A proof of the Heine-Borel Theorem
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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).
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
Un = R\
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.

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