15-02-2011, 12:28 PM

A proof of the Heine-Borel Theorem

Theorem (Heine-Borel Theorem). A subset S of Ris compact if and only if S is closed and bounded.

Proof. First we suppose that S is compact. To see that S is bounded is fairly simple: Let In = (−n, n).

Then

1

[

n=1

In = R.

Therefore S is covered by the collection of {In}. Hence, since S is compact, finitely many will suffice.

S (In1 [ · · · [ Ink ) = Im,

where m = max{n1, . . . , nk}. Therefore |x| m for all x 2 S, and S is bounded.

Now we will show that S is closed. Suppose not. Then there is some point p 2 (cl S) \ S. For each n,

define the neighborhood around p of radius 1/n, Nn = N(p, 1/n). Take the complement of the closure of

Nn, Un = R\ clNn. Then Un is open (since its complement is closed), and we have

1

[

n=1

Un = R\

1

\

n=1

clNn = R\ {p} S.

Therefore, {Un} is an open cover for S. Since S is compact, there is a finite subcover {Un1 , · · · ,Unk}

for S. Furthermore, by the way they are constructed, Ui Uj if i j. It follows that S Um where

m = max{n1, . . . , nk}. But then S \ N(p, 1/m) = ?, which contradicts our choice of p 2 (cl S) \ S.

Conversely, we want to show that if S is closed and bounded, then S is compact. Let Fbe an open

cover for S. For each x 2 R, define the set

Sx = S \ (−1, x],

and let

B = {x : Sx is covered by a finite subcover of F}.

Since S is closed and bounded, our lemma tells us that S has both a maximum and a minimum. Let

d = min S. Then Sd = {d} and this is certainly covered by a finite subcover of F. Therefore, d 2 B and B is

nonempty. If we can show that B is not bounded above, then it will contain a number p greater than max S.

But then, Sp = S so we can conclude that S is covered by a finite subcover, and is therefore compact.

Toward this end, suppose that B is bounded above and let m = supB. We shall show that m 2 S and

m /2 S both lead to contradictions.

If m 2 S, then since Fis an open cover of S, there exists F0 in Fsuch that m 2 F0. Since F0 is open,

there exists an interval [x1, x2] in F0 such that

x1 < m < x2.

Since x1 < m and m = supB, there exists F1, . . . , Fk in Fthat cover Sx1 . But then F0, F1, . . . , Fk cover

Sx2 , so that x2 2 B. But this contradicts m = supB.

If m /2 S, then since S is closed there exists ε > 0 such that N(m, ε) \ S = ?. But then

Sm−" = Sm+".

Since m− ε 2 B we have m+ ǫ 2 B, which again contradicts m = supB.

Therefore, either way, if B is bounded above, we get a contradiction. We conclude that B is not bounded

above, and S must be compact.

download full report

http://math.utah.edu/~bobby/3210/heine-borel.pdf