proof of Heine-Borel theorem
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).
Compact implies closed and bounded.
Reduction for closed and bounded implies compact.
The case : the closed interval is compact..
Let be an arbitrary cover of by open sets in . Define
Set ; then .
We first note that the supremum is attained: that is, there is a finite subcollection of that covers . Obviously we can choose one open set from that covers the point . This must also cover the open interval for some . But by the definition of , the interval has a finite subcover. When this finite subcover is put together with , we obtain a finite subcover for .
It is easy to see that cannot be less than . For the same open set above covers the interval , so we have a finite subcover for . If is the supremum then it has go to be at the right endpoint of the interval , i.e. . ∎
For the case , apply Tychonoff’s Theorem, in the finite case (http://planetmath.org/ProofOfTychonoffsTheoremInFiniteCase), which states:
If are each compact, then in the product topology is compact.
Here we set to be the closed intervals; then the general closed rectangle in 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 . And it is again a standard exercise to show that the product topology on is the same as the norm topology on as a vector space.
Proof by a bisection argument
There is another proof of the Heine-Borel theorem for 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 into subrectangles. Suppose the open cover of has no finite subcover. Then one of the subrectangles — call it — must have no finite subcover by . We can subdivide into pieces; since has no finite subcover, one of the new subrectangles of also has no finite subcover. And continue dividing to get a nested sequence of rectangles whose side lengths approach zero, and possessing no finite subcover.
By the nested interval theorem, the “limit rectangle” must consist of a sole point , and this obviously has a finite subcover by an open set . But must contain a small rectangle with centre , which for large enough, contradicts 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.
|Title||proof of Heine-Borel theorem|
|Date of creation||2013-03-22 12:57:40|
|Last modified on||2013-03-22 12:57:40|
|Last modified by||stevecheng (10074)|