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proof of Heine-Borel theorem
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(Proof)
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To shorten the proofs, we will be using certain concepts and facts from general point-set topology, such as the very first assertion below. Even if you are not familiar with them, these assertions are all easily proven from the definitions (left as an exercise).
Proof. [Reduction for closed and bounded implies compact] To prove that if $A$ is closed and bounded implies it is compact, it is only necessary to prove that the rectangle $[a,b]^n$ is compact. For general $A$ since $A$ is bounded, it is contained in some such compact rectangle $[a, b]^n$ Since $A$ is closed and contained in a compact
set, $A$ is also compact. 
Proof. [The case $n=1$ the closed interval is compact.] Let $\mathcal{C}$ be an arbitrary cover of $[a, b]$ by open sets in $\real^1$ Define $$ S = \{ x \in [a,b] : \textrm{some finite subcollection of $\mathcal{C}$ covers $[a, x]$} \}\,. $$ Set $t = \sup S$ then $a \leq t \leq b < \infty$
We first note that the supremum $t$ is attained: that is, there is a finite subcollection of $\mathcal{C}$ that covers $[a,t]$ Obviously we can choose one open set $U$ from $\mathcal{C}$ that covers the point $t$ This $U$ must also cover the open interval $(t-\epsilon, t+\epsilon)$ for some $\epsilon > 0$ But by the definition of $t$ the interval $[a, t-\epsilon]$ has a finite subcover. When this finite subcover is put together with $U$ we obtain a finite subcover for $[a, t]$
It is easy to see that $t$ cannot be less than $b$ For the same open set $U$ above covers the interval $(t, t+\epsilon)$ so we have a finite subcover for $[a, t+\epsilon/2]$ If $t$ is the supremum then it has go to be at the right endpoint of the interval $[a, b]$ i.e. $t = b$ 
For the case $n > 1$ apply Tychonoff's Theorem, in the finite case, which states:
If $X_1, \dotsc, X_n$ are each compact, then $X_1 \times \dotsm \times X_n$ in the product topology is compact.
Here we set $X_i$ to be the closed intervals; then the general closed rectangle in $\real^n$ is compact. It is not hard to see that the product topology for the closed rectangle is the same as its subspace topology in the product topology $\real^n = \real \times \dotsm \times \real$ And it is again a standard exercise to show that the product topology on $\real^n$ is the same as the norm topology on $\real^n$ as a vector space.
This completes the proof of the Heine-Borel theorem.
There is another proof of the Heine-Borel theorem for $\real^n$ without resorting to Tychonoff's Theorem. It goes by bisecting the rectangle along each of its sides. At the first stage, we divide up the rectangle $A$ into $2^n$ subrectangles. Suppose the open cover $\mathcal{C}$ of $A$ has no finite subcover. Then one of the subrectangles -- call it $A_1$ -- must have no finite subcover by $\mathcal{C}$ We can subdivide $A_1$ into $2^n$ pieces; since
$A_1$ has no finite subcover, one of the new subrectangles of $A_1$ also has no finite subcover. And continue dividing to get a nested sequence of rectangles $A \supset A_1 \supset A_2 \supset \dotsb$ whose side lengths approach zero, and possessing no finite subcover.
By the nested interval theorem, the ``limit rectangle'' $\bigcap_{i=1}^\infty A_i$ must consist of a sole point $x$ and this obviously has a finite subcover by an open set $U \in \mathcal{C}$ But $U$ must contain a small rectangle with centre $x$ which for $i$ large enough, contradicts $A_i$ having no finite subcover.
Of course, in both proofs of the Heine-Borel theorem, the completeness of the reals (the least upper bound property) enters in an essential way.
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"proof of Heine-Borel theorem" is owned by stevecheng. [ full author list (2) | owner history (2) ]
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Cross-references: least upper bound property, reals, centre, contain, nested interval theorem, lengths, sequence, divide, sides, Tychonoff's theorem, Heine-Borel theorem, completes, vector space, norm, subspace topology, product topology, endpoint, right, easy to see, subcover, interval, open interval, point, finite, supremum, open sets, cover, closed interval, compact set, compact rectangle, rectangle, necessary, compact, bounded set, contained, finite cover, open cover, compact subset, Hausdorff space, bounded, closed, implies, definitions, even, topology, proofs
This is version 20 of proof of Heine-Borel theorem, born on 2002-08-21, modified 2005-08-03.
Object id is 3328, canonical name is ProofOfHeineBorelTheorem.
Accessed 15331 times total.
Classification:
| AMS MSC: | 54D30 (General topology :: Fairly general properties :: Compactness) |
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Pending Errata and Addenda
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