\in [0,1]:[0,x] \mbox{ has finite subcover from } \mathcal{O}\}$.

(2) Then $A$ is non-empty, bounded above by $1$; let $\alpha$ be its supremum.

(3) Since $\mathcal{O}$ is open cover of $[0,1]$, $\alpha$ is in some $U$ from $\mathcal{O}$.

(4) There is an open interval $J$, $\alpha\in J\subseteq U$ s.t.all points of $J$ to the left of $\alpha$ are also in $U$.

(5) Since $\alpha$ is supremum of $A$, there is an $x\in J$ such that $x\in A$. How?

(6) Then $[0,x]$ is covered by finite subcover; this together with $U$ covers $[0,\alpha]$; so $\alpha\in A$.

(7) One tries to prove that $\alpha=1$, and proof will complete.

Q.1 It is in step 5, which I don't understand.

Q.2 Are there different proofs of this theorem? (I don't find other in 5-6 standard books than this).

  • $\begingroup$ Another proof of compactness. $\endgroup$ – Sahiba Arora Aug 16 '17 at 10:26
  • $\begingroup$ The theorem in question goes by the name of Heine Borel Theorem and is 8ne of the ways to express the completeness property of real numbers. There are many other (easier) ways to express completeness and each of them is equivalent to another (see math.stackexchange.com/a/1787254/72031). So you do have multiple proofs of this theorem. $\endgroup$ – Paramanand Singh Aug 16 '17 at 11:52
  • $\begingroup$ Another proof which is tricky is based on the existence of a positive number $\delta$ such that any subinterval of $[0,1]$ with length less than or equal to $\delta$ is contained in one of the intervals in $\mathcal{O} $. The existence of $\delta$ can be ascertained by use of contradiction and Bolzano Weierstrass theorem. $\endgroup$ – Paramanand Singh Aug 16 '17 at 13:03
  • $\begingroup$ I have added the tag (proof-explanation), since your question seems to be mainly about this specific proof and explanation of some particular steps. (If it is not the case, feel free to remove the tag - in which case it would probably quality as a duplicate of: How to prove every closed interval in R is compact? Other posts linked there might be of interst, too.) $\endgroup$ – Martin Sleziak Aug 16 '17 at 13:51


Suppose the contrary, that is, assume that there is no finite subcover that covers $[0,1]$. Then, at least one of the intervals $[0,1/2]$ and $[1/2, 1]$ can't be covered with a finite subcover. Pick this interval and call it $I_1$. This is the base of an inductive process.

Now assume that $I_n$ has length $1/2^n$ and can't be covered with a finite subcover. Divide $I_n$ into two intervals of the same length. One of these intervals can't be covered with a finite subcover. Pick this one and call it $I_{n+1}$ This new interval has length $1/2^{n+1}$. This completes the inductive process, which yields a sequence of intervals $I_n$. Each interval of the sequence is contained and has a length that is a half of the preceeding one. And none of them can be covered with a finite subcover.

Since $I_{n+1}\subset I_n$, $\bigcap_{n=1}^\infty I_n$ is not empty. And since the length of $I_n$ tends to $0$ it has only one point. That point is in some open set from the cover, and this open set alone covers some $I_m$. Contradiction.

Remark: this idea can be used to show that a finite cartesian product of closed, bounded invervals is compact.

  • 1
    $\begingroup$ Nice proof. I prefer Spivak's because he doesn't have to choose infinitely many intervals. $\endgroup$ – Gribouillis Aug 16 '17 at 10:36
  • 1
    $\begingroup$ The proof can be upgraded by choosing the leftmost interval in the case both intervals can't be finitely covered. This would meet Spivak's proof in converging to the smallest point that does not belong to any finite cover. $\endgroup$ – Gribouillis Aug 16 '17 at 10:54
  • $\begingroup$ This proof seems to be flawed? It assumes there's only finitely many subintervals that can be infinitely decomposed like this, which isn't necessarily true. $\endgroup$ – BlueRaja - Danny Pflughoeft Nov 4 '19 at 7:06
  • $\begingroup$ @BlueRaja-DannyPflughoeft At every stage, only one interval of length $2^{-n}$ is considered. That is divided into two halves, and by the assumption that the interval cannot be covered by a finite subfamily, at least one of the two halves also has that property. As the next interval to be considered we pick such a half. (And if we state that we always pick the left one if admissible, we don't even use any choice.) $\endgroup$ – Daniel Fischer Nov 15 '19 at 19:02


Since $\alpha = \sup(A)$, for any $\epsilon > 0$, there is an $x\in A$ such that $\alpha - \epsilon < x\le \alpha$. By choosing $\epsilon$ small enough, one has $(\alpha -\epsilon, \alpha]\subset J$, which proves (5).


A set $A$ is sequentially compact if every $x_n \in A$ has a convergent subsequence.

$Theorem:$ Every sequentially compact metric space $X$ is compact.

For a proof see this:


Now let $x_n \in [0,1] \Rightarrow x_n$ is bounded thus from Bolzano-Weierstrass theorem is has a convergent subsequence therefore $[0,1]$ is sequentially compact.

From the above theorem $[0,1]$ is compact.


From a general topology point of view, you could also apply Alexander's subbase lemma (see my answer on Topology Atlas here where the final application is the fact that in ordered space $X$, $X$ is compact iff every subset of $X$ has a supremum. So that's a different proof, but it also relies (as it must) on the completeness of the reals.

You could also prove it's complete and totally bounded (a uniform space approach).


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