Heine Borel - Sequences and Series, CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

Theorem (Heine-Borel 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{n₁, . . . , 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{n₁, . . . , 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 F₀ in F such that m ∈ 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, . . . , F_k in F that cover S_x1 . But then F₀ , F₁ , . . . , F_k cover S_x2 , so that x₂ ∈ 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.