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Theorem (HeineBorel Theorem): A subset S of R is 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 I_{n} = (−n, n). Then
Therefore S is covered by the collection of {In }. Hence, since S is compact, ﬁnitely many will suﬃce.
where m = max{n_{1}, . . . , n_{k}}. Therefore x ≤ m for all x ∈ S , and S is bounded.
Now we will show that S is closed. Suppose not. Then there is some point p ∈ (cl S) \ S . For each n, deﬁne the neighborhood around p of radius 1/_{n}, N_{n} = N (p, 1/n). Take the complement of the closure of N_{n}, U_{n} = R \ cl N_{n} . Then U_{n} is open (since its complement is closed), and we have
Therefore, {U_{n}} is an open cover for S . Since S is compact, there is a ﬁnite subcover {U_{n1} , � � � , U_{nk}} for S . Furthermore, by the way they are constructed, U_{i} ⊆ U_{j} if i ≤ j . It follows that S ⊆ U_{m} where m = max{n_{1}, . . . , n_{k}}. But then S ∩ N (p, 1/m) = �, which contradicts our choice of p ∈ (cl S) \ S .
Conversely, we want to show that if S is closed and bounded, then S is compact. Let F be an open cover for S . For each x ∈ R, deﬁne the set
S_{x} = S ∩ (−∞, 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 S_{d} = {d} and this is certainly covered by a ﬁnite subcover of F. Therefore, d ∈ 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 = sup B. We shall show that m ∈ S and m S both lead to contradictions.
If m ∈ S , then since F is is an open cover of S , there exists F_{0} in F such that m ∈ F_{0} . Since F_{0} is open there exists an interval [x_{1} , x_{2}] in F_{0} such that
x_{1} < m < x_{2}.
Since x_{1} < m and m = supB , there exists F_{1}, . . . , F_{k} in F that cover S_{x1} . But then F_{0} , F_{1} , . . . , F_{k} cover S_{x2} , so that x_{2} ∈ B. But this contradicts m = sup B.
If m ∉ S , then since S is closed there exists ε > 0 such that N (m, ε) ∩ S = �. But then
S_{m−ε} = S_{m+ε}.
Since m − ε ∈ B we have m + ε ∈ B, which again contradicts m = sup B.
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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